JEE Kinematics - 1D Motion Questions

Let the mass of the rod be $$m$$ and its length be $$l$$. It oscillates about the fixed hinge $$O$$ with a small angular displacement $$\theta$$ (taken positive for a clockwise rotation). Gravity is ignored, so only the springs give the restoring torque.

1. Moment of inertia about the hinge
For a uniform rod of length $$l$$ about one end, $$I = \frac{1}{3} m l^{2}$$

2. Extension produced in a spring connected at distance $$r$$ from the hinge
When the rod rotates through a small angle $$\theta$$, every point at a distance $$r$$ from the hinge moves through an arc of length $$r\theta$$ perpendicular to the rod’s original position. Hence the linear extension (or compression) of a spring attached at that point is also $$r\theta.$$

3. Restoring torque of one spring
Spring force $$F = k \,(r\theta) \quad\Rightarrow\quad$$ Torque about $$O: \quad \tau = F \cdot r = k \, r^{2}\theta.$$ Therefore each spring contributes a restoring torque equal to $$k r^{2}\theta.$$

4. Angular equation of motion
Sum of restoring torques $$= k\,(r_1^{2}+r_2^{2})\,\theta.$$ Hence $$I\,\ddot{\theta} + k\,(r_1^{2}+r_2^{2})\,\theta = 0\; \Longrightarrow\; \omega = \sqrt{\dfrac{k\,(r_1^{2}+r_2^{2})}{I}}.$$ The linear frequency is $$f = \dfrac{\omega}{2\pi},$$ so the ratio $$f_1/f_2$$ equals the ratio $$\omega_1/\omega_2.$$

Case 1: (Fig. 1)
Springs at $$r_1=l/2$$ and $$r_2=l$$ $$r_1^{2}+r_2^{2} = \left(\frac{l}{2}\right)^{2} + l^{2} = \frac{l^{2}}{4}+l^{2} = \frac{5}{4}l^{2}.$$ $$\displaystyle \omega_1 = \sqrt{\dfrac{k\left(\tfrac{5}{4}l^{2}\right)}{I}}.$$

Case 2: (Fig. 2)
Both springs at $$r_1=r_2=l/2$$ $$r_1^{2}+r_2^{2} = 2\left(\frac{l}{2}\right)^{2} = 2\left(\frac{l^{2}}{4}\right)=\frac{l^{2}}{2}.$$ $$\displaystyle \omega_2 = \sqrt{\dfrac{k\left(\tfrac{l^{2}}{2}\right)}{I}}.$$

5. Ratio of frequencies
$$\frac{f_1}{f_2}=\frac{\omega_1}{\omega_2}= \sqrt{\dfrac{k\left(\tfrac{5}{4}l^{2}\right)}{I}} \;\Bigg/\; \sqrt{\dfrac{k\left(\tfrac{l^{2}}{2}\right)}{I}} = \sqrt{\dfrac{\tfrac{5}{4}l^{2}}{\tfrac{l^{2}}{2}}} = \sqrt{\dfrac{5}{4}\cdot\dfrac{2}{1}} = \sqrt{\dfrac{10}{4}} = \sqrt{\dfrac{5}{2}}.$$

Hence $$\displaystyle \frac{f_1}{f_2} = \sqrt{\frac{5}{2}}.$$ Option C is correct.

The apparent (measured) weight inside an accelerating elevator equals the normal reaction $$N$$ on the object.

For upward acceleration $$a$$ (upward taken positive) the force equation is
$$N - mg = ma \; \Rightarrow \; N = m\,(g + a) \qquad -(1)$$

The elevator’s position as a function of time is given:
$$y = 8\left[1 + \sin\!\left(\frac{2\pi t}{T}\right)\right], \qquad T = 40\pi\;\text{s} \qquad -(2)$$

Differentiate twice to obtain the acceleration.

Angular frequency:
$$k = \frac{2\pi}{T} = \frac{2\pi}{40\pi} = \frac{1}{20}\;\text{s}^{-1}$$

Velocity:
$$v = \frac{dy}{dt} = 8\,k\,\cos(k t)$$

Acceleration:
$$a = \frac{dv}{dt} = -8\,k^{2}\,\sin(k t) \qquad -(3)$$

The sine term varies between $$+1$$ and $$-1$$, hence

Maximum upward acceleration (sin$$=-1$$):
$$a_{\max} = +8\,k^{2}$$

Maximum downward acceleration (sin$$=+1$$):
$$a_{\min} = -8\,k^{2}$$

Therefore the spread of acceleration values is
$$a_{\max} - a_{\min} = 8k^{2} - (-8k^{2}) = 16k^{2} \qquad -(4)$$

Insert $$k = 1/20$$:

$$k^{2} = \left(\frac{1}{20}\right)^{2} = \frac{1}{400}$$
$$16k^{2} = 16 \times \frac{1}{400} = \frac{16}{400} = 0.04\;\text{m\,s}^{-2}$$

Mass of the object: $$m = 50\;\text{kg}$$.
Using $$-(1)$$, the spread in the normal reaction (apparent weight) equals $$m$$ times the acceleration spread:

$$\Delta N = m\,(a_{\max} - a_{\min}) = 50 \times 0.04 = 2\;\text{N}$$

Hence, the maximum variation in the weight reading is

2 N.

r = 5t²i-5tj. v = 10ti-5j. At t=2: v = 20i-5j. |v| = √(400+25) = √425 = 5√17.

Direction: angle with -ve y-axis: tan θ = 20/5 = 4. So θ = tan⁻¹(4) with -ve y-axis.

The correct answer is Option 4: 5√17 m/s, making angle tan⁻¹4 with -ve Y-axis.

The transmitter-receiver pair behaves like two identical simple pendulums of length $$L = 8\;{\rm m}$$ that execute small-angle oscillations in the vertical plane containing the $$X$$-axis. Both start simultaneously from angular displacement $$\pm\theta_0$$ (with $$\theta_0 = \cos^{-1}(0.9)$$) on opposite sides of their respective equilibrium positions.

Step 1 : Maximum linear speed of either bob
At the lowest point the loss in gravitational potential energy equals the gain in kinetic energy: $$\tfrac12 m v_{\max}^2 = m g L\,(1-\cos\theta_0).$$ Hence $$v_{\max}= \sqrt{2gL\,(1-\cos\theta_0)} = \sqrt{2 \times 10 \times 8 \times (1-0.9)} = \sqrt{16}\;{\rm m\,s^{-1}} = 4\;{\rm m\,s^{-1}}.$$ So each pendulum bob attains an extreme speed of $$u = 4\;{\rm m\,s^{-1}}$$ along the $$X$$-axis.

Step 2 : Situations that give the greatest Doppler shift
For the Doppler effect (with air at rest) $$f' = f\;\frac{v + v_o}{v - v_s},$$ where $$v = 330\;{\rm m\,s^{-1}},\qquad v_s$$ = component of the transmitter’s velocity along the line joining it to the receiver (positive when the source moves away), $$v_o$$ = corresponding component of the receiver’s velocity (positive when the observer moves toward the source).

The two bobs always have equal magnitudes of velocity, but their directions keep changing. The largest possible value of the factor $$(v + v_o)/(v - v_s)$$ occurs when

• the transmitter moves directly away from the receiver with speed $$u$$  ($$v_s = +u$$), and
• simultaneously the receiver moves toward the transmitter with the same speed $$u$$  ($$v_o = +u$$).

(Both bobs then travel in the same physical direction along the $$X$$-axis; this condition is realised twice in every oscillation.)

Step 3 : Maximum and minimum audible frequencies
Maximum frequency: $$f_{\max}= 660\, \frac{330 + 4}{330 - 4} = 660 \times \frac{334}{326} \approx 676.4\;{\rm Hz}.$$ Minimum frequency (interchanging the velocity directions, i.e. $$v_s = -u,\, v_o = -u$$): $$f_{\min}= 660\, \frac{330 - 4}{330 + 4} = 660 \times \frac{326}{334} \approx 643.6\;{\rm Hz}.$$

Step 4 : Maximum variation in the received frequency
$$\Delta f_{\max}= f_{\max}-f_{\min} \approx 676.4 - 643.6 \approx 32.8\;{\rm Hz}.$$

Because the data are given only to two significant figures, any answer in the range 26 - 33 Hz is acceptable.

Final answer : about $$32\;{\rm Hz}$$  (maximum variation).

Let’s check each graph using basic rules of 1D motion.

Key idea: for any physical motion, at a given time there must be a unique value of position/velocity/etc. Also, graphs must satisfy physical definitions (like velocity = dx/dt, distance can’t decrease, etc.)

(A) Phase vs Time (straight line)
Phase increasing linearly with time is possible (like uniform motion in SHM phase representation).
No contradiction → valid.

(B) Velocity vs Displacement (circle)
This can represent motion like SHM where

$$v^2\propto(A^2-x^2)$$

which gives a circular/elliptic relation.
For each displacement, two velocities (forward/backward) are allowed → physically valid.

(C) Velocity vs Time (circle)
This is not possible.
At a given time, velocity must have a single value.
But a circle gives two velocities for the same time → violates definition.
So invalid.

(D) Total distance vs Time
Total distance must:

• always increase (never decrease)
• can be constant temporarily (if particle stops)

The graph shown is non-decreasing → valid.

Final answer:
A, B and D only
(C is not possible because it gives multiple velocities at the same time)

First, remember:

• Slope of x-t graph = velocity
• Straight line → constant velocity
• Horizontal line → zero velocity
• Average velocity = (final position − initial position) / time

(A) Average velocity from 0 to 3 s

From the graph:

$$At\ t=0,x=0$$

$$At\ t=3,x=5$$

So,

$$v_{avg}=\frac{5-0}{3}=\frac{5}{3}\approx1.67\text{ m/s}$$

Given statement says $$10m/s,$$ which is clearly wrong.

So, (A) is false

(B) Average velocity from 3 to 5 s

From graph:

• $$At\ t=3,x=5$$
• $$At\ t=5,x=5$$

So displacement does not change

$$v_{avg}=\frac{5-5}{5-3}=0$$

This means the object is at rest.

So, (B) is true

(C) Instantaneous velocity at t=2 s

At t=2, the graph lies on the straight line between:

• $$(1,-5)and(3,5)$$

Slope (velocity) of this line:

$$v=\frac{5-(-5)}{3-1}=\frac{10}{2}=5\text{ m/s}$$

Since it’s a straight line, velocity is constant in that region.

So, (C) is true

(D) Compare velocities

Average velocity from 5 to 7:

• $$At\ t=5,x=5$$
• $$At t=7,x=0$$

$$v_{avg}=\frac{0-5}{7-5}=\frac{-5}{2}=-2.5$$

Instantaneous velocity at 6.5 s:

This lies between t=6 and t=7, where graph is a straight line from:

• (6,10)to (7,0)

$$v=\frac{0-10}{7-6}=-10$$

Clearly:

$$-2.5\ne-10$$

So, (D) is false

(E) Average velocity from 0 to 9 s

• At t=0, x=0
• At t=9, x=0

$$v_{avg}=\frac{0-0}{9-0}=0$$

So, (E) is true

Final Answer:

(B), (C), (E) only

The displacement of the particle is given as
$$x = c_0\,(t^{2}-2) + c\,(t-2)^{2} \qquad -(1)$$

Step 1 - Find velocity
Velocity is the time-derivative of displacement:
$$v = \frac{dx}{dt}$$

Differentiating $$-(1)$$ term by term:

• For $$c_0\,(t^{2}-2)$$: $$\frac{d}{dt}\big(c_0\,(t^{2}-2)\big)=c_0\,(2t)=2c_0t$$
• For $$c\,(t-2)^{2}$$: $$\frac{d}{dt}\big(c\,(t-2)^{2}\big)=c\,(2\,(t-2))=2c\,(t-2)$$

Therefore
$$v = 2c_0t + 2c\,(t-2) \qquad -(2)$$

Step 2 - Find acceleration
Acceleration is the time-derivative of velocity:
$$a = \frac{dv}{dt}$$

Differentiating $$-(2)$$:

• $$\frac{d}{dt}\big(2c_0 t\big)=2c_0$$
• $$\frac{d}{dt}\big(2c\,(t-2)\big)=2c$$

Hence
$$a = 2c_0 + 2c = 2\,(c_0 + c) \qquad -(3)$$

Acceleration is a constant equal to $$2(c_0 + c)$$.

Step 3 - Check other statements

Initial velocity (at $$t=0$$) from $$-(2)$$:
$$v(0)=2c_0\,(0) + 2c\,(0-2) = -4c$$

Thus the initial velocity is $$-4c$$, not $$4c$$.

Conclusion
From $$-(3)$$, the only correct statement is: “the acceleration of the particle is $$2(c + c_0)$$,” which corresponds to Option D.

Distance travelled :A TO B,B TO A, A TO B

A TO B =$$\pi\ $$r
B to A =$$\pi\ $$r
A TO B =$$\pi\ $$r

Total distance =3*$$\pi\ $$r
Displacement : A TO B = 2r

Let us divide this problem into 2 parts:-

First 2 seconds where the velocity changes , which means acceleration is present

Second part from 2 seconds to 30.5 (28.5 seconds in total) seconds where Velocity is constant

We know that distance s:-

$$s=ut+\ \frac{\ 1}{2}at^2$$

For first 2 seconds :-

u=$$200\ \ \frac{\ m}{s}$$

a=$$\ \frac{\ v_f=v_i}{t}$$

a=$$\ \frac{\ 400-200}{2}$$

a=100 $$\ \frac{\ m}{s^2}$$

Now $$s=200\times\ 2\ +\ \ \frac{\ 1}{2}\times\ 100\times\ \left(2\right)^2$$

$$s=600\ m$$

For next 28.5 s a=0(constant velocity), Here u=400 $$\ \frac{\ m}{s}$$

so s=$$ut$$

s=$$400\times\ 28.5$$

s=11400 m

now adding both 

$$s_{total}$$ = 600+ 11400

$$s_{total}$$= 12000 m = 12 Km

Total distance covered  $$S = x + \frac{3}{2}x = \frac{5}{2}x$$.

Time taken in the first part  $$t_1 = \frac{\text{distance}}{\text{speed}} = \frac{x}{v_1}$$.

Given  $$v_1 = 5 \text{ m/s}$$, so  $$t_1 = \frac{x}{5}$$.

Time taken in the second part  $$t_2 = \frac{\frac{3}{2}x}{v_2} = \frac{3x}{2v_2}$$.

Total time  $$T = t_1 + t_2 = \frac{x}{5} + \frac{3x}{2v_2}$$.

The average velocity formula is
$$\text{Average velocity} = \frac{\text{Total distance}}{\text{Total time}}$$.

According to the question,
$$\frac{\frac{5}{2}x}{\frac{x}{5} + \frac{3x}{2v_2}} = \frac{50}{7}$$.

Cancel the common factor $$x$$ from numerator and denominator:
$$\frac{\frac{5}{2}}{\frac{1}{5} + \frac{3}{2v_2}} = \frac{50}{7}$$.

Cross-multiply:
$$\frac{5}{2} \times \frac{7}{50} = \frac{1}{5} + \frac{3}{2v_2}$$.

Simplify the left side:
$$\frac{5}{2} \times \frac{7}{50} = \frac{35}{100} = \frac{7}{20}$$.

Thus,
$$\frac{7}{20} = \frac{1}{5} + \frac{3}{2v_2}$$.

Convert $$\frac{1}{5}$$ to a denominator of 20:
$$\frac{1}{5} = \frac{4}{20}$$.

Subtract $$\frac{4}{20}$$ from both sides:
$$\frac{7}{20} - \frac{4}{20} = \frac{3}{2v_2}$$
$$\frac{3}{20} = \frac{3}{2v_2}$$.

Cancel the common factor 3:
$$\frac{1}{20} = \frac{1}{2v_2}$$.

Cross-multiply to find $$v_2$$:
$$2v_2 = 20 \; \Longrightarrow \; v_2 = 10 \text{ m/s}.$$

Hence, the required velocity $$v_2$$ is 10 m/s.

Let the tower height be $$h$$, initial speed be $$u$$, and $$g$$ be acceleration due to gravity.

Projected upward: Taking downward as positive:

$$h = -ut_1 + \frac{1}{2}gt_1^2$$ ... (1)

Projected downward:

$$h = ut_2 + \frac{1}{2}gt_2^2$$ ... (2)

Dropped (u = 0):

$$h = \frac{1}{2}gt^2$$ ... (3)

From (1): $$h = \frac{1}{2}gt_1^2 - ut_1$$
From (2): $$h = \frac{1}{2}gt_2^2 + ut_2$$

Adding (1) and (2): $$2h = \frac{1}{2}g(t_1^2 + t_2^2) + u(t_2 - t_1)$$

Subtracting (1) from (2): $$0 = \frac{1}{2}g(t_2^2 - t_1^2) + u(t_1 + t_2)$$

$$u = \frac{g(t_1^2 - t_2^2)}{2(t_1 + t_2)} = \frac{g(t_1 - t_2)}{2}$$

Substituting back into (2): $$h = \frac{g(t_1-t_2)}{2} \cdot t_2 + \frac{1}{2}gt_2^2 = \frac{g}{2}(t_1 t_2 - t_2^2 + t_2^2) = \frac{g t_1 t_2}{2}$$

From (3): $$t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \cdot \frac{g t_1 t_2}{2}}{g}} = \sqrt{t_1 t_2}$$.

The correct answer is Option A: $$\sqrt{t_1 t_2}$$.

$$S_1 = \frac{1}{2}a(p-1)^2$$, $$S_2 = \frac{1}{2}ap^2$$.

$$S_1 + S_2 = \frac{a}{2}[(p-1)^2 + p^2] = \frac{a}{2}(2p^2 - 2p + 1)$$

If this equals $$\frac{1}{2}at^2$$: $$t^2 = 2p^2 - 2p + 1$$, $$t = \sqrt{2p^2-2p+1}$$.

The answer corresponds to Option (2).

Let the initial velocity be $$u$$. After travelling 4 cm, velocity = $$\frac{2u}{3}$$.

Using $$v^2 = u^2 - 2as$$ (deceleration $$a$$):

$$\frac{4u^2}{9} = u^2 - 2a(0.04)$$

$$2a(0.04) = u^2 - \frac{4u^2}{9} = \frac{5u^2}{9}$$

$$a = \frac{5u^2}{9 \times 0.08} = \frac{5u^2}{0.72}$$

For the bullet to come to rest from $$v = \frac{2u}{3}$$:

$$0 = \frac{4u^2}{9} - 2a \cdot D'$$

$$D' = \frac{4u^2}{18a} = \frac{4u^2}{18 \times \frac{5u^2}{0.72}} = \frac{4 \times 0.72}{18 \times 5} = \frac{2.88}{90} = 0.032 \text{ m} = 32 \times 10^{-3} \text{ m}$$

So $$D = 32$$.

The answer is $$32$$, which corresponds to Option (1).

We need to find the velocity of a particle when its acceleration is zero, given $$x = t^3 - 6t^2 + 20t + 15$$.

Find the velocity.

$$v = \frac{dx}{dt} = 3t^2 - 12t + 20$$

Find the acceleration.

$$a = \frac{dv}{dt} = 6t - 12$$

Find when acceleration is zero.

$$6t - 12 = 0 \implies t = 2$$ s

Find velocity at $$t = 2$$.

$$v(2) = 3(4) - 12(2) + 20 = 12 - 24 + 20 = 8$$ m/s

The correct answer is Option (2): 8 m/s.

Let the total distance be $$2d$$.

First half (distance $$d$$): Speed = 6 m/s, so time = $$\frac{d}{6}$$.

Second half (distance $$d$$): Covered in two equal time intervals. Let each interval be $$t$$.

Distance in first interval = $$9t$$, distance in second = $$15t$$.

Total: $$9t + 15t = 24t = d$$, so $$t = \frac{d}{24}$$.

Total time for second half = $$2t = \frac{d}{12}$$.

Total time = $$\frac{d}{6} + \frac{d}{12} = \frac{2d + d}{12} = \frac{3d}{12} = \frac{d}{4}$$.

Average speed = $$\frac{\text{Total distance}}{\text{Total time}} = \frac{2d}{d/4} = 8$$ m/s.

The correct answer is Option 2: 8 m/s.

First, convert the speeds from km/h to m/s using the conversion factor: $$1 \text{ km/h} = \frac{5}{18} \text{ m/s}$$.

For Train A moving north:
Speed $$v_A = 72 \text{ km/h} = 72 \times \frac{5}{18} = 20 \text{ m/s}$$ (north direction).

For Train B moving south:
Speed $$v_B = 108 \text{ km/h} = 108 \times \frac{5}{18} = 30 \text{ m/s}$$ (south direction).

Assign directions: Let north be positive and south be negative.
Thus, $$v_A = +20 \text{ m/s}$$ and $$v_B = -30 \text{ m/s}$$.

Velocity of train B with respect to A ($$v_{BA}$$):
The relative velocity formula is $$v_{BA} = v_B - v_A$$.
Substitute values: $$v_{BA} = (-30) - (20) = -50 \text{ m/s}$$.

Velocity of ground with respect to B ($$v_{GB}$$):
The ground is stationary, so $$v_G = 0 \text{ m/s}$$.
The relative velocity formula is $$v_{GB} = v_G - v_B$$.
Substitute values: $$v_{GB} = 0 - (-30) = +30 \text{ m/s}$$.

Therefore, the velocities are $$-50 \text{ m/s}$$ and $$30 \text{ m/s}$$, which corresponds to option C.

Two cars travel towards each other at 20 m/s each, initially 300 m apart, both braking with retardation 2 m/s$$^2$$. We need the distance between them when they stop.

Using the kinematic equation $$v^2 = u^2 - 2as$$ (where $$v = 0$$ at stop):

$$ 0 = (20)^2 - 2(2)s $$

$$ 0 = 400 - 4s $$

$$ s = \frac{400}{4} = 100 \text{ m} $$

Each car travels 100 m before stopping.

Both cars move toward each other, so the total distance closed between them is:

$$ s_{total} = 100 + 100 = 200 \text{ m} $$

Initial separation was 300 m. After both cars stop:

$$ d = 300 - 200 = 100 \text{ m} $$

The distance between them when they come to rest is 100 m.

The correct answer is Option (2): 100 m.

Let A be at distance $$s$$ from the starting point. The body covers 80 m from A to B in 2 seconds.

Distance from start to A: $$s = \frac{1}{2}gt_1^2$$ where $$t_1$$ is time to reach A.

Distance from start to B: $$s + 80 = \frac{1}{2}g(t_1 + 2)^2$$

$$80 = \frac{1}{2}g[(t_1+2)^2 - t_1^2] = \frac{1}{2}(10)(4t_1 + 4) = 5(4t_1 + 4) = 20t_1 + 20$$

$$20t_1 = 60$$

$$t_1 = 3$$ s

$$s = \frac{1}{2}(10)(9) = 45$$ m

The answer is $$\boxed{45}$$ m.

A body starts from rest on a frictionless plane. We are given the formula for the ratio $$\frac{S_{n-1}}{S_n} = \left(1 - \frac{2}{x}\right)$$ for $$n = 10$$, and we need to find $$x$$.

We first recall the distance covered in the $$n$$-th second under uniform acceleration.

For a body starting from rest (initial velocity $$u = 0$$) with uniform acceleration $$a$$, the distance covered from $$t = 0$$ to $$t = n$$ is:

$$ S(n) = \frac{1}{2}an^2 $$

The distance covered in the $$n$$-th second (between $$t = n-1$$ and $$t = n$$) is:

$$ S_n = S(n) - S(n-1) = \frac{1}{2}an^2 - \frac{1}{2}a(n-1)^2 $$

Expanding $$(n-1)^2 = n^2 - 2n + 1$$:

$$ S_n = \frac{a}{2}\left[n^2 - (n^2 - 2n + 1)\right] = \frac{a}{2}(2n - 1) $$

Similarly, replacing $$n$$ with $$n-1$$ in the formula gives:

$$ S_{n-1} = \frac{a}{2}(2(n-1) - 1) = \frac{a}{2}(2n - 3) $$

Next, we compute the ratio $$\frac{S_{n-1}}{S_n}$$.

$$ \frac{S_{n-1}}{S_n} = \frac{\frac{a}{2}(2n-3)}{\frac{a}{2}(2n-1)} = \frac{2n - 3}{2n - 1} $$

This fraction can be rewritten as:

$$ \frac{2n - 3}{2n - 1} = \frac{(2n-1) - 2}{2n - 1} = 1 - \frac{2}{2n - 1} $$

Since $$\frac{S_{n-1}}{S_n} = 1 - \frac{2}{x}$$, equating this to $$1 - \frac{2}{2n - 1}$$ gives:

$$ \frac{2}{x} = \frac{2}{2n - 1} $$

$$ x = 2n - 1 $$

Finally, substituting $$n = 10$$ gives:

$$ x = 2(10) - 1 = 20 - 1 = 19 $$

The value of $$x$$ is $$\boxed{19}$$.

v=72 km/h=20 m/s. t=4s. a=-20/4=-5 m/s². s=ut+½at²=20(4)+½(-5)(16)=80-40=40 m.

The answer is $$\boxed{40}$$.

Find the acceleration of a particle with velocity $$v = 4\sqrt{x}$$ m/s.

Acceleration can be expressed as $$ a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = \frac{dv}{dx} \cdot v $$ because by the chain rule, $$\frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt}$$ and $$\frac{dx}{dt} = v$$.

Since $$v = 4\sqrt{x} = 4x^{1/2}$$, it follows that $$ \frac{dv}{dx} = 4 \times \frac{1}{2}x^{-1/2} = \frac{2}{\sqrt{x}} $$.

Substituting into the expression for acceleration gives $$ a = \frac{dv}{dx} \cdot v = \frac{2}{\sqrt{x}} \times 4\sqrt{x} = \frac{2 \times 4\sqrt{x}}{\sqrt{x}} = 8 \text{ m/s}^2 $$. Note that $$\sqrt{x}$$ cancels, making the acceleration constant and independent of position.

The answer is 8 m/s$$^2$$.

Given $$x = -3t^3 + 18t^2 + 16t$$, we want to find the velocity when the acceleration is zero.

Differentiating the position with respect to time gives the velocity: $$ v = \frac{dx}{dt} = \frac{d}{dt}(-3t^3 + 18t^2 + 16t) = -9t^2 + 36t + 16 $$.

Further differentiating the velocity yields the acceleration: $$ a = \frac{dv}{dt} = \frac{d}{dt}(-9t^2 + 36t + 16) = -18t + 36 $$.

Setting the acceleration to zero gives $$ -18t + 36 = 0 \implies 18t = 36 \implies t = 2 \text{ s} $$.

Substituting $$t = 2$$ s into the velocity expression gives $$ v(2) = -9(2)^2 + 36(2) + 16 = -9(4) + 72 + 16 = -36 + 72 + 16 = 52 \text{ m/s} $$.

The answer is 52 m/s.

$$x^2 = 1+t^2$$.

$$2x\dot{x} = 2t \Rightarrow \dot{x} = t/x$$.

$$\ddot{x} = (x-t\dot{x})/x^2 = (x-t^2/x)/x^2 = (x^2-t^2)/x^3 = 1/x^3$$.

So $$n = 3$$.

The answer is 3.

We are given that in the time interval from $$t$$ to $$(t+1)$$ s, the displacement is $$125$$ m and the increase in velocity is $$50$$ m/s.

Since the increase in velocity in 1 second equals the acceleration:

$$a = 50 \text{ m/s}^2$$

The displacement in the interval from $$t$$ to $$(t+1)$$ s can be written as:

$$s = v_t \cdot (1) + \frac{1}{2}a(1)^2 = v_t + \frac{a}{2}$$

where $$v_t$$ is the velocity at time $$t$$.

$$125 = v_t + \frac{50}{2} = v_t + 25$$

$$v_t = 100 \text{ m/s}$$

The velocity at time $$(t+1)$$:

$$v_{t+1} = v_t + a = 100 + 50 = 150 \text{ m/s}$$

The distance travelled in the $$(t+2)$$-th second (i.e., from $$(t+1)$$ to $$(t+2)$$ s) is:

$$s_{(t+2)\text{th}} = v_{t+1} + \frac{a}{2} = 150 + \frac{50}{2} = 150 + 25 = 175 \text{ m}$$

The correct answer is $$175$$ m.

Let the ball start from the origin with speed $$v_0$$ and projection angle $$\theta_0$$ (measured from the +x-axis).
Its velocity components are $$v_{bx}=v_0\cos\theta_0$$ and $$v_{by}=v_0\sin\theta_0$$.

The stone is thrown simultaneously from $$(L,0)$$ at an angle $$(180^\circ-\theta_1)$$ with the +x-axis. Hence its horizontal component is toward the negative x-direction:
$$v_{sx}=-v_s\cos\theta_1$$, $$v_{sy}=v_s\sin\theta_1$$, where $$v_s$$ is the stone’s launch speed.

Suppose they meet after time $$t$$ (same $$t$$ for both because they start together).

Horizontal co-ordinate equality
$$x_{\text{ball}} = x_{\text{stone}}$$ gives $$v_0\cos\theta_0\,t = L + (-v_s\cos\theta_1)\,t$$ $$\Rightarrow t\,(v_0\cos\theta_0+v_s\cos\theta_1)=L \quad -(1)$$

Vertical co-ordinate equality
Both experience the same downward acceleration $$g$$, so the $$-\dfrac12 g t^2$$ term cancels: $$v_0\sin\theta_0\,t = v_s\sin\theta_1\,t$$ Since $$t\gt 0$$, $$v_0\sin\theta_0 = v_s\sin\theta_1 \quad -(2)$$

From $$(2)$$, express the stone’s speed:
$$v_s = \dfrac{v_0\sin\theta_0}{\sin\theta_1} \quad -(3)$$

Insert $$(3)$$ into $$(1)$$ to obtain the common time of flight:

$$t = \dfrac{L}{v_0\cos\theta_0 + \left(\dfrac{v_0\sin\theta_0}{\sin\theta_1}\right)\cos\theta_1 } = \dfrac{L}{v_0\left(\cos\theta_0+\sin\theta_0\cot\theta_1\right)} \quad -(4)$$

Thus, for any pair $$(\theta_0,\theta_1)$$, $$T(\theta_0,\theta_1)=\dfrac{L}{v_0\left(\cos\theta_0+\sin\theta_0\cot\theta_1\right)}$$

Case 1: $$(\theta_0,\theta_1)=(45^\circ,45^\circ)$$ $$\cos45^\circ=\sin45^\circ=\dfrac{\sqrt2}{2}, \;\cot45^\circ=1$$
Denominator in $$(4)$$: $$\cos45^\circ+\sin45^\circ\cot45^\circ = \dfrac{\sqrt2}{2}+\dfrac{\sqrt2}{2}= \sqrt2$$ Hence $$T_1=\dfrac{L}{v_0\sqrt2} \quad -(5)$$

Case 2: $$(\theta_0,\theta_1)=(60^\circ,30^\circ)$$ $$\cos60^\circ=\dfrac12,\; \sin60^\circ=\dfrac{\sqrt3}{2},\; \cot30^\circ=\sqrt3$$
Denominator in $$(4)$$: $$\dfrac12+\dfrac{\sqrt3}{2}\,\sqrt3 = \dfrac12+\dfrac{3}{2}=2$$ Thus $$T_2=\dfrac{L}{2v_0} \quad -(6)$$

Required ratio
$$\left(\dfrac{T_1}{T_2}\right)^2 =\left(\dfrac{\,L/(v_0\sqrt2)\,}{\,L/(2v_0)\,}\right)^2 =\left(\dfrac{2}{\sqrt2}\right)^2 =\left(\sqrt2\right)^2 =2$$

Therefore, $$(T_1/T_2)^2 = 2$$.

A body travels 102.5 m in the $$n$$-th second and 115.0 m in the $$(n+2)$$-th second. We need to find the acceleration.

For a body with initial velocity $$u$$ and constant acceleration $$a$$, the distance covered in the $$n$$-th second (i.e., between $$t = n-1$$ and $$t = n$$) is given by $$ s_n = u + \frac{a}{2}(2n - 1)\,. $$ This formula follows from $$s_n = S(n) - S(n-1) = \left(un + \frac{1}{2}an^2\right) - \left(u(n-1) + \frac{1}{2}a(n-1)^2\right) = u + \frac{a}{2}(2n-1)\,. $$

Using the given data, for the $$n$$-th second we have $$ s_n = u + \frac{a}{2}(2n - 1) = 102.5\,, $$ and for the $$(n+2)$$-th second $$ s_{n+2} = u + \frac{a}{2}(2(n+2) - 1) = u + \frac{a}{2}(2n + 3) = 115.0\,. $$

Subtracting these two equations gives $$ s_{n+2} - s_n = \frac{a}{2}[(2n+3) - (2n-1)]\,, $$ so $$ 115.0 - 102.5 = \frac{a}{2}[2n + 3 - 2n + 1]\,, $$ which simplifies to $$ 12.5 = \frac{a}{2} \times 4 \quad\Longrightarrow\quad 12.5 = 2a\,. $$

Therefore, $$ a = \frac{12.5}{2} = 6.25 \text{ m/s}^2\,. $$ The correct answer is Option (1): 6.25 m/s2.

The time-distance relation is given as $$t = \alpha x^{2} + \beta x$$, where $$\alpha$$ and $$\beta$$ are constants.

First differentiate with respect to $$x$$:
$$\frac{dt}{dx} = 2\alpha x + \beta$$ $$-(1)$$

Velocity is $$v = \frac{dx}{dt} = \frac{1}{\dfrac{dt}{dx}} = \frac{1}{2\alpha x + \beta}$$ $$-(2)$$

Acceleration is $$a = \frac{dv}{dt}$$.
Using the chain rule, $$a = \frac{dv}{dx}\,\frac{dx}{dt} = v \,\frac{dv}{dx}$$ $$-(3)$$

Differentiate $$v$$ from $$(2)$$ with respect to $$x$$:
$$\frac{dv}{dx} = \frac{d}{dx}\left[(2\alpha x + \beta)^{-1}\right] = -1 \cdot (2\alpha)\,(2\alpha x + \beta)^{-2} = -\frac{2\alpha}{(2\alpha x + \beta)^{2}}$$ $$-(4)$$

Substitute $$(2)$$ and $$(4)$$ into $$(3)$$:
$$a = v\left(-\frac{2\alpha}{(2\alpha x + \beta)^{2}}\right) = -\frac{2\alpha\,v}{(2\alpha x + \beta)^{2}}$$ $$-(5)$$

From $$(2)$$, $$v = \dfrac{1}{2\alpha x + \beta} \;\Rightarrow\; 2\alpha x + \beta = \frac{1}{v}$$ $$-(6)$$

Insert $$(6)$$ into $$(5)$$ to eliminate $$x$$:
$$a = -\frac{2\alpha\,v}{\left(\dfrac{1}{v}\right)^{2}} = -2\alpha\,v \, v^{2} = -2\alpha\,v^{3}$$

Thus the required relation is $$a = -2\alpha v^{3}$$.

Option A is correct.

An artillery piece of mass $$M_1$$ fires a shell of mass $$M_2$$ horizontally, and we seek the ratio of the kinetic energy of the artillery to that of the shell immediately after firing.

Before firing, both the artillery and the shell are at rest, so the total momentum is zero. By conservation of momentum, the total momentum after firing must also be zero:
$$ M_1 v_1 + M_2 v_2 = 0 $$ where $$v_1$$ is the recoil velocity of the artillery and $$v_2$$ is the velocity of the shell. It follows that
$$ M_1 v_1 = -M_2 v_2 $$ and taking magnitudes gives
$$ |v_1| = \frac{M_2 |v_2|}{M_1} $$.

The kinetic energy of the artillery is
$$ KE_1 = \frac{1}{2}M_1 v_1^2 \quad (\text{artillery}) $$ and that of the shell is
$$ KE_2 = \frac{1}{2}M_2 v_2^2 \quad (\text{shell}) $$.

Their ratio is
$$ \frac{KE_1}{KE_2} = \frac{\frac{1}{2}M_1 v_1^2}{\frac{1}{2}M_2 v_2^2} = \frac{M_1}{M_2} \cdot \frac{v_1^2}{v_2^2} $$. Substituting $$v_1 = \frac{M_2 v_2}{M_1}$$ gives
$$ \frac{KE_1}{KE_2} = \frac{M_1}{M_2} \cdot \frac{M_2^2 v_2^2}{M_1^2 v_2^2} = \frac{M_1}{M_2} \cdot \frac{M_2^2}{M_1^2} = \frac{M_2}{M_1} $$.

An alternative derivation notes that both objects have the same magnitude of momentum ($$p = M_1|v_1| = M_2|v_2|$$), and since $$KE = \frac{p^2}{2M}$$ one obtains
$$ \frac{KE_1}{KE_2} = \frac{p^2/(2M_1)}{p^2/(2M_2)} = \frac{M_2}{M_1} $$.

The correct answer is Option (2): $$\frac{M_2}{M_1}$$.

A particle starts with initial velocity 10.0 m/s and accelerates uniformly at 2.0 m/s$$^2$$. We need to find the time to reach 60.0 m/s.

First, identify the known quantities.

Initial velocity: $$u = 10.0$$ m/s

Acceleration: $$a = 2.0$$ m/s$$^2$$

Final velocity: $$v = 60.0$$ m/s

Time: $$t = ?$$

Next, select the appropriate equation of motion.

Since we know $$u$$, $$v$$, and $$a$$, and need to find $$t$$, we use the first equation of motion:

$$ v = u + at $$

This equation relates velocity to time for uniformly accelerated motion. It follows directly from the definition of acceleration: $$a = \frac{v - u}{t}$$.

Now, substitute and solve for $$t$$.

$$ 60.0 = 10.0 + 2.0 \times t $$

$$ 2.0t = 60.0 - 10.0 = 50.0 $$

$$ t = \frac{50.0}{2.0} = 25\;\text{s} $$

The time taken is 25 s.

The correct answer is Option 1: 25 s.

Average velocity for equal distances: $$v = \frac{2x}{x/v_1 + x/v_2} = \frac{2v_1v_2}{v_1 + v_2}$$

This gives: $$\frac{2}{v} = \frac{v_1 + v_2}{v_1v_2} = \frac{1}{v_1} + \frac{1}{v_2}$$

The correct answer is Option 4.

A car travels a distance $$x$$ with speed $$v_1$$ and then the same distance $$x$$ with speed $$v_2$$ in the same direction. We need to find the average speed.

Calculate the total distance.

Total distance = $$x + x = 2x$$

Calculate the total time.

Time for first part: $$t_1 = \frac{x}{v_1}$$

Time for second part: $$t_2 = \frac{x}{v_2}$$

Total time: $$t_1 + t_2 = \frac{x}{v_1} + \frac{x}{v_2} = x\left(\frac{v_1 + v_2}{v_1 v_2}\right)$$

Calculate the average speed.

$$ v_{avg} = \frac{\text{Total distance}}{\text{Total time}} = \frac{2x}{x\left(\frac{v_1 + v_2}{v_1 v_2}\right)} = \frac{2v_1 v_2}{v_1 + v_2} $$

This is the harmonic mean of the two speeds, which is the correct formula when equal distances are covered at different speeds.

The correct answer is Option D: $$\frac{2v_1 v_2}{v_1 + v_2}$$.

$$v_{avg} = \frac{\text{Total Distance}}{\text{Total Time}}$$

$$AB = BC = x$$

$$AD = 3AB = 3x$$

$$CD = AD - (AB + BC) = 3x - 2x = x$$

$$t_1 = \frac{x}{v_1}, \quad t_2 = \frac{x}{v_2}, \quad t_3 = \frac{x}{v_3}$$

$$v_{avg} = \frac{x + x + x}{t_1 + t_2 + t_3}$$

$$v_{avg} = \frac{3x}{\frac{x}{v_1} + \frac{x}{v_2} + \frac{x}{v_3}}$$

$$v_{avg} = \frac{3}{\frac{1}{v_1} + \frac{1}{v_2} + \frac{1}{v_3}}$$

$$v_{avg} = \frac{3v_1v_2v_3}{v_1v_2 + v_2v_3 + v_3v_1}$$

The shaded region under a $$v\!-\!t$$ graph always gives displacement when the sign (above / below the time-axis) is taken into account and gives distance when only the magnitude of the area is added.

From the given graph it is seen that:

• For the first $$20\ \text{s}$$ the velocity is constant and positive (the horizontal line at height $$v_0$$).
   Area (rectangle) $$A_1 = v_0 \times 20$$

• For the next $$5\ \text{s}$$ (from $$t = 20\ \text{s}$$ to $$t = 25\ \text{s}$$) the velocity is the same in magnitude but negative (the horizontal line at depth $$-v_0$$).
   Area (rectangle) $$A_2 = (-\,v_0) \times 5$$

Distance travelled in 25 s
Add magnitudes of both areas:$$|A_1| + |A_2| = v_0(20) + v_0(5) = 25\,v_0$$

Displacement in 25 s
Add the signed areas:$$A_1 + A_2 = v_0(20) + (-v_0)(5) = v_0(20 - 5) = 15\,v_0$$

Required ratio
$$\text{distance} \; : \; \text{displacement} = \frac{25\,v_0}{15\,v_0} = \frac{5}{3}$$

Hence the ratio of distance to displacement in the first $$25\ \text{s}$$ of motion is $$\frac{5}{3}$$.
Option C is correct.

We need to analyze each statement carefully.

For Statement I, it says the area under the velocity-time graph gives the distance travelled. However, the area under the velocity-time graph actually gives the displacement, not the distance. Distance is the total path length and equals the area only when velocity does not change sign. If velocity becomes negative, the area below the time axis is negative, giving displacement. To get distance, we would need the area under the $$|v|$$ vs $$t$$ graph. So Statement I is incorrect in general.

Now consider Statement II, which says the area under the acceleration-time graph equals the change in velocity. We know that $$a = \frac{dv}{dt}$$, so

$$\int_{t_1}^{t_2} a \, dt = v(t_2) - v(t_1) = \Delta v$$

The area under the acceleration-time graph indeed gives the change in velocity (since integrating acceleration over time recovers the velocity change). So Statement II is true.

Hence, Statement I is incorrect but Statement II is true. Hence, the correct answer is Option 4.

Key idea:

$$v=\frac{dx}{dt}=\text{slope of }x\text{-}t\text{ graph}$$

Graph A:-

• Slope increases with time → velocity increases with time
• That means accelerated motion

Matches II (v increases linearly with time)

So:
A → II

Graph B:-

• Initially slope is large (negative direction), then becomes less steep
• So velocity starts negative and approaches zero
• Velocity increasing towards zero (but still negative initially)

Matches IV (velocity increasing from negative towards zero)

So:
B → IV

Graph C:-

• First slope positive → velocity positive
• Then slope negative → velocity negative
• Sudden change → velocity jumps from +v to −v

Matches III (step from +v to −v)

So:
C → III

Graph D:-

• Constant slope → constant velocity

Matches I (constant velocity)

So:
D → I

Final answer:

A→II,B→IV,C→III,D→I

Given: $$x = 4t^2$$

Velocity: $$v = \frac{dx}{dt} = 8t$$

At $$t = 5$$ s:

$$v = 8 \times 5 = 40$$ m/s

The correct answer is Option 1: 40 m s⁻¹.

look at the graph slowly and interpret each thing correctly

horizontal axis → time
vertical axis → position (distance from school)

both students start from school (origin)

now key ideas
slope of line = speed
final height = how far their home is
final time coordinate = time taken

1. time taken

both lines end at the same time coordinate on x-axis
so both students reach home at the same time

so A does NOT take lesser time
(C) is wrong

2.who is faster

compare slopes

line B is steeper than A
steeper line → greater slope → higher speed

so B is faster
(E) is correct
(D) is wrong

3. who lives closer

compare final heights (end points on y-axis)

B’s endpoint is higher → B has travelled more distance
so B lives farther

A’s endpoint is lower → A lives closer

so (A) is correct
(B) is wrong

final answer: A and E

The velocity field is $$\vec v = \alpha\bigl(y\,\hat x + 2x\,\hat y\bigr)$$.
Hence

$$v_x = \alpha y , \qquad v_y = 2\alpha x$$

Acceleration is the material (total) derivative of velocity:

$$\vec a = \frac{d\vec v}{dt} = v_x\frac{\partial\vec v}{\partial x} + v_y\frac{\partial\vec v}{\partial y}$$

Component $$a_x$$:

$$a_x = v_x\frac{\partial v_x}{\partial x} + v_y\frac{\partial v_x}{\partial y}$$

Since $$v_x = \alpha y$$, we have $$\frac{\partial v_x}{\partial x}=0$$ and $$\frac{\partial v_x}{\partial y}=\alpha$$.

Therefore $$a_x = v_x(0) + v_y(\alpha) = (2\alpha x)(\alpha) = 2\alpha^{2}x$$.

Component $$a_y$$:

$$a_y = v_x\frac{\partial v_y}{\partial x} + v_y\frac{\partial v_y}{\partial y}$$

Because $$v_y = 2\alpha x$$, we have $$\frac{\partial v_y}{\partial x}=2\alpha$$ and $$\frac{\partial v_y}{\partial y}=0$$.

Thus $$a_y = v_x(2\alpha) + v_y(0) = (\alpha y)(2\alpha) = 2\alpha^{2}y$$.

Combining the two components:

$$\vec a = 2\alpha^{2}\bigl(x\,\hat x + y\,\hat y\bigr)$$

Using Newton’s second law $$\vec F = m\vec a$$:

$$\vec F = 2m\alpha^{2}\bigl(x\,\hat x + y\,\hat y\bigr)$$

Therefore, the correct option is:
Option A which is: $$\vec{F} = 2m\alpha^{2}(x\hat{x} + y\hat{y})$$.

When the ball is dropped from a height of $$9.8$$ m, its velocity just before hitting the floor is found using $$v^2 = 2gh$$: $$v = \sqrt{2 \times 10 \times 9.8} = \sqrt{196} = 14$$ m/s (downward).

After rebounding to a height of $$5.0$$ m, the velocity just after leaving the floor is $$v' = \sqrt{2 \times 10 \times 5} = \sqrt{100} = 10$$ m/s (upward).

Taking upward as positive, the velocity changes from $$-14$$ m/s (downward, just before contact) to $$+10$$ m/s (upward, just after contact). The average acceleration during the $$0.2$$ s contact time is $$a = \dfrac{v' - v}{\Delta t} = \dfrac{10 - (-14)}{0.2} = \dfrac{24}{0.2} = \boxed{120}$$ m s$$^{-2}$$.

A body is dropped from height $$h_1$$ and rebounds to height $$h_2$$. The ratio of velocities just before and after hitting the ground is $$4$$.

The velocity just before hitting the ground is $$v_1 = \sqrt{2gh_1}$$ and the velocity just after rebounding is $$v_2 = \sqrt{2gh_2}$$, satisfying $$\dfrac{v_1}{v_2} = 4$$.

Squaring gives $$\dfrac{v_1^2}{v_2^2} = 16$$, so the ratio of kinetic energies after and before impact is $$\dfrac{KE_{\text{after}}}{KE_{\text{before}}} = \dfrac{v_2^2}{v_1^2} = \dfrac{1}{16}$$.

The percentage loss in kinetic energy is $$\% \text{ loss} = \left(1 - \dfrac{v_2^2}{v_1^2}\right)\times 100 = \left(1 - \dfrac{1}{16}\right)\times 100 = \dfrac{15}{16}\times 100 = 93.75\%$$.

Since $$\dfrac{x}{4} = 93.75$$, it follows that $$x = 93.75 \times 4 = 375$$.

The value of $$x$$ is $$\boxed{375}$$.

A horse rider covers half the distance at 5 m/s, then the remaining half distance with speeds 10 and 15 m/s for equal time intervals.

Set up variables.

Let the total distance = $$2d$$.

Time for first half.

$$t_1 = \frac{d}{5}$$

Time for second half.

Let the time for each speed be $$t_2$$. Distance in second half:

$$d = 10t_2 + 15t_2 = 25t_2 \implies t_2 = \frac{d}{25}$$

Total time for second half = $$2t_2 = \frac{2d}{25}$$

Calculate mean speed.

Total time = $$\frac{d}{5} + \frac{2d}{25} = \frac{5d + 2d}{25} = \frac{7d}{25}$$

Mean speed = $$\frac{2d}{\frac{7d}{25}} = \frac{50}{7}$$ m/s

Therefore, $$x = \boxed{50}$$.

Scenario 1: Full stopping distance

Initial velocity $$u = 20 \text{ m/s}$$

Final velocity $$v_1 = 0 \text{ m/s}$$

Displacement $$s_1 = 500 \text{ m}$$

$$v_1^2 - u^2 = 2as_1$$

$$0^2 - (20)^2 = 2a(500)$$

$$-400 = 1000a$$

$$a = -0.4 \text{ m/s}^2$$

Scenario 2: Half the stopping distance

The brakes are now applied at half the distance, meaning the train travels $$s_2 = 250 \text{ m}$$ while decelerating. We want to find its speed $$v_2$$ as it crosses the station.

Initial velocity $$u = 20 \text{ m/s}$$

Acceleration $$a = -0.4 \text{ m/s}^2$$

Displacement $$s_2 = 250 \text{ m}$$

$$v_2^2 - u^2 = 2as_2$$

$$v_2^2 - (20)^2 = 2(-0.4)(250)$$

$$v_2^2 - 400 = -0.8(250)$$

$$v_2^2 - 400 = -200$$

$$v_2^2 = 200$$

$$v_2 = \sqrt{200} \text{ m/s}$$

We have a particle of mass $$m = 10$$ g $$= 10^{-2}$$ kg moving in a straight line with retardation $$2x$$, where $$x$$ is the displacement in metres. Since it is retardation, the acceleration is $$a = -2x$$ m/s$$^2$$.

The retarding force is $$F = ma = -2mx$$. Using the work-energy theorem, the work done by this force over a displacement from 0 to $$x$$ is:

$$W = \int_0^x (-2mx) \, dx = -2m \left[\frac{x^2}{2}\right]_0^x = -mx^2$$

Since $$W = KE_f - KE_i$$, the loss of kinetic energy is:

$$\text{Loss of KE} = -W = mx^2 = 10^{-2} \times x^2 \text{ J}$$

We can verify this using kinematics. Since $$a = v\dfrac{dv}{dx} = -2x$$, integrating both sides:

$$\int_{v_0}^{v} v \, dv = \int_0^x (-2x) \, dx$$

$$\frac{v^2 - v_0^2}{2} = -x^2$$

or $$v_0^2 - v^2 = 2x^2$$

So the loss of kinetic energy is $$\frac{1}{2}m(v_0^2 - v^2) = \frac{1}{2} \times 10^{-2} \times 2x^2 = 10^{-2} \times x^2$$ J, which confirms our result.

Comparing with $$10^{-n} \cdot x^2$$, we get $$n = 2$$.

Hence, the answer is $$2$$.

A ball is thrown vertically upward with an initial speed of 150 m/s, and we are asked to determine the value of x such that the ratio of its velocities at t = 3 s and t = 5 s equals $$\frac{x+1}{x}$$.

Using the relation $$v = u - gt$$ with g = 10 m/s2 and upward taken as positive, we find the velocity at t = 3 s and t = 5 s as follows:

$$

v_3 = 150 - 10(3) = 150 - 30 = 120\;\text{m/s}

$$

$$

v_5 = 150 - 10(5) = 150 - 50 = 100\;\text{m/s}

$$

Since the required ratio is $$\frac{v_3}{v_5} = \frac{120}{100} = \frac{6}{5}$$, setting this equal to $$\frac{x+1}{x}$$ leads to

$$

\frac{x+1}{x} = \frac{6}{5}

$$

$$

5(x+1) = 6x \implies 5x + 5 = 6x \implies x = 5

$$

Hence, the correct answer is Option 4: 5.

$$x = 5t^2 - 4t + 5$$

Velocity: $$v = \frac{dx}{dt} = 10t - 4$$

At $$t = 2$$ s: $$v = 10(2) - 4 = 20 - 4 = 16$$ m/s

The magnitude of velocity is $$\mathbf{16}$$ m/s.

look at the v-t graph as areas of rectangles

area above time axis → positive displacement
area below time axis → negative displacement

now take each interval one by one

from 0 to 2 s
velocity = +8 m/s (constant)
so displacement = area = 8 × 2 = 16

from 2 to 4 s
velocity = −4 m/s
so displacement = −4 × 2 = −8

from 4 to 8 s
velocity = +4 m/s
so displacement = 4 × 4 = 16

from 8 to 10 s
velocity = −4 m/s
so displacement = −4 × 2 = −8

now combine properly

net displacement = 16 − 8 + 16 − 8 = 16

distance is different → here we don’t cancel signs
we take total path length

distance = 16 + 8 + 16 + 8 = 48

so finally

ratio = displacement / distance = 16 / 48 = 1/3

idea to remember
displacement = signed area
distance = total area (ignore sign)

A toy starts from rest under constant acceleration and travels 10 m in $$t$$ seconds. We need to find the distance in the next $$t$$ seconds.

Using $$s = ut + \frac{1}{2}at^2$$ with $$u = 0$$:

$$10 = \frac{1}{2}at^2 \quad \ldots (1)$$

$$s_{2t} = \frac{1}{2}a(2t)^2 = \frac{1}{2}a \cdot 4t^2 = 4 \times \frac{1}{2}at^2 = 4 \times 10 = 40 \text{ m}$$

$$s_{\text{next}} = s_{2t} - s_t = 40 - 10 = 30 \text{ m}$$

Hence, the correct answer is Option C: 30 m.

We are given the positions of two buses:

$$x_P(t) = \alpha t + \beta t^2$$

$$x_Q(t) = ft - t^2$$

The velocity of each bus is the derivative of its position with respect to time.

Velocity of bus P:

$$v_P = \frac{dx_P}{dt} = \alpha + 2\beta t$$

Velocity of bus Q:

$$v_Q = \frac{dx_Q}{dt} = f - 2t$$

For same velocity, we set $$v_P = v_Q$$:

$$\alpha + 2\beta t = f - 2t$$

$$2\beta t + 2t = f - \alpha$$

$$t(2\beta + 2) = f - \alpha$$

$$t = \frac{f - \alpha}{2(\beta + 1)} = \frac{f - \alpha}{2(1 + \beta)}$$

Hence, the correct answer is Option D.

We have a ball thrown vertically upward reaching maximum height $$h$$. At the top, all kinetic energy is converted to potential energy, so $$\frac{1}{2}mu^2 = mgh$$, giving $$u = \sqrt{2gh}$$.

Using the equation $$s = ut - \frac{1}{2}gt^2$$ with $$s = \frac{h}{3}$$, we get $$\frac{h}{3} = ut - \frac{1}{2}gt^2$$. Since $$u = \sqrt{2gh}$$ and $$h = \frac{u^2}{2g}$$, we substitute $$\frac{h}{3} = \frac{u^2}{6g}$$. The equation becomes $$\frac{u^2}{6g} = ut - \frac{1}{2}gt^2$$, which rearranges to $$3g^2t^2 - 6gtu + u^2 = 0$$. Dividing through by $$g^2$$ and writing $$T_0 = \frac{u}{g}$$ (total time to reach the top), we get $$3t^2 - 6T_0 t + T_0^2 = 0$$.

Applying the quadratic formula: $$t = \frac{6T_0 \pm \sqrt{36T_0^2 - 12T_0^2}}{6} = \frac{6T_0 \pm \sqrt{24T_0^2}}{6} = \frac{6T_0 \pm 2\sqrt{6}\,T_0}{6} = T_0\left(\frac{3 \pm \sqrt{6}}{3}\right)$$.

The ball is at height $$\frac{h}{3}$$ at two times: $$t_1 = T_0\left(\frac{3 - \sqrt{6}}{3}\right)$$ (going up) and $$t_2 = T_0\left(\frac{3 + \sqrt{6}}{3}\right)$$ (coming down).

The required ratio is $$\frac{t_1}{t_2} = \frac{3 - \sqrt{6}}{3 + \sqrt{6}}$$. Rationalising by multiplying numerator and denominator by $$(3 - \sqrt{6})$$: $$\frac{(3 - \sqrt{6})^2}{9 - 6} = \frac{9 - 6\sqrt{6} + 6}{3} = \frac{15 - 6\sqrt{6}}{3} = 5 - 2\sqrt{6}$$.

Now we check Option B: $$\frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}$$. Rationalising: $$\frac{(\sqrt{3} - \sqrt{2})^2}{3 - 2} = (3 - 2\sqrt{6} + 2) = 5 - 2\sqrt{6}$$.

This matches our result exactly. Hence, the correct answer is Option B.

We need to identify the correct velocity-time graph for a bullet shot vertically downward. Initially, the velocity $$u = 100$$ m/s downward; taking downward as positive, the velocity after time $$t$$ is given by $$v = u + gt = 100 + 10 \times 10 = 200\text{ m/s}$$ at $$t = 10\text{ s}$$. Thus the velocity increases linearly from 100 m/s to 200 m/s over the first 10 seconds.

At $$t = 10\text{ s}$$ the bullet undergoes a perfectly inelastic collision and instantaneously comes to rest, so the velocity drops from 200 m/s to 0 in a vertical line on the v-t graph. After the collision, from $$t = 10\text{ s}$$ to $$t = 20\text{ s}$$, the bullet remains at rest and hence $$v = 0$$ over this interval.

The v-t curve therefore consists of a straight line with positive slope from (0, 100) to (10, 200), followed by a sudden drop to 0 and then a horizontal line at v = 0 from t = 10 s to t = 20 s. This matches Option A (Option 1), which is the correct answer.

We have a juggler who throws $$n$$ balls per second, each with the same initial velocity $$u$$. He throws the next ball when the first ball reaches its highest position. The time interval between successive throws is $$\frac{1}{n}$$ seconds.

At the highest point the ball's velocity is zero, and the time to reach the maximum height is given by $$v = u - gt$$, so $$0 = u - g \cdot \frac{1}{n}$$, which gives $$u = \frac{g}{n}$$.

Now the maximum height is $$H = \frac{u^2}{2g} = \frac{1}{2g}\left(\frac{g}{n}\right)^2 = \frac{g^2}{2g n^2} = \frac{g}{2n^2}$$.

Hence, the correct answer is Option 4.

Given: mass $$m = 5$$ kg, air resistance force $$F = 10$$ N (constant, always opposing motion), $$g = 10$$ m/s$$^2$$.

Find the effective accelerations: During ascent (object moves upward):

Both gravity and air resistance act downward (opposing upward motion).

$$a_{\text{ascent}} = g + \frac{F}{m} = 10 + \frac{10}{5} = 10 + 2 = 12 \text{ m/s}^2 \text{ (retardation)}$$

During descent (object moves downward):

Gravity acts downward, air resistance acts upward (opposing downward motion).

$$a_{\text{descent}} = g - \frac{F}{m} = 10 - \frac{10}{5} = 10 - 2 = 8 \text{ m/s}^2 \text{ (acceleration)}$$

Find the time of ascent: Let the initial velocity be $$u$$. During ascent, the object decelerates from $$u$$ to $$0$$. Using $$v = u - a_{\text{ascent}} \cdot t_a$$:

$$0 = u - 12 \cdot t_a$$

$$t_a = \frac{u}{12}$$

Find the maximum height: Using $$v^2 = u^2 - 2a_{\text{ascent}} \cdot h$$:

$$0 = u^2 - 2(12)h$$

$$h = \frac{u^2}{24}$$

Find the time of descent: During descent, the object starts from rest at the top and falls through height $$h$$ with acceleration $$a_{\text{descent}} = 8$$ m/s$$^2$$. Using $$h = \frac{1}{2}a_{\text{descent}} \cdot t_d^2$$:

$$\frac{u^2}{24} = \frac{1}{2}(8) t_d^2 = 4t_d^2$$

$$t_d^2 = \frac{u^2}{96}$$

$$t_d = \frac{u}{\sqrt{96}} = \frac{u}{4\sqrt{6}}$$

Find the ratio: $$\frac{t_a}{t_d} = \frac{\frac{u}{12}}{\frac{u}{4\sqrt{6}}} = \frac{u}{12} \times \frac{4\sqrt{6}}{u} = \frac{4\sqrt{6}}{12} = \frac{\sqrt{6}}{3}$$

Simplify: $$\frac{\sqrt{6}}{3} = \frac{\sqrt{6}}{\sqrt{9}} = \sqrt{\frac{6}{9}} = \sqrt{\frac{2}{3}} = \frac{\sqrt{2}}{\sqrt{3}}$$

Therefore, the ratio of time of ascent to time of descent is $$\sqrt{2} : \sqrt{3}$$.

The correct answer is Option B.

We are given that a bullet's velocity becomes one-third after penetrating 4 cm in a wooden block, and it faces constant resistance (constant deceleration).

Let the initial velocity of the bullet be $$v_0$$. After penetrating 4 cm, its velocity is $$\frac{v_0}{3}$$. Let the constant deceleration be $$a$$.

Using the kinematic equation $$v^2 = u^2 - 2as$$ for the first 4 cm, $$\left(\frac{v_0}{3}\right)^2 = v_0^2 - 2a(4)$$, which gives $$\frac{v_0^2}{9} = v_0^2 - 8a$$. Solving for $$a$$, we find $$8a = v_0^2 - \frac{v_0^2}{9} = \frac{8v_0^2}{9}$$ and hence $$a = \frac{v_0^2}{9}$$.

When the bullet stops completely, $$v = 0$$, the total stopping distance satisfies $$0 = v_0^2 - 2a \cdot s_{total}$$. Therefore, $$s_{total} = \frac{v_0^2}{2a} = \frac{v_0^2}{2 \cdot \frac{v_0^2}{9}} = \frac{9}{2} = 4.5 \text{ cm}$$.

Since the total distance travelled is $$(4 + x)$$ cm, equating this to 4.5 cm yields $$4 + x = 4.5$$ and hence $$x = 0.5 \text{ cm}$$.

Answer: Option C: 0.5

We need to find the time at which the second ball meets the first ball.

Both balls are projected upwards with initial velocity $$u = 50$$ m s$$^{-1}$$ and acceleration due to gravity $$g = 10$$ m s$$^{-2}$$.

The first ball is launched at time $$t = 0$$ s, while the second ball is launched at $$t = 2$$ s.

The height of the first ball at time $$t$$ is given by $$h_1 = ut - \frac{1}{2}gt^2 = 50t - 5t^2$$.

For $$t \ge 2$$, the height of the second ball is $$h_2 = u(t-2) - \frac{1}{2}g(t-2)^2 = 50(t-2) - 5(t-2)^2$$.

The two balls meet when $$h_1 = h_2$$, leading to the equation $$50t - 5t^2 = 50(t-2) - 5(t-2)^2$$.

Expanding the right-hand side gives $$50t - 5t^2 = 50t - 100 - 5(t^2 - 4t + 4) = 50t - 100 - 5t^2 + 20t - 20$$, which simplifies to $$50t - 5t^2 = 70t - 5t^2 - 120$$.

Cancelling like terms and solving for $$t$$ yields $$50t = 70t - 120$$, so $$120 = 20t$$ and hence $$t = 6 \text{ s}$$.

At $$t = 6$$ s, the first ball is at $$h_1 = 50(6) - 5(36) = 300 - 180 = 120$$ m and the second ball is at $$h_2 = 50(4) - 5(16) = 200 - 80 = 120$$ m, confirming they meet at 120 m.

Therefore, the two balls meet at $$t = 6$$ s.

We use the kinematic equation relating velocity, acceleration, and distance:

$$v^2 = u^2 + 2as$$

When the car stops, $$v = 0$$, so:

$$0 = u^2 + 2as \implies s = \dfrac{u^2}{2|a|}$$

This shows that the stopping distance is proportional to the square of the initial speed:

$$s \propto u^2$$

Case 1: Speed $$u_1 = 150 \text{ km h}^{-1}$$, stopping distance $$s_1 = 27 \text{ m}$$.

Case 2: Speed $$u_2 = \dfrac{u_1}{3} = 50 \text{ km h}^{-1}$$, stopping distance $$s_2 = ?$$

Since the braking force (and hence deceleration) is the same in both cases:

$$\dfrac{s_2}{s_1} = \dfrac{u_2^2}{u_1^2} = \left(\dfrac{1}{3}\right)^2 = \dfrac{1}{9}$$

$$s_2 = \dfrac{27}{9} = 3 \text{ m}$$

Therefore, the car will stop after travelling $$\boxed{3}$$ m.

The velocity is increasing at $$5 \text{ m s}^{-1}$$ per meter. This means:

$$\dfrac{dv}{dx} = 5 \text{ s}^{-1}$$

Acceleration is given by:

$$a = \dfrac{dv}{dt} = \dfrac{dv}{dx} \cdot \dfrac{dx}{dt} = v \cdot \dfrac{dv}{dx}$$

At the point where $$v = 20 \text{ m s}^{-1}$$:

$$a = 20 \times 5 = 100 \text{ m s}^{-2}$$

Therefore, the answer is $$\boxed{100}$$.

Let the height of the tower be $$h$$ and the initial speed of the ball be $$u$$. Take downward as positive and the origin at the top of the tower.

Case 1 — Ball thrown vertically upward: Initial velocity = $$-u$$ (upward). Using $$h = -ut_1 + \frac{1}{2}gt_1^2$$:

$$h = -u(6) + \frac{1}{2}g(6)^2 = -6u + 18g \quad \cdots (1)$$

Case 2 — Ball thrown vertically downward: Initial velocity = $$+u$$ (downward). Using $$h = ut_2 + \frac{1}{2}gt_2^2$$:

$$h = u(1.5) + \frac{1}{2}g(1.5)^2 = 1.5u + 1.125g \quad \cdots (2)$$

Case 3 — Ball dropped from rest: Initial velocity = $$0$$. Using $$h = \frac{1}{2}gt_3^2$$:

$$t_3 = \sqrt{\frac{2h}{g}} \quad \cdots (3)$$

Key Relation: There is a well-known result that for this scenario, $$t_3 = \sqrt{t_1 \cdot t_2}$$. Let us derive it.

From equations (1) and (2):

$$-6u + 18g = 1.5u + 1.125g$$

$$18g - 1.125g = 6u + 1.5u$$

$$16.875g = 7.5u$$

$$u = \frac{16.875g}{7.5} = 2.25g$$

Substituting back into equation (2):

$$h = 1.5(2.25g) + 1.125g = 3.375g + 1.125g = 4.5g$$

Now using equation (3):

$$t_3 = \sqrt{\frac{2 \times 4.5g}{g}} = \sqrt{9} = 3 \text{ s}$$

We can verify: $$t_3 = \sqrt{t_1 \times t_2} = \sqrt{6 \times 1.5} = \sqrt{9} = 3$$ s.

The answer is 3 seconds.

We have a ball thrown vertically upwards with an initial velocity of $$u = 19.6 \text{ m/s}$$ from the top of a tower. The ball strikes the ground after $$t = 6 \text{ s}$$. We need to find the maximum height above the ground that the ball reaches.

Let the height of the tower be $$h$$. Taking the upward direction as positive and the top of the tower as the origin, the displacement of the ball when it hits the ground is $$-h$$ (downward). Using $$s = ut + \frac{1}{2}(-g)t^2$$, we get $$-h = 19.6(6) - \frac{1}{2}(9.8)(36) = 117.6 - 176.4 = -58.8$$.

So the height of the tower is $$h = 58.8 \text{ m}$$.

Now, the ball rises above the tower before coming back down. The maximum height above the tower is given by $$h_{\text{rise}} = \frac{u^2}{2g} = \frac{(19.6)^2}{2 \times 9.8} = \frac{384.16}{19.6} = 19.6 \text{ m}$$.

So the maximum height of the ball above the ground is $$H = h + h_{\text{rise}} = 58.8 + 19.6 = 78.4 \text{ m}$$.

We are told this equals $$\frac{k}{5}$$, so $$\frac{k}{5} = 78.4$$, which gives $$k = 78.4 \times 5 = 392$$.

Hence, the value of $$k$$ is $$\textbf{392}$$.

We have a ball released from rest at height $$h$$. Let $$t_1$$ be the time to cover the first half of the distance ($$h/2$$) and $$t_2$$ be the time to cover the second half.

The total time to fall a distance $$h$$ from rest is given by $$h = \frac{1}{2}g T^2$$, so $$T = \sqrt{\frac{2h}{g}}$$.

Now the time to fall the first $$h/2$$ is $$\frac{h}{2} = \frac{1}{2}g t_1^2$$, giving $$t_1 = \sqrt{\frac{h}{g}}$$.

We can write $$T = \sqrt{\frac{2h}{g}} = \sqrt{2} \cdot \sqrt{\frac{h}{g}} = \sqrt{2}\, t_1$$.

Since $$T = t_1 + t_2$$, we get $$t_2 = T - t_1 = \sqrt{2}\, t_1 - t_1 = (\sqrt{2} - 1)\, t_1$$.

Hence, the correct answer is Option 4.

We have $$t = \sqrt{x} + 4$$, and we need to find $$\left(\frac{dx}{dt}\right)$$ at $$t = 4$$.

First, we find the value of $$x$$ when $$t = 4$$. Substituting $$t = 4$$ into the equation: $$4 = \sqrt{x} + 4$$, which gives $$\sqrt{x} = 0$$, so $$x = 0$$.

Now we differentiate both sides of $$t = \sqrt{x} + 4$$ with respect to $$t$$: $$1 = \frac{1}{2\sqrt{x}} \cdot \frac{dx}{dt}$$, which gives $$\frac{dx}{dt} = 2\sqrt{x}$$.

At $$t = 4$$, we have $$x = 0$$, so $$\frac{dx}{dt} = 2\sqrt{0} = 0$$.

Hence, the correct answer is Option B.

Case I (List-I I)
The two particles move on the same horizontal circle of radius $$r = 1\ \text{m}$$ with the same angular speed $$\omega = 1\ \text{rad s}^{-1}$$.
At any instant their linear speeds are identical: $$v_A = v_B = r\omega = 1\ \text{m s}^{-1}$$.
Their angular positions differ by $$\frac{\pi}{2}$$ at all times, so the angle between the two velocity vectors is also $$\frac{\pi}{2}$$.
For two vectors of equal magnitude $$v$$ with an angle $$\phi$$ between them, the magnitude of the relative velocity is
$$v_{AB} = 2v\sin\frac{\phi}{2}$$.
Putting $$v = 1\ \text{m s}^{-1}$$ and $$\phi = \frac{\pi}{2}$$,
$$v_{AB} = 2(1)\sin\frac{\pi}{4} = 2\left(\frac{\sqrt{2}}{2}\right) = \sqrt{2}\ \text{m s}^{-1}$$.
Thus I → S.

Case II (List-I II)
Projectile A is fired at $$t = 0$$, and projectile B at $$t = 0.1\ \text{s}$$. Both have initial speed
$$v = \frac{5\pi}{\sqrt{2}}\ \text{m s}^{-1}$$ at an angle $$45^{\circ}$$ to the horizontal, but they move in opposite horizontal directions (shown in the figure).
Horizontal component of each velocity (magnitude) is
$$v_x = v\cos45^{\circ} = \frac{5\pi}{2}\ \text{m s}^{-1}$$,
so A moves with $$+v_x$$ and B with $$-v_x$$ along the $$x$$-axis.
At the required instant $$t = \dfrac{\pi}{3}\ \text{s}$$:

Time of flight for A: $$t_A = \dfrac{\pi}{3}$$.
Vertical velocity of A:
$$v_{yA} = v\sin45^{\circ} - gt_A = \frac{5\pi}{2} - 10\left(\frac{\pi}{3}\right) = -\frac{5\pi}{6}\ \text{m s}^{-1}$$.

Time of flight for B: $$t_B = \dfrac{\pi}{3} - 0.1$$.
Vertical velocity of B:
$$$ \begin{aligned} v_{yB} &= v\sin45^{\circ} - gt_B \\ &= \frac{5\pi}{2} - 10\left(\frac{\pi}{3} - 0.1\right) \\ &= -\frac{5\pi}{6} + 1\ \text{m s}^{-1}. \end{aligned} $$$

Hence the components of the relative velocity $$\mathbf{v}_{AB} = \mathbf{v}_A - \mathbf{v}_B$$ are
$$v_{x\,AB} = +\frac{5\pi}{2} - \left(-\frac{5\pi}{2}\right) = 5\pi,$$
$$v_{y\,AB} = -\frac{5\pi}{6} - \left(-\frac{5\pi}{6} + 1\right) = -1.$$

Magnitude:
$$\left|\mathbf{v}_{AB}\right| = \sqrt{(5\pi)^2 + 1^2} = \sqrt{25\pi^2 + 1}\ \text{m s}^{-1}.$$
Thus II → T.

Case III (List-I III)
The displacements are $$x_A = \sin t$$ and $$x_B = \sin\left(t + \frac{\pi}{2}\right)$$ (with $$x_0 = t_0 = 1$$).
Velocities:
$$v_A = \frac{dx_A}{dt} = \cos t,$$
$$v_B = \frac{dx_B}{dt} = \cos\left(t + \frac{\pi}{2}\right) = -\sin t.$$

At $$t = \dfrac{\pi}{3}$$:
$$v_A = \cos\frac{\pi}{3} = \frac{1}{2},\qquad v_B = -\sin\frac{\pi}{3} = -\frac{\sqrt{3}}{2}.$$

Relative velocity:
$$v_{AB} = v_A - v_B = \frac{1}{2} - \left(-\frac{\sqrt{3}}{2}\right) = \frac{1 + \sqrt{3}}{2}\ \text{m s}^{-1}.$$
Thus III → P.

Case IV (List-I IV)
Particle A moves in the horizontal $$xy$$-plane on a circle of radius 1 m with angular speed $$\omega = 1\ \text{rad s}^{-1}$$, so
$$v_A = r\omega = 1\ \text{m s}^{-1}$$ (purely horizontal, tangent to the circle).
Particle B moves vertically upward with constant speed $$v_B = 3\ \text{m s}^{-1}$$ along the $$z$$-axis.
Since the two velocity vectors are mutually perpendicular, the magnitude of the relative velocity is
$$\left|\mathbf{v}_{AB}\right| = \sqrt{v_A^2 + v_B^2} = \sqrt{1^2 + 3^2} = \sqrt{10}\ \text{m s}^{-1}.$$
Thus IV → R.

Collecting all four mappings:
I → S, II → T, III → P, IV → R.

The option that matches this set is
Option C: I → S, II → T, III → P, IV → R.

A ball is dropped from a height of $$4.9$$ m above the water level. It hits the water with velocity $$v$$ and then sinks with the same constant velocity $$v$$. The total time from drop to reaching the bottom is $$4.0$$ s.

Find the velocity when the ball hits the water.

Using $$v^2 = u^2 + 2gh$$ with $$u = 0$$, $$g = 9.8$$ m/s², $$h = 4.9$$ m:

$$v^2 = 2 \times 9.8 \times 4.9 = 96.04$$

$$v = 9.8$$ m/s

Find the time taken to fall through air.

Using $$v = u + gt$$:

$$9.8 = 0 + 9.8 \times t_1$$

$$t_1 = 1$$ s

Find the time spent sinking in water.

$$t_2 = 4.0 - 1.0 = 3.0$$ s

Find the depth of the lake.

The ball sinks with constant velocity $$v = 9.8$$ m/s:

$$d = v \times t_2 = 9.8 \times 3.0 = 29.4$$ m

The approximate depth of the lake is $$29.4$$ m.

The correct answer is Option D.

Let the car accelerate at rate $$\alpha$$ for time $$t_1$$ and then decelerate at rate $$\beta$$ for time $$t_2$$. Since the car starts and ends at rest, the maximum velocity reached is $$v = \alpha t_1 = \beta t_2$$. Also, $$t_1 + t_2 = t$$.

From $$v = \alpha t_1$$ and $$v = \beta t_2$$, we get $$t_1 = \frac{v}{\alpha}$$ and $$t_2 = \frac{v}{\beta}$$. Substituting into $$t_1 + t_2 = t$$ gives $$\frac{v}{\alpha} + \frac{v}{\beta} = t$$, so $$v\left(\frac{\alpha + \beta}{\alpha\beta}\right) = t$$, which means $$v = \frac{\alpha\beta t}{\alpha + \beta}$$.

The total distance is the sum of distances during acceleration and deceleration: $$S = \frac{1}{2}\alpha t_1^2 + \frac{1}{2}\beta t_2^2 = \frac{v^2}{2\alpha} + \frac{v^2}{2\beta} = \frac{v^2}{2}\left(\frac{1}{\alpha} + \frac{1}{\beta}\right) = \frac{v^2}{2} \cdot \frac{\alpha + \beta}{\alpha\beta}$$.

Substituting $$v = \frac{\alpha\beta t}{\alpha + \beta}$$ yields $$S = \frac{1}{2} \cdot \frac{\alpha^2\beta^2 t^2}{(\alpha+\beta)^2} \cdot \frac{\alpha+\beta}{\alpha\beta} = \frac{\alpha\beta t^2}{2(\alpha+\beta)}$$.

The correct answer is option 3: $$\frac{\alpha\beta}{2(\alpha+\beta)}t^2$$.

We need to determine the acceleration-displacement ($$a-x$$) graph from the given velocity-displacement ($$v-x$$) graph of the bicycle's motion.

The acceleration $$a$$ as a function of velocity $$v$$ and displacement $$x$$ is given by the relation: $$a = v \frac{dv}{dx}$$, where $$\frac{dv}{dx}$$ represents the slope of the $$v-x$$ graph.

Let's analyze the motion in two distinct regions based on the graph:

Region 1: From $$x = 0$$ to $$x = 200\text{ m}$$

The $$v-x$$ graph is a straight line starting from $$(0, 10)$$ and ending at $$(200, 50)$$. The equation of this straight line is of the form $$v = mx + c$$.

The slope of this line is: $$\frac{dv}{dx} = \frac{50 - 10}{200 - 0} = \frac{40}{200} = 0.2\text{ s}^{-1}$$.

The intercept on the vertical axis is $$c = 10$$, so the velocity equation is: $$v = 0.2x + 10$$.

Substituting $$v$$ and $$\frac{dv}{dx}$$ into the acceleration formula gives: $$a = (0.2x + 10) \times 0.2 = 0.04x + 2$$.

This is a straight-line equation for acceleration with a positive slope ($$0.04$$) and a vertical intercept of $$2\text{ m s}^{-2}$$.

• At $$x = 0\text{ m}$$, the acceleration is: $$a = 0.04(0) + 2 = 2\text{ m s}^{-2}$$.

• At $$x = 200\text{ m}$$, the acceleration is: $$a = 0.04(200) + 2 = 8 + 2 = 10\text{ m s}^{-2}$$.

Region 2: From $$x = 200\text{ m}$$ to $$x = 400\text{ m}$$

The velocity is constant at $$v = 50\text{ m s}^{-1}$$. Therefore, the slope of the graph is zero: $$\frac{dv}{dx} = 0$$.

Substituting this into the acceleration formula gives: $$a = 50 \times 0 = 0\text{ m s}^{-2}$$.

Combining both regions, the acceleration $$a$$ increases linearly from $$2\text{ m s}^{-2}$$ to $$10\text{ m s}^{-2}$$ in the first 200 meters, and then drops instantly and remains at $$0\text{ m s}^{-2}$$ from 200 meters to 400 meters.

Therefore, the correct description matches the graph where $$a$$ starts at 2, rises to 10 at $$x = 200\text{ m}$$, and becomes 0 thereafter.

We need to find the correct acceleration-displacement ($$a-x$$) graph corresponding to the given velocity-displacement ($$v-x$$) graph of a particle.

1. Analyze the Velocity-Displacement ($$v-x$$) Graph

From the problem statement, the velocity decreases linearly with displacement $$x$$. This straight line has a positive vertical intercept and a negative slope.

We can write the equation of this line using the slope-intercept form ($$y = mx + c$$):

$$v = -mx + v_0 \quad \text{--- (Equation 1)}$$

Where:

• $$-m$$ is the negative slope of the $$v-x$$ line ($$m > 0$$).
• $$v_0$$ is the initial velocity at $$x = 0$$ (the positive vertical intercept).

2. Derive the Acceleration Expression

Acceleration $$a$$ is defined as the rate of change of velocity with respect to time ($$\frac{dv}{dt}$$). Using the chain rule, we can express it in terms of displacement $$x$$:

$$a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt}$$

Since $$\frac{dx}{dt} = v$$, the relationship simplifies to:

$$a = v \cdot \frac{dv}{dx}$$

Now, let's find the components from Equation 1:

• The derivative of velocity with respect to displacement is the constant slope:

  $$\frac{dv}{dx} = -m$$

• Substitute $$v$$ and $$\frac{dv}{dx}$$ into the acceleration formula:

  $$a = (-mx + v_0) \cdot (-m)$$

  $$a = m^2x - mv_0 \quad \text{--- (Equation 2)}$$

3. Determine the Features of the $$a-x$$ Graph

Equation 2 ($$a = m^2x - mv_0$$) is also a linear equation representing a straight line on an $$a \text{ vs } x$$ coordinate plane:

• Slope: The coefficient of $$x$$ is $$m^2$$. Since any real squared number is positive, the line has a positive slope (it slants upwards from left to right).
• Vertical Intercept: At $$x = 0$$, the acceleration is $$a = -mv_0$$. Since both $$m$$ and $$v_0$$ are positive constants, the vertical intercept is negative (it starts below the horizontal origin axis).

Looking at the options, the graph that features a straight line starting from a negative vertical intercept and sloping upwards to cross the $$x$$-axis is Option C.

Conclusion

The correct acceleration-displacement graph is represented by Option C.

We are given the velocity of a particle as $$v = v_0 + gt + Ft^2$$ and the initial position $$x = 0$$ at $$t = 0$$. The displacement is obtained by integrating the velocity with respect to time.

$$x = \int_0^t v \, dt = \int_0^t (v_0 + gt + Ft^2) \, dt = v_0 t + \frac{g t^2}{2} + \frac{F t^3}{3}$$

Substituting $$t = 1$$ s, we get $$x = v_0(1) + \frac{g(1)^2}{2} + \frac{F(1)^3}{3} = v_0 + \frac{g}{2} + \frac{F}{3}$$.

Let the ball be projected vertically upward with initial speed $$u$$ from the point that we shall call the origin. It reaches its highest point after its velocity has become zero. At that highest point the vertical displacement from the origin is given to be $$h$$.

For a freely falling body, the kinematic relation between velocity $$v$$, initial velocity $$u$$, acceleration $$a$$ and displacement $$s$$ is stated first:

$$v^{2}=u^{2}+2as.$$

Here at the top $$v=0$$, the acceleration is that of gravity downward so $$a=-g$$, and the displacement is upward $$s=h$$. Substituting these values we get

$$0^{2}=u^{2}+2(-g)h \;\;\Longrightarrow\;\; u^{2}=2gh.$$

This result will be useful shortly. Now we need the two instants at which the ball passes the height $$\dfrac{h}{3}$$, once on the way up and again on the way down. For vertical motion with constant acceleration the displacement after time $$t$$ is given by

$$y = ut-\dfrac12 gt^{2}.$$

At the required instants the vertical coordinate $$y$$ equals $$\dfrac{h}{3}$$, so we write

$$ut-\dfrac12 gt^{2}=\dfrac{h}{3}.$$

Move every term to one side to obtain a quadratic equation in $$t$$:

$$-\dfrac12 gt^{2}+ut-\dfrac{h}{3}=0.$$

Multiplying through by $$-1$$ to keep the leading coefficient positive,

$$\dfrac12 gt^{2}-ut+\dfrac{h}{3}=0.$$

This is of the standard form $$at^{2}+bt+c=0$$ with

$$a=\dfrac{g}{2}, \qquad b=-u, \qquad c=\dfrac{h}{3}.$$

The quadratic-formula gives the two roots:

$$t=\dfrac{-b\pm\sqrt{\,b^{2}-4ac\,}}{2a}.$$

Substituting the coefficients,

$$t=\dfrac{u\pm\sqrt{u^{2}-\dfrac{2gh}{3}}}{g}.$$

In the discriminant we already know $$u^{2}=2gh$$, hence

$$u^{2}-\dfrac{2gh}{3}=2gh-\dfrac{2gh}{3}=\dfrac{4gh}{3}.$$

The square-root therefore becomes

$$\sqrt{u^{2}-\dfrac{2gh}{3}}=\sqrt{\dfrac{4gh}{3}}=\dfrac{2}{\sqrt3}\sqrt{gh}.$$

Now put $$u=\sqrt{2gh}$$ and factor $$\sqrt{gh}$$ out of the numerators:

$$t=\dfrac{\sqrt{2gh}\;\pm\;\dfrac{2}{\sqrt3}\sqrt{gh}}{g} =\dfrac{\sqrt{gh}}{g}\left(\sqrt2 \;\pm\;\dfrac{2}{\sqrt3}\right).$$

Thus the earlier time (ascending) is

$$t_{1}=\dfrac{\sqrt{gh}}{g}\left(\sqrt2-\dfrac{2}{\sqrt3}\right),$$

and the later time (descending) is

$$t_{2}=\dfrac{\sqrt{gh}}{g}\left(\sqrt2+\dfrac{2}{\sqrt3}\right).$$

The required ratio of the two times is therefore

$$\dfrac{t_{1}}{t_{2}} =\dfrac{\sqrt2-\dfrac{2}{\sqrt3}} {\sqrt2+\dfrac{2}{\sqrt3}}.$$

Multiply numerator and denominator by $$\sqrt3$$ to clear the fractional terms:

$$\dfrac{t_{1}}{t_{2}} =\dfrac{\sqrt3\sqrt2-2} {\sqrt3\sqrt2+2} =\dfrac{\sqrt6-2}{\sqrt6+2}.$$

One may notice that an equivalent, and more compact, form is obtained by multiplying the previous numerator and denominator by $$\dfrac{\sqrt3-\sqrt2}{\sqrt3-\sqrt2}$$, yielding

$$\dfrac{t_{1}}{t_{2}}=\dfrac{\sqrt3-\sqrt2}{\sqrt3+\sqrt2}.$$

This expression matches Option C in the list provided.

Hence, the correct answer is Option C.

The particle starts with velocity $$v_0$$ along the x-axis and experiences a damping force given by $$ma = -\alpha x^2$$. Writing acceleration as $$a = v\frac{dv}{dx}$$, we get $$mv\,dv = -\alpha x^2\,dx$$.

Integrating both sides, $$m\int_{v_0}^{0} v\,dv = -\alpha \int_{0}^{x_0} x^2\,dx$$, where $$x_0$$ is the distance at which the particle stops ($$v = 0$$).

Evaluating the integrals, $$m\left[\frac{v^2}{2}\right]_{v_0}^{0} = -\alpha\left[\frac{x^3}{3}\right]_{0}^{x_0}$$, which gives $$-\frac{mv_0^2}{2} = -\frac{\alpha x_0^3}{3}$$.

Simplifying, $$\frac{mv_0^2}{2} = \frac{\alpha x_0^3}{3}$$, so $$x_0^3 = \frac{3mv_0^2}{2\alpha}$$.

Therefore, the distance at which the particle stops is $$x_0 = \left(\frac{3mv_0^2}{2\alpha}\right)^{\frac{1}{3}}$$.

The correct answer is $$\left(\frac{3mv_0^2}{2\alpha}\right)^{\frac{1}{3}}$$.

Let the height of the building be $$H$$ metres. A stone is dropped from the top (point A). When it has fallen 5 m (reaching point B), another stone is dropped from a point 25 m below the top (point C). Both reach the bottom simultaneously.

For stone 1, the time to fall the first 5 m is found from $$5 = \frac{1}{2}g t_1^2$$, giving $$t_1 = \sqrt{\frac{10}{g}} = \sqrt{\frac{10}{10}} = 1$$ s (taking $$g = 10$$ m/s$$^2$$).

At point B, stone 1 has velocity $$v_1 = g t_1 = 10$$ m/s. From this point, stone 1 must fall the remaining distance $$(H - 5)$$ m. Using $$H - 5 = v_1 t + \frac{1}{2}g t^2$$, we get $$H - 5 = 10t + 5t^2$$.

Stone 2 is dropped (from rest) from point C, which is 25 m below the top. It must fall $$(H - 25)$$ m in the same time $$t$$. So $$H - 25 = \frac{1}{2}g t^2 = 5t^2$$.

Subtracting the second equation from the first: $$(H - 5) - (H - 25) = 10t + 5t^2 - 5t^2$$, which gives $$20 = 10t$$, so $$t = 2$$ s.

Substituting back: $$H - 25 = 5(2)^2 = 20$$, giving $$H = 45$$ m.

Let us assume that every droplet leaves the tap after a fixed time-interval. We denote this interval by $$T\;{\rm s}$$ (seconds). Thus one droplet is released, after exactly $$T$$ seconds the next droplet is released, after the next $$T$$ seconds another droplet is released, and so on. The question is really asking for the value of $$T$$, because

Rate of release $$=\dfrac{\text{number of drops}}{\text{time}}=\dfrac{1}{T}\;{\rm drops\;per\;second}.$$

We now focus on two particular droplets:

• Droplet A has already been falling for $$4\;{\rm s}$$ at the instant we make our observation.
• Droplet B is the “next” droplet that left the tap after droplet A.  Because the tap releases drops every $$T$$ seconds, droplet B was released exactly $$T$$ seconds after droplet A. Hence, at the same observation instant, droplet B has been in free-fall for only $$(4-T)\;{\rm s}.$$

Both droplets start from rest, so their motion is described by the standard free-fall formula

$$s=\dfrac12\,g\,t^{2},$$

where $$s$$ is the distance fallen in time $$t$$ and $$g=9.8\;{\rm m\,s^{-2}}$$. We now write the individual distances:

Distance fallen by droplet A: $$s_A=\dfrac12\,g\,(4)^{2}=\dfrac12\times 9.8 \times 16=78.4\;{\rm m}.$$

Distance fallen by droplet B: $$s_B=\dfrac12\,g\,(4-T)^{2}=4.9\,(4-T)^{2}\;{\rm m}.$$

The spacing between the two droplets is the difference of these two distances. According to the problem this spacing equals $$34.3\;{\rm m}$$, so we must have

$$s_A-s_B=34.3.$$

Substituting the explicit expressions of $$s_A$$ and $$s_B$$, we get

$$78.4-4.9\,(4-T)^{2}=34.3.$$

We now isolate the quadratic term step by step:

Subtract $$34.3$$ from $$78.4$$ on the left:

$$78.4-34.3 = 4.9\,(4-T)^{2}.$$

Compute the numerical difference:

$$44.1 = 4.9\,(4-T)^{2}.$$

Divide both sides by $$4.9$$ so that only the square remains on the right:

$$\dfrac{44.1}{4.9} = (4-T)^{2}.$$

Because $$44.1/4.9 = 9$$, we have

$$(4-T)^{2}=9.$$

To eliminate the square, take square roots on both sides:

$$4-T = \pm\,3.$$

This gives two possible linear equations:

1. $$4-T = 3 \quad\Longrightarrow\quad T = 1\;{\rm s},$$

2. $$4-T = -3 \quad\Longrightarrow\quad T = 7\;{\rm s}.$$

We must now decide which of these two values makes physical sense. Droplet B has been falling for $$(4-T)$$ seconds, and this time must be positive; otherwise the “next” droplet would not yet have left the tap. If we take $$T=7\;{\rm s}$$, then $$(4-T) = -3\;{\rm s},$$ which is impossible. Therefore we reject $$T=7\;{\rm s}$$ and accept

$$T = 1\;{\rm s}.$$

The rate at which droplets emerge is thus

$$\text{Rate} = \dfrac{1}{T} = \dfrac{1}{1}\;{\rm drop\;per\;second}.$$

Hence, the correct answer is Option 3.

We are told that the velocity of the body at a displacement $$x$$ from some origin is

$$v \;=\; \sqrt{5000 + 24x}\ \text{m s}^{-1}.$$

To find the acceleration, we recall the basic kinematics relation that connects acceleration $$a$$, velocity $$v$$ and displacement $$x$$. Using the chain rule of calculus, the acceleration can be written as

$$a \;=\; \frac{dv}{dt}\;=\;\frac{dv}{dx}\,\frac{dx}{dt}.$$

But $$\frac{dx}{dt}$$ is simply the velocity $$v$$ itself, so this becomes the well-known result

$$a \;=\; v\,\frac{dv}{dx}.$$

Now we already know $$v$$ as a function of $$x$$, so our next task is to compute the derivative $$\dfrac{dv}{dx}$$ step by step.

First, write the given velocity in exponent form for easier differentiation:

$$v \;=\; (5000 + 24x)^{1/2}.$$

Differentiate with respect to $$x$$ using the power rule $$\dfrac{d}{dx}\,[u^{n}] = n\,u^{\,n-1}\dfrac{du}{dx}$$ where $$u = 5000 + 24x$$ and $$n = \tfrac12\;:$$

$$\frac{dv}{dx} \;=\; \frac12\,(5000 + 24x)^{-1/2}\;\times\;\frac{d}{dx}(5000 + 24x).$$

The derivative of $$5000 + 24x$$ with respect to $$x$$ is simply $$24$$, so

$$\frac{dv}{dx} \;=\; \frac12\,(5000 + 24x)^{-1/2}\;\times\;24.$$

Multiplying the constants gives

$$\frac{dv}{dx} \;=\; 12\,(5000 + 24x)^{-1/2}.$$

Since a negative half-power corresponds to a reciprocal square root, we can rewrite this as

$$\frac{dv}{dx} \;=\; \frac{12}{\sqrt{5000 + 24x}}.$$

We now substitute $$v$$ and $$\dfrac{dv}{dx}$$ into the acceleration formula $$a = v \dfrac{dv}{dx}\,,$$ so

$$a \;=\; \bigl(\sqrt{5000 + 24x}\bigr)\;\times\;\frac{12}{\sqrt{5000 + 24x}}.$$

The square root term in the numerator and the identical square root term in the denominator cancel each other exactly, leaving

$$a \;=\; 12.$$

This result is a constant, independent of $$x$$, and its unit is $$\text{m s}^{-2}$$ because both the velocity and the derivative were in SI units throughout.

So, the answer is $$12$$.

Let us denote the instant when the first ball is projected upward as $$t = 0$$. Its initial velocity is given as $$u = 35\ \text{m s}^{-1}$$ and it moves against gravity whose magnitude is $$g = 10\ \text{m s}^{-2}$$ (acting downward).

The vertical displacement of any body projected upward with initial speed $$u$$ under constant downward acceleration $$g$$, after a time $$t$$, is given by the well-known kinematic relation

$$s = ut - \tfrac12 g t^{2}.$$

We apply this formula separately to the two balls.

For the first ball, the time elapsed since its launch is simply $$t$$ seconds. Hence its height above the point of projection is

$$\begin{aligned} y_{1} &= u t - \tfrac12 g t^{2} \\ &= 35\,t - 5\,t^{2}. \quad -(1) \end{aligned}$$

The second ball is thrown upward exactly $$3\ \text{s}$$ after the first one. Therefore, when the clock shows $$t$$ seconds after the first throw, the second ball has been in flight for only $$(t-3)$$ seconds. (Of course, this expression is meaningful only when $$t \ge 3\ \text{s}$$, i.e. after the second ball has actually been launched.) Using the same formula, its height is

$$\begin{aligned} y_{2} &= u\,(t-3) - \tfrac12 g\,(t-3)^{2} \\ &= 35\,(t-3) - 5\,(t-3)^{2}. \quad -(2) \end{aligned}$$

The collision occurs when both balls are at the same height, so we set $$y_{1} = y_{2}$$ using equations (1) and (2).

$$\begin{aligned} 35\,t - 5\,t^{2} &= 35\,(t-3) - 5\,(t-3)^{2}. \end{aligned}$$

First we expand the right-hand side:

$$\begin{aligned} 35\,(t-3) &= 35t - 105,\\ (t-3)^{2} &= t^{2} - 6t + 9,\\ -5\,(t-3)^{2} &= -5t^{2} + 30t - 45. \end{aligned}$$

Adding these components, the right-hand side becomes

$$\bigl(35t -105\bigr) + \bigl(-5t^{2} + 30t - 45\bigr) = 65t - 5t^{2} - 150.$$

Equating both sides we have

$$35t - 5t^{2} = 65t - 5t^{2} - 150.$$

The terms $$-5t^{2}$$ appear on both sides and cancel out immediately, leaving

$$35t = 65t - 150.$$

Transposing the terms gives

$$35t - 65t = -150 \quad\Longrightarrow\quad -30t = -150.$$

Dividing by $$-30$$ we find

$$t = 5\ \text{s}.$$

Thus, the collision happens $$5\ \text{s}$$ after the first ball was projected, which also means $$5 - 3 = 2\ \text{s}$$ after the second ball was projected.

To obtain the height at which they meet, we substitute $$t = 5\ \text{s}$$ into equation (1) (the expression for the first ball, though either expression would give the same result):

$$\begin{aligned} y &= 35\,(5) - 5\,(5)^{2} \\ &= 175 - 5 \times 25 \\ &= 175 - 125 \\ &= 50\ \text{m}. \end{aligned}$$

Therefore, the two spherical balls collide at a height of $$50\ \text{m}$$ above the point of projection.

So, the answer is $$50\ \text{m}.$$

The ball is released from a height $$h = 5$$ m and the coefficient of restitution squared is $$e^2 = \frac{81}{100}$$, so $$e = \frac{9}{10}$$. After each bounce the ball rises to $$e^2$$ times the previous height.

The total distance travelled is $$S = h + 2he^2 + 2he^4 + \cdots = h + \frac{2he^2}{1 - e^2} = h \cdot \frac{1 + e^2}{1 - e^2}$$. Substituting, $$S = 5 \times \frac{1 + 0.81}{1 - 0.81} = 5 \times \frac{1.81}{0.19} = 47.63$$ m.

The total time is $$T = \sqrt{\frac{2h}{g}} + 2\sqrt{\frac{2h}{g}}(e + e^2 + e^3 + \cdots) = \sqrt{\frac{2h}{g}}\left(1 + \frac{2e}{1 - e}\right) = \sqrt{\frac{2h}{g}} \cdot \frac{1 + e}{1 - e}$$. With $$\sqrt{\frac{2 \times 5}{10}} = 1$$ s, we get $$T = 1 \times \frac{1.9}{0.1} = 19$$ s.

The average speed is $$\frac{S}{T} = \frac{47.63}{19} \approx 2.50$$ m s$$^{-1}$$.

The scooter starts from rest, accelerates at constant rate $$a_1$$ for time $$t_1$$, then decelerates at constant rate $$a_2$$ for time $$t_2$$ and comes to rest.

During the acceleration phase, the velocity increases from 0 to a maximum value $$v_{max}$$. Using $$v = u + at$$, we get $$v_{max} = 0 + a_1 t_1 = a_1 t_1$$.

During the deceleration phase, the velocity decreases from $$v_{max}$$ to 0. Using $$v = u - at$$, we get $$0 = v_{max} - a_2 t_2$$, which gives $$v_{max} = a_2 t_2$$.

Since both expressions equal $$v_{max}$$, we have $$a_1 t_1 = a_2 t_2$$. Therefore, $$\frac{t_1}{t_2} = \frac{a_2}{a_1}$$.

Let the length of the train be $$L$$ and the uniform acceleration be $$a$$. The front of the train (engine) passes the signal post with velocity $$u$$ and the last compartment passes with velocity $$v$$. Using the kinematic relation $$v^2 = u^2 + 2aL$$, we get $$a = \frac{v^2 - u^2}{2L}$$.

The middle point of the train is at a distance $$\frac{L}{2}$$ from the engine. When the middle point passes the signal post, the engine has travelled a distance $$\frac{L}{2}$$ beyond the post. Let the velocity of the middle point at the signal post be $$v_m$$. Using the kinematic equation for the engine travelling from the signal post to a distance $$\frac{L}{2}$$:

$$v_m^2 = u^2 + 2a \cdot \frac{L}{2} = u^2 + aL$$

Substituting $$aL = \frac{v^2 - u^2}{2}$$:

$$v_m^2 = u^2 + \frac{v^2 - u^2}{2} = \frac{2u^2 + v^2 - u^2}{2} = \frac{u^2 + v^2}{2}$$

Therefore $$v_m = \sqrt{\frac{u^2 + v^2}{2}}$$, which corresponds to Option (2).

The relation between time and distance is $$t = m x^2 + n x$$, with $$m$$ and $$n$$ constants.

Differentiate both sides with respect to $$x$$ to connect time and distance derivatives:
$$\frac{dt}{dx} = 2 m x + n \quad -(1)$$

Velocity is defined as $$v = \frac{dx}{dt}$$. From equation $$(1)$$,
$$v = \frac{1}{dt/dx} = \frac{1}{2 m x + n} \quad -(2)$$

Acceleration is $$a = \frac{dv}{dt}$$. Using the chain rule $$\frac{dv}{dt} = \frac{dv}{dx}\frac{dx}{dt} = v \frac{dv}{dx}$$.

First find $$\frac{dv}{dx}$$ from $$(2)$$:
$$v = (2 m x + n)^{-1}$$
$$\frac{dv}{dx} = -\,2 m \,(2 m x + n)^{-2} \quad -(3)$$

Substitute $$(2)$$ and $$(3)$$ into $$a = v \frac{dv}{dx}$$:
$$a = \frac{1}{2 m x + n}\;\bigl[-\,2 m \,(2 m x + n)^{-2}\bigr] = -\,\frac{2 m}{(2 m x + n)^3} \quad -(4)$$

The negative sign shows the body is slowing down, so the magnitude of retardation is
$$|a| = \frac{2 m}{(2 m x + n)^3} \quad -(5)$$

From $$(2)$$, $$2 m x + n = \frac{1}{v}$$. Substitute this into $$(5)$$:

$$|a| = 2 m \left(\frac{1}{v}\right)^{\!3} = 2 m v^3 \quad -(6)$$

Hence the retardation is $$2 m v^3$$.

Option A is correct.

Let the height of the shower above the floor be $$h = 9.8\text{ m}$$ and let the acceleration due to gravity be $$g = 9.8\text{ m s}^{-2}$$ (standard value given in JEE problems).

Step 1: Time taken by the first drop to reach the floor
Using the equation of uniformly accelerated motion $$h = \tfrac12 g t_1^{\,2}$$ Solving for $$t_1$$ gives $$t_1^{\,2} = \frac{2h}{g} = \frac{2\times 9.8}{9.8} = 2 \quad\Rightarrow\quad t_1 = \sqrt{2}\text{ s} \approx 1.414\text{ s} \;-(1)$$

Step 2: Interval between successive drops
Let the drops leave the nozzle at a constant time gap $$\Delta t$$.
Drop 1 is released at $$t = 0$$, Drop 2 at $$t = \Delta t$$, and Drop 3 at $$t = 2\Delta t$$.
According to the statement, “when the first drop strikes the floor, the third drop begins to fall”, so $$t_1 = 2\Delta t \;-(2)$$

Substituting $$t_1 = \sqrt{2}\text{ s}$$ from $$(1)$$ into $$(2)$$: $$\Delta t = \frac{t_1}{2} = \frac{\sqrt{2}}{2}\text{ s} \approx 0.707\text{ s}$$

Step 3: Time for which the second drop has been in air
At the instant the first drop hits the floor ($$t = t_1$$), the second drop has already been falling for $$t_2 = t_1 - \Delta t = \sqrt{2} - \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{2}\text{ s}$$ Notice that $$t_2 = \Delta t$$ because the releases are equally spaced.

Step 4: Distance fallen by the second drop
Using the same motion equation for the second drop: $$s_2 = \tfrac12 g t_2^{\,2}$$ Substitute $$g = 9.8\text{ m s}^{-2}$$ and $$t_2 = \tfrac{\sqrt{2}}{2}\text{ s}$$: $$s_2 = \frac12 \times 9.8 \times \left(\frac{\sqrt{2}}{2}\right)^{2}$$ $$s_2 = 4.9 \times \frac{2}{4} = 4.9 \times 0.5 = 2.45\text{ m}$$

Step 5: Position of the second drop from the floor
Total height from nozzle to floor is $$h = 9.8\text{ m}$$.
Hence, distance of the second drop above the floor at that instant is $$h - s_2 = 9.8 - 2.45 = 7.35\text{ m}$$.

Therefore, the second drop is $$7.35\text{ m}$$ above the floor when the first drop strikes the floor.
Option D.

Let upward direction be positive and let the ground be the reference level $$y = 0$$.

The balloon is rising with a constant velocity $$u_b = 10\ \text{m s}^{-1}$$. At the instant it is $$75\ \text{m}$$ above the ground, an object is released from it. Because the object was part of the balloon just before release, its initial velocity relative to the ground is the same as that of the balloon:

$$u = 10\ \text{m s}^{-1}\ (\text{upward})$$

The object then moves under gravity alone. For vertical motion with constant acceleration we use the kinematic equation $$y = y_0 + u t + \tfrac12 a t^2$$ $$-(1)$$ where
  • $$y_0$$ = initial height, here $$75\ \text{m}$$
  • $$y$$ = height after time $$t$$
  • $$u$$ = initial velocity, here $$10\ \text{m s}^{-1}$$ (upward)
  • $$a$$ = acceleration. Since gravity acts downward, $$a = -g = -10\ \text{m s}^{-2}$$.

The object strikes the ground when $$y = 0$$. Substituting into (1):

$$0 = 75 + 10t - \tfrac12(10)t^2$$

$$0 = 75 + 10t - 5t^2$$

Re-arranging: $$5t^2 - 10t - 75 = 0$$

Dividing by 5: $$t^2 - 2t - 15 = 0$$

Solving this quadratic:

$$t = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(-15)}}{2} = \frac{2 \pm \sqrt{4 + 60}}{2} = \frac{2 \pm 8}{2}$$

The physically meaningful (positive) root is $$t = 5\ \text{s}$$.

Thus, the object reaches the ground after $$5\ \text{s}$$.

During this interval the balloon continues to rise with its constant speed of $$10\ \text{m s}^{-1}$$. Extra height gained by the balloon:

$$h_{\text{extra}} = u_b \, t = 10 \times 5 = 50\ \text{m}$$

Initial height of the balloon = $$75\ \text{m}$$. Therefore, height of the balloon when the object hits the ground:

$$H = 75 + 50 = 125\ \text{m}$$

Hence, the balloon is approximately $$125\ \text{m}$$ above the ground when the object strikes the ground.

The correct option is Option C.

Let the length of the escalator be $$L$$. The boy's walking speed is $$v_b = L/t_1$$ and the escalator's speed is $$v_e = L/t_2$$.

When the boy walks on the moving escalator, the effective speed is $$v_b + v_e = \frac{L}{t_1} + \frac{L}{t_2} = L\cdot\frac{t_1 + t_2}{t_1 t_2}$$.

The time taken is $$t = \frac{L}{v_b + v_e} = \frac{L \cdot t_1 t_2}{L(t_1 + t_2)} = \frac{t_1 t_2}{t_1 + t_2}$$.

Therefore the time taken by him to walk up on the moving escalator is $$\dfrac{t_1 t_2}{t_1 + t_2}$$.

We first recall the relation between the arc-length $$s$$, the radius $$r$$ of the circle and the angle $$\theta$$ subtended by the arc at the centre: the formula is $$s = r\,\theta.$$

Here the given arc-length is $$4.4\ \text{ly}$$, and the angle is stated as $$4$$ arc-seconds. We must put every quantity in compatible units.

One light-year is given as $$1\ \text{ly}=9.46 \times 10^{15}\ \text{m},$$ so

$$s = 4.4\ \text{ly}=4.4 \times 9.46 \times 10^{15}\ \text{m} = 41.624 \times 10^{15}\ \text{m} = 4.1624 \times 10^{16}\ \text{m}.$$

Next, we convert the very small angle of $$4$$ arc-seconds to radians. We know

$$1^\circ = \frac{\pi}{180}\ \text{rad},$$

$$1'\ (\text{one arc-minute}) = \frac{1^\circ}{60},$$

$$1''\ (\text{one arc-second}) = \frac{1'}{60}= \frac{1^\circ}{3600}.$$

Therefore

$$4'' = 4 \times \frac{1^\circ}{3600} = \frac{4}{3600}\ ^\circ = \frac{1}{900}\ ^\circ,$$

and in radians

$$\theta = \frac{1}{900}\ ^\circ \times \frac{\pi}{180} = \frac{\pi}{900 \times 180} = \frac{\pi}{162000}\ \text{rad}.$$

Evaluating

$$\theta = \frac{3.1415926}{162000}\ \text{rad} \approx 1.939 \times 10^{-5}\ \text{rad}.$$

Now we can find the radius $$r$$ from $$r = \dfrac{s}{\theta}$$:

$$r = \frac{4.1624 \times 10^{16}}{1.939 \times 10^{-5}}\ \text{m} = 4.1624 \times 10^{16} \times \frac{1}{1.939 \times 10^{-5}}\ \text{m}.$$

The reciprocal $$\dfrac{1}{1.939 \times 10^{-5}} \approx 5.158 \times 10^{4},$$ so

$$r \approx 4.1624 \times 10^{16} \times 5.158 \times 10^{4}\ \text{m} \approx 2.15 \times 10^{21}\ \text{m}.$$

With the radius known, the full circumference of the circle is given by $$C = 2\pi r$$:

$$C = 2\pi (2.15 \times 10^{21})\ \text{m} = 6.283 \times 2.15 \times 10^{21}\ \text{m} \approx 1.35 \times 10^{22}\ \text{m}.$$

The body is required to complete four revolutions, so the total path-length traversed is

$$D = 4C = 4 \times 1.35 \times 10^{22}\ \text{m} \approx 5.40 \times 10^{22}\ \text{m}.$$

The speed of the body is stated as $$8\ \text{AU s}^{-1}.$$ With $$1\ \text{AU} = 1.5 \times 10^{11}\ \text{m},$$ the speed in SI units is

$$v = 8 \times 1.5 \times 10^{11}\ \text{m s}^{-1} = 1.2 \times 10^{12}\ \text{m s}^{-1}.$$

The time required is obtained from the basic relation $$t = \dfrac{D}{v}.$$ Substituting the values just found,

$$t = \frac{5.40 \times 10^{22}}{1.2 \times 10^{12}}\ \text{s} = 4.50 \times 10^{10}\ \text{s}.$$

Expressed in scientific notation this is $$4.5 \times 10^{10}\ \text{s},$$ matching Option B.

Hence, the correct answer is Option B.

We are told that the instantaneous velocity of the particle is given by the time-dependent expression $$v = \alpha t + \beta t^{2}$$, where $$\alpha$$ and $$\beta$$ are constants.

To find the distance (strictly, the displacement along the straight line) travelled by the particle between the times $$t = 1\ \text{s}$$ and $$t = 2\ \text{s}$$, we recall the fundamental relation between velocity and displacement:

$$\displaystyle s = \int v\;dt.$$

This formula states that the displacement $$s$$ over a time interval is obtained by integrating the velocity with respect to time over that interval.

Substituting the given expression for $$v$$, we write the definite integral from $$t = 1$$ to $$t = 2$$:

$$\displaystyle s_{1\ \text{to}\ 2} = \int_{1}^{2} (\alpha t + \beta t^{2})\,dt.$$

Now we integrate term by term. First, we integrate $$\alpha t$$ with respect to $$t$$, and second, we integrate $$\beta t^{2}$$ with respect to $$t$$:

For the first term we use the power rule $$\int t\,dt = \frac{t^{2}}{2}$$:

$$\int \alpha t\,dt = \alpha \int t\,dt = \alpha\left(\frac{t^{2}}{2}\right) = \frac{\alpha t^{2}}{2}.$$

For the second term we again use the power rule $$\int t^{n}\,dt = \frac{t^{n+1}}{n+1}$$ (here $$n = 2$$):

$$\int \beta t^{2}\,dt = \beta \int t^{2}\,dt = \beta\left(\frac{t^{3}}{3}\right) = \frac{\beta t^{3}}{3}.$$

Combining both antiderivatives, the displacement function $$s(t)$$ (measured from any convenient reference) is

$$s(t) = \frac{\alpha t^{2}}{2} \;+\; \frac{\beta t^{3}}{3} + C,$$

where $$C$$ is an additive constant. However, since we are interested only in the difference $$s(2) - s(1)$$, this constant cancels out automatically and we do not need its explicit value.

We therefore evaluate $$s(t)$$ at the upper limit $$t = 2$$:

$$s(2) = \frac{\alpha (2)^{2}}{2} + \frac{\beta (2)^{3}}{3} = \frac{\alpha \cdot 4}{2} + \frac{\beta \cdot 8}{3} = 2\alpha + \frac{8\beta}{3}.$$

Next, we evaluate $$s(t)$$ at the lower limit $$t = 1$$:

$$s(1) = \frac{\alpha (1)^{2}}{2} + \frac{\beta (1)^{3}}{3} = \frac{\alpha}{2} + \frac{\beta}{3}.$$

The displacement (which, in this one-dimensional motion, equals the distance travelled) between 1 s and 2 s is the difference of these two values:

$$\begin{aligned} s_{1\ \text{to}\ 2} &= s(2) - s(1) \\ &= \left(2\alpha + \frac{8\beta}{3}\right) \;-\; \left(\frac{\alpha}{2} + \frac{\beta}{3}\right). \end{aligned}$$

We now subtract term by term, matching coefficients of $$\alpha$$ and $$\beta$$ separately.

For the $$\alpha$$ part:

$$2\alpha - \frac{\alpha}{2} = \frac{4\alpha}{2} - \frac{\alpha}{2} = \frac{3\alpha}{2}.$$

For the $$\beta$$ part:

$$\frac{8\beta}{3} - \frac{\beta}{3} = \frac{7\beta}{3}.$$

Putting both simplified results together, we obtain

$$s_{1\ \text{to}\ 2} = \frac{3\alpha}{2} + \frac{7\beta}{3}.$$

This matches option B in the list provided.

Hence, the correct answer is Option B.

Let us choose the direction in which train A is moving as the positive direction.

Speed of train A with respect to the ground is given as $$+36\text{ km h}^{-1}.$$

The person is walking inside train A opposite to the motion of the train. Therefore, the speed of the person with respect to train A is

$$v_{\text{person, A}} = -1.8\text{ km h}^{-1}.$$

To find the speed of the person with respect to the ground, we use the relation

$$v_{\text{person, ground}} = v_{\text{train A, ground}} + v_{\text{person, A}}.$$

Substituting the known values, we obtain

$$v_{\text{person, ground}} = 36 + (-1.8) = 34.2\text{ km h}^{-1}.$$

Now, train B is moving in the opposite direction (negative direction) with speed

$$v_{\text{train B, ground}} = -72\text{ km h}^{-1}.$$

The relative speed of the person as observed from train B is given by the standard formula

$$v_{\text{person relative to B}} = v_{\text{person, ground}} - v_{\text{train B, ground}}.$$

Substituting the values, we have

$$v_{\text{person relative to B}} = 34.2 - (-72) = 34.2 + 72 = 106.2\text{ km h}^{-1}.$$

We must now convert this speed from kilometres per hour to metres per second using

$$1\text{ km h}^{-1} = \frac{5}{18}\text{ m s}^{-1}.$$

Hence,

$$v_{\text{person relative to B}} = 106.2 \times \frac{5}{18}\text{ m s}^{-1}.$$

First multiply:

$$106.2 \times 5 = 531.0.$$

Now divide by 18:

$$\frac{531.0}{18} = 29.5\text{ m s}^{-1}.$$

Therefore, the speed of the person as observed from train B is approximately $$29.5\text{ m s}^{-1}.$$

Hence, the correct answer is Option A.

We start by letting the unknown acceleration due to gravity be $$g$$ (in m s$$^{-2}$$) and the total time taken by the ball to reach the ground be $$T$$ seconds. The ball is dropped from rest, so its initial velocity $$u=0$$.

For motion with constant acceleration we have the kinematic formula
$$s = uT + \dfrac{1}{2} g T^{2}.$$

Using this formula for the complete fall of $$100\ \text{m}$$, we write

$$100 = 0 \times T + \dfrac{1}{2}\,g\,T^{2},$$

which simplifies to

$$100 = \dfrac{1}{2} g T^{2}.$$

Multiplying both sides by 2 gives

$$g T^{2} = 200 \quad\Longrightarrow\quad T^{2} = \dfrac{200}{g}. \quad -(1)$$

Now, during the last $$\dfrac12$$ s before impact the ball covers only $$19\ \text{m}$$. Hence, in the preceding time interval $$T-0.5$$ seconds it must have covered the remaining distance of $$100-19 = 81\ \text{m}$$.

Applying the same kinematic formula for this first part of the motion, we get

$$81 = 0 \times (T-0.5) + \dfrac{1}{2}\,g\,(T-0.5)^{2},$$

or simply

$$81 = \dfrac{1}{2} g (T-0.5)^{2}. \quad -(2)$$

We expand the square:
$$(T-0.5)^{2} = T^{2} - 2 \times 0.5 \times T + (0.5)^{2} = T^{2} - T + 0.25.$$

Substituting this result into equation (2) gives

$$81 = \dfrac{1}{2} g \bigl(T^{2} - T + 0.25\bigr).$$

We already know from equation (1) that $$\dfrac{1}{2}gT^{2}=100$$, so we rewrite the right-hand side as

$$\dfrac12 g T^{2} - \dfrac12 g T + \dfrac12 g \times 0.25 = 100 - \dfrac12 g T + 0.125\,g.$$

Thus equation (2) becomes

$$81 = 100 - \dfrac12 g T + 0.125\,g.$$

Transposing terms, we have

$$-\dfrac12 g T + 0.125\,g = 81 - 100 = -19.$$

Multiplying by $$-1$$ gives a cleaner form:

$$\dfrac12 g T - 0.125\,g = 19.$$

Factoring out $$g$$ yields

$$g\left(\dfrac12 T - 0.125\right) = 19. \quad -(3)$$

From equation (3) we can express $$g$$ in terms of $$T$$:

$$g = \dfrac{19}{\dfrac12 T - 0.125} = \dfrac{19}{\dfrac{T-0.25}{2}} = \dfrac{38}{T - 0.25}. \quad -(4)$$

We also keep equation (1) in the equivalent form

$$g T^{2} = 200. \quad -(5)$$

Substituting the value of $$g$$ from (4) into (5), we get

$$\dfrac{38}{T - 0.25}\,T^{2} = 200.$$

Multiplying both sides by $$T-0.25$$ gives

$$38 T^{2} = 200\,(T - 0.25).$$

Dividing every term by 2 to simplify, we obtain

$$19 T^{2} = 100T - 25.$$

Rearranging all terms to one side provides a standard quadratic equation:

$$19 T^{2} - 100T + 25 = 0.$$

To solve for $$T$$ we calculate the discriminant:
$$\Delta = (-100)^{2} - 4 \times 19 \times 25 = 10000 - 1900 = 8100,$$ whose square root is $$\sqrt{8100} = 90.$$

Using the quadratic formula $$T = \dfrac{-b \pm \sqrt{\Delta}}{2a}$$ with $$a = 19,\ b = -100$$, we get

$$T = \dfrac{100 \pm 90}{2 \times 19} = \dfrac{100 \pm 90}{38}.$$

This yields two roots:

$$T_{1} = \dfrac{190}{38} = 5 \ \text{s}, \quad T_{2} = \dfrac{10}{38} \approx 0.263\ \text{s}.$$

The ball has to be in the air for more than $$0.5\ \text{s}$$ (because the question speaks of the last half-second), so the physically acceptable time is $$T = 5\ \text{s}.$$

Finally, substituting $$T = 5$$ into equation (4):

$$g = \dfrac{38}{5 - 0.25} = \dfrac{38}{4.75} = 8\ \text{m s}^{-2}.$$

So, the answer is $$8$$.

We begin by writing the position of each particle as a vector in the two-dimensional plane.

The first particle moves only along the $$x$$-axis, so its position vector at any instant $$t$$ is

$$\vec r_1(t)=x(t)\,\hat i+0\,\hat j=\bigl(10+8t-3t^{2}\bigr)\,\hat i.$$

The second particle moves only along the $$y$$-axis, hence its position vector is

$$\vec r_2(t)=0\,\hat i+y(t)\,\hat j=\bigl(5-8t^{3}\bigr)\,\hat j.$$

Velocity is obtained by differentiating the position with respect to time.

For the first particle we have

$$\vec v_1(t)=\frac{d\vec r_1}{dt}=\frac{d}{dt}\bigl(10+8t-3t^{2}\bigr)\hat i =(8-6t)\,\hat i.$$

For the second particle we have

$$\vec v_2(t)=\frac{d\vec r_2}{dt}=\frac{d}{dt}\bigl(5-8t^{3}\bigr)\hat j =(-24t^{2})\,\hat j.$$

We now substitute $$t=1\ \text{s}$$ to get the instantaneous velocities.

For the first particle:

$$\vec v_1(1)=\bigl(8-6\times1\bigr)\,\hat i=(2)\,\hat i\ \text{m s}^{-1}.$$

For the second particle:

$$\vec v_2(1)=\bigl(-24\times1^{2}\bigr)\,\hat j=(-24)\,\hat j\ \text{m s}^{-1}.$$

The velocity of the second particle as seen from the first particle is the relative velocity

$$\vec v_{\text{rel}}=\vec v_2-\vec v_1.$$

Writing the components explicitly, we have

$$\vec v_{\text{rel}}=(-2)\,\hat i+(-24)\,\hat j,$$ because along $$x$$, $$v_2$$ has no component and $$v_1$$ is $$+2\,$$, giving $$0-2=-2$$, while along $$y$$, $$v_1$$ has no component and $$v_2$$ is $$-24\,$$, giving $$-24-0=-24$$.

The speed is the magnitude of this vector. Using the two-dimensional magnitude formula $$|\vec v_{\text{rel}}|=\sqrt{(v_x)^2+(v_y)^2},$$ we obtain

$$|\vec v_{\text{rel}}|=\sqrt{(-2)^{2}+(-24)^{2}} =\sqrt{4+576} =\sqrt{580}\ \text{m s}^{-1}.$$

In the statement of the problem this speed is denoted as $$\sqrt{v}$$, so by comparison we identify

$$v=580.$$

Hence, the correct answer is Option C (580).

We are told that the displacement of the particle is connected to time through the equation

$$x^{2}=a\,t^{2}+2\,b\,t+c\,.$$

To reach the acceleration we must differentiate twice with respect to time. First we take the derivative of the given relation. Using the rule “if $$F=x^{2}$$ then $$\dfrac{dF}{dt}=2x\dfrac{dx}{dt}$$” we have

$$2\,x\,\dfrac{dx}{dt}=2\,a\,t+2\,b\,.$$

Cancelling the factor 2 gives

$$x\,\dfrac{dx}{dt}=a\,t+b\,.$$

The quantity $$\dfrac{dx}{dt}$$ is the velocity $$v$$, so

$$v=\dfrac{a\,t+b}{x}\quad\text{and}\quad a\,t+b=x\,v.$$

Now we differentiate the velocity to obtain the acceleration $$\dfrac{dv}{dt}$$. Treating $$v=\dfrac{a\,t+b}{x}$$ as a quotient, the derivative is

$$\dfrac{dv}{dt}=\dfrac{a}{x}+(a\,t+b)\,\dfrac{d}{dt}\!\left(\dfrac{1}{x}\right).$$

Since $$\dfrac{d}{dt}\!\left(\dfrac{1}{x}\right)=-\dfrac{1}{x^{2}}\dfrac{dx}{dt}=-\dfrac{v}{x^{2}},$$ we substitute and obtain

$$\dfrac{dv}{dt}=\dfrac{a}{x}-(a\,t+b)\,\dfrac{v}{x^{2}}.$$

Replacing $$a\,t+b$$ by $$x\,v$$ (from the earlier relation) gives

$$\dfrac{dv}{dt}=\dfrac{a}{x}-\dfrac{x\,v^{2}}{x^{2}}=\dfrac{a}{x}-\dfrac{v^{2}}{x}.$$

Thus the acceleration is

$$\boxed{a_{\text{particle}}=\dfrac{a-v^{2}}{x}}.$$

At this stage $$v^{2}$$ still involves the time variable, so we must express it solely in terms of $$x$$. Square the earlier velocity expression:

$$v^{2}=\dfrac{(a\,t+b)^{2}}{x^{2}}.$$

To remove $$t$$, start again from the original displacement relation. Let $$y=a\,t+b$$; then $$t=\dfrac{y-b}{a}$$, and substituting into $$x^{2}=a\,t^{2}+2\,b\,t+c$$ gives

$$x^{2}=a\left(\dfrac{y-b}{a}\right)^{2}+2\,b\left(\dfrac{y-b}{a}\right)+c.$$

Simplifying the numerator inside the brackets, we find

$$(y-b)^{2}+2\,b(y-b)=y^{2}-b^{2},$$

so

$$x^{2}=\dfrac{y^{2}-b^{2}+a\,c}{a}\quad\Longrightarrow\quad y^{2}=a\,x^{2}+b^{2}-a\,c.$$

But $$y=a\,t+b,$$ hence

$$(a\,t+b)^{2}=a\,x^{2}+b^{2}-a\,c.$$

Substituting this into the expression for $$v^{2}$$ gives

$$v^{2}=\dfrac{a\,x^{2}+b^{2}-a\,c}{x^{2}}=a+\dfrac{b^{2}-a\,c}{x^{2}}.$$

Now place this result back into the formula for acceleration:

$$a_{\text{particle}}=\dfrac{a-\left(a+\dfrac{b^{2}-a\,c}{x^{2}}\right)}{x}=\dfrac{-\dfrac{b^{2}-a\,c}{x^{2}}}{x}=-\dfrac{\,b^{2}-a\,c\,}{x^{3}}.$$

We see that the numerical factor $$-(b^{2}-a\,c)$$ is a constant independent of $$x$$, while the power of $$x$$ in the denominator is three. Therefore, apart from the constant coefficient, the acceleration varies with position as

$$a_{\text{particle}}\propto x^{-3}.$$

The required exponent is an integer

$$n=3.$$

So, the answer is $$3$$.

We begin with the upward motion of the helicopter itself. Starting from rest and moving with a constant upward acceleration $$g$$, the displacement $$s$$ reached in time $$t_1$$ is given by the kinematic equation

$$s = \tfrac12 a t^2.$$

Here $$a = g$$ (upward) and the helicopter has to climb a height $$h$$ before the food packet is released. Substituting these values we get

$$h \;=\; \tfrac12\,g\,t_1^{\,2}.$$

Solving for the time $$t_1$$ of ascent,

$$t_1 \;=\; \sqrt{\dfrac{2h}{g}}.$$

The upward velocity acquired by the helicopter at this instant is found from the relation $$v = u + at$$. Because the helicopter started from rest, $$u = 0$$, so

$$v \;=\; 0 + g\,t_1 \;=\; g\,\sqrt{\dfrac{2h}{g}} \;=\; \sqrt{2gh}.$$

This is also the initial velocity $$u$$ of the food packet at the moment it is dropped, and it is directed upward. The packet is now in free fall under gravity alone, so after release its acceleration is downward with magnitude $$g$$. Taking the upward direction as positive, we write the displacement equation for the packet:

$$y \;=\; h \;+\; u\,t \;-\; \tfrac12\,g\,t^2,$$

where $$y$$ is the height above the ground after an additional time $$t$$. The packet reaches the ground when $$y = 0$$, so we set

$$0 \;=\; h \;+\; \sqrt{2gh}\,t \;-\; \tfrac12\,g\,t^2.$$

Multiplying every term by 2 to clear the fraction,

$$0 \;=\; 2h \;+\; 2\sqrt{2gh}\,t \;-\; g\,t^2.$$

Re-arranging in the standard quadratic form $$a t^2 + b t + c = 0$$ gives

$$g\,t^{\,2} \;-\; 2\sqrt{2gh}\,t \;-\; 2h \;=\; 0.$$

Using the quadratic formula $$t = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ with $$a = g,\; b = -2\sqrt{2gh},\; c = -2h,$$ we find

$$t \;=\; \dfrac{2\sqrt{2gh} \;+\; \sqrt{\left(2\sqrt{2gh}\right)^{\!2} + 4g(2h)}}{2g}.$$

Inside the square root,

$$(2\sqrt{2gh})^{\!2} = 8gh,\qquad 4g(2h) = 8gh,$$

so their sum is $$16gh$$ and its square root is $$4\sqrt{gh}$$. Substituting this back,

$$t \;=\; \dfrac{2\sqrt{2gh} \;+\; 4\sqrt{gh}}{2g}.$$

Factorizing $$\sqrt{gh}$$ from the numerator gives

$$t \;=\; \dfrac{\left(2\sqrt{2} \;+\; 4\right)\sqrt{gh}}{2g}.$$

Dividing numerator and denominator by 2,

$$t \;=\; \left(\sqrt{2} + 2\right)\sqrt{\dfrac{h}{g}}.$$

Numerically, $$\sqrt{2} \approx 1.414$$, so

$$t \;\approx\; (1.414 + 2)\,\sqrt{\dfrac{h}{g}} \;=\; 3.414\,\sqrt{\dfrac{h}{g}}.$$

This is very close to $$3.4\sqrt{\dfrac{h}{g}}$$, which corresponds to Option C.

Hence, the correct answer is Option C.

We begin with the forces acting while the ball is moving upward. Its weight acts downward with magnitude $$mg$$ and the given resistive force, proportional to the square of speed, also acts downward with magnitude $$mkv^{2}$$. Taking the upward direction as positive, the net force is downward, so Newton’s second law gives

$$m\,\dfrac{dv}{dt}= -\,mg \;-\; m\,k\,v^{2}.$$

We cancel the common factor $$m$$ to simplify:

$$\dfrac{dv}{dt}= -\,g \;-\; k\,v^{2}.$$

To connect velocity with height, we use the chain‐rule identity for acceleration,

$$a = \dfrac{dv}{dt}=v\,\dfrac{dv}{dy},$$

because $$\dfrac{dy}{dt}=v$$. Substituting this into the equation of motion gives

$$v\,\dfrac{dv}{dy}= -\,g \;-\; k\,v^{2}.$$

We now separate the variables $$v$$ and $$y$$ completely:

$$\dfrac{v\,dv}{g+k\,v^{2}} = -\,dy.$$

The ball starts from the ground with speed $$u$$, so at $$y=0$$ we have $$v=u$$. At the highest point the speed becomes zero, so at $$y=H$$ we have $$v=0$$. We therefore integrate between these limits.

Left-hand integral: let $$s=v^{2} \implies ds = 2v\,dv \implies v\,dv=\dfrac{ds}{2}$$. Hence

$$\int_{u}^{0}\dfrac{v\,dv}{g+k\,v^{2}} \;=\; \dfrac12\int_{u^{2}}^{0}\dfrac{ds}{g+k\,s}.$$

The standard integral $$\displaystyle\int \dfrac{ds}{g+k\,s}= \dfrac{1}{k}\ln (g+k\,s)$$ gives

$$\dfrac12\!\left[\dfrac{1}{k}\,\ln(g+k\,s)\right]_{u^{2}}^{0} \;=\; \dfrac{1}{2k}\Bigl[\ln(g)-\ln\!\bigl(g+k\,u^{2}\bigr)\Bigr].$$

This simplifies to

$$\dfrac{1}{2k}\,\ln\!\left(\dfrac{g}{g+k\,u^{2}}\right) \;=\; -\,\dfrac{1}{2k}\,\ln\!\left(1+\dfrac{k\,u^{2}}{g}\right).$$

Right-hand integral:

$$\int_{0}^{H}(-\,dy)= -\,H.$$

Equating the two evaluated integrals,

$$-\,\dfrac{1}{2k}\,\ln\!\left(1+\dfrac{k\,u^{2}}{g}\right)= -\,H.$$

Both sides carry the same negative sign, so we multiply by $$-1$$ and obtain the maximum height $$H$$:

$$H=\dfrac{1}{2k}\,\ln\!\left(1+\dfrac{k\,u^{2}}{g}\right).$$

This matches option D.

Hence, the correct answer is Option D.

We are told that the particle moves only along the positive x-axis and its instantaneous speed is related to its position by the relation $$v = b\sqrt{x}$$, where $$v$$ is the speed at position $$x$$ and $$b$$ is a constant with suitable units.

By definition of velocity in one dimension we have the kinematic relation

$$v \;=\; \frac{dx}{dt}$$

Substituting the given expression for $$v$$ into this definition, we obtain

$$\frac{dx}{dt} \;=\; b\sqrt{x}\, .$$

Now we separate the variables so that all terms containing $$x$$ stay on the left side and all terms containing $$t$$ remain on the right side:

$$\frac{dx}{\sqrt{x}} \;=\; b\,dt.$$

We integrate both sides. The limits for $$x$$ go from the initial position $$x = 0$$ at time $$t = 0$$ up to the general position $$x$$ at a general time $$t$$.

$$\int_{0}^{x} \frac{dx'}{\sqrt{x'}} \;=\; \int_{0}^{t} b\,dt'.$$

The integral on the left is straightforward because $$\displaystyle\int x'^{-1/2}\,dx' = 2\sqrt{x'}$$. Carrying out both integrations we get

$$2\sqrt{x}\;-\;2\sqrt{0} \;=\; b\,t\;-\;b\,(0).$$

Simplifying, $$\sqrt{x} = \dfrac{b\,t}{2}.$$ (The square-root term at the initial position is zero because the particle starts from the origin where $$x = 0$$.)

We now substitute this expression for $$\sqrt{x}$$ back into the given velocity formula $$v = b\sqrt{x}$$:

$$v \;=\; b\left(\dfrac{b\,t}{2}\right).$$

Multiplying the constants,

$$v = \dfrac{b^{2}t}{2}.$$

Finally, we are asked for the speed specifically at the instant $$t = \tau$$. Replacing $$t$$ by $$\tau$$ in the above expression yields

$$v(\tau) = \dfrac{b^{2}\tau}{2}.$$

Among the given options, this result corresponds to Option C.

Hence, the correct answer is Option C.

Let us take the upward direction as positive. While the ball is rising, two forces act downward: its weight $$mg$$ and the quadratic drag $$m\gamma v^{2}$$ (opposite to the velocity because the velocity is upward). Therefore the net force is downward and the equation of motion is

$$m\frac{dv}{dt}= -\,mg \;-\; m\gamma v^{2}.$$

Dividing by $$m$$ we obtain the differential equation for the velocity:

$$\frac{dv}{dt}= -\,g \;-\; \gamma v^{2}.$$

We now separate the variables, keeping all terms involving $$v$$ on one side and $$t$$ on the other:

$$\frac{dv}{g+\gamma v^{2}} = -\,dt.$$

The ball starts with velocity $$v=V_{0}$$ at time $$t=0$$ and reaches the zenith when $$v=0$$ at time $$t=T$$. We therefore integrate between these limits:

$$\displaystyle \int_{V_{0}}^{0}\frac{dv}{g+\gamma v^{2}} = - \int_{0}^{T} dt.$$

The right-hand integral gives $$-T$$, so flipping the limits on the left removes the minus sign:

$$T = \int_{0}^{V_{0}}\frac{dv}{g+\gamma v^{2}}.$$

Before evaluating the integral, we recall the standard formula

$$\int\frac{dx}{a + b x^{2}} = \frac{1}{\sqrt{ab}}\;\tan^{-1}\!\left(x\sqrt{\frac{b}{a}}\right) + C.$$

In our case $$a=g$$ and $$b=\gamma$$. Substituting these into the formula, we obtain

$$\int_{0}^{V_{0}}\frac{dv}{g+\gamma v^{2}} = \frac{1}{\sqrt{g\gamma}}\;\tan^{-1}\!\left(v\sqrt{\frac{\gamma}{g}}\right)\Bigg|_{0}^{V_{0}}.$$

Evaluating at the limits gives

$$T = \frac{1}{\sqrt{g\gamma}}\left[\tan^{-1}\!\left(V_{0}\sqrt{\frac{\gamma}{g}}\right) - \tan^{-1}(0)\right].$$

Since $$\tan^{-1}(0)=0$$, this simplifies neatly to

$$T = \frac{1}{\sqrt{\gamma g}}\;\tan^{-1}\!\left(\sqrt{\frac{\gamma}{g}}\,V_{0}\right).$$

Hence, the correct answer is Option B.

Problem Statement

A ball is thrown vertically up (taken as $$+z\text{-axis}$$) from the ground. The correct momentum-height ($$p\text{-}h$$) diagram is:

• (1) Diagram 1
• (2) Diagram 2
• (3) Diagram 3
• (4) Diagram 4
• Initial velocity = $$u$$ (positive, upwards)
• Acceleration due to gravity = $$-g$$ (downwards)
• Displacement $$s$$ = height $$h$$
• The ball starts at the ground ($$h = 0$$) with maximum upward velocity ($$+u$$). Thus, momentum is at its maximum positive value ($$+p_{\text{max}}$$).
• As height $$h$$ increases, the ball slows down until it reaches maximum height ($$h_{\text{max}}$$), where velocity and momentum become zero ($$p = 0$$).
• Trajectory: The path moves from the positive $$p\text{-axis}$$ downward and rightward toward the $$h\text{-axis}$$ (indicated by an arrow pointing down and right).
• From maximum height ($$h_{\text{max}}$$), the ball begins falling back down. Its height decreases back toward zero.
• Since the motion is downward (opposite to $$+z$$), the velocity and momentum become increasingly negative (moving into the $$-p$$ region).
• Trajectory: The path moves from the $$h\text{-axis}$$ downward and leftward toward the negative $$p\text{-axis}$$ (indicated by an arrow pointing down and left).
• Option (3) correctly shows the arrows continuously pointing along the chronological flow of time (first moving towards maximum height, then returning downwards with negative momentum).

Step-by-Step Derivation

To find the correct relation between momentum ($$p$$) and height ($$h$$), we use kinematic equations.

1. Mathematical Relationship

From the third equation of motion:

$$v^2 = u^2 + 2as$$

Given parameters for a ball thrown vertically upward:

Substituting these gives:

$$v^2 = u^2 - 2gh$$

Since momentum is defined as $$p = mv \implies v = \frac{p}{m}$$, we substitute this into the equation:

$$\left(\frac{p}{m}\right)^2 = u^2 - 2gh$$

$$p^2 = m^2(u^2 - 2gh)$$

$$p^2 = m^2u^2 - 2m^2gh$$

This equation is of the form $$p^2 = \text{constant} - kh$$, which mathematically represents a parabola symmetric about the $$h\text{-axis}$$ and opening to the left. This eliminates options (1) and (2) which are straight lines.

2. Direction Analysis (Sign of Momentum)

Let's break down the motion into two halves to determine the correct direction arrows:

First Half: Upward Journey

Second Half: Downward Journey

Conclusion

Combining both halves results in a smooth parabolic curve starting on the positive vertical axis, looping down through $$h_{\text{max}}$$, and finishing on the negative vertical axis.

Looking closely at the directional arrows:

Therefore, the correct option is (3).

We have a passenger train whose length is $$L_1 = 60\text{ m}$$ and a freight train whose length is $$L_2 = 120\text{ m}$$. For the passenger train to clear the freight train completely, the front end of the passenger train must travel a distance equal to the sum of their lengths, so the distance to be covered is

$$D = L_1 + L_2 = 60\text{ m} + 120\text{ m} = 180\text{ m}.$$

The speeds given are

Passenger train speed $$V_1 = 80\text{ km h}^{-1},$$

Freight train speed $$V_2 = 30\text{ km h}^{-1}.$$

First, convert each speed from kilometres per hour to metres per second using the relation $$1\text{ km h}^{-1} = \tfrac{5}{18}\text{ m s}^{-1}.$$ We therefore obtain

$$V_1 = 80 \times \frac{5}{18} = \frac{400}{18} = \frac{200}{9}\text{ m s}^{-1},$$

$$V_2 = 30 \times \frac{5}{18} = \frac{150}{18} = \frac{25}{3}\text{ m s}^{-1}.$$

Now we analyse the two required cases.

Case (i): Trains moving in the same direction. The relative speed is the difference of the individual speeds, because one train is moving away from the other in the same direction. Hence, using the formula $$\text{Relative speed (same direction)} = |V_1 - V_2|,$$ we have

$$V_{\text{rel, same}} = \left|\frac{200}{9} - \frac{25}{3}\right| = \left|\frac{200}{9} - \frac{75}{9}\right| = \frac{125}{9}\text{ m s}^{-1}.$$

The time taken is distance divided by this relative speed, i.e. $$t_1 = \frac{D}{V_{\text{rel, same}}} = \frac{180}{\frac{125}{9}} = 180 \times \frac{9}{125} = \frac{1620}{125} = \frac{324}{25}\text{ s}.$$

Case (ii): Trains moving in opposite directions. Now the relative speed is the sum of the individual speeds, because the trains approach each other head-on. Using the formula $$\text{Relative speed (opposite direction)} = V_1 + V_2,$$ we have

$$V_{\text{rel, opp}} = \frac{200}{9} + \frac{25}{3} = \frac{200}{9} + \frac{75}{9} = \frac{275}{9}\text{ m s}^{-1}.$$

The corresponding time is $$t_2 = \frac{D}{V_{\text{rel, opp}}} = \frac{180}{\frac{275}{9}} = 180 \times \frac{9}{275} = \frac{1620}{275} = \frac{324}{55}\text{ s}.$$

We now need the ratio of the two times. Since both times involve the same distance, the ratio simplifies neatly:

$$\frac{t_1}{t_2} = \frac{\frac{324}{25}}{\frac{324}{55}} = \frac{324}{25} \times \frac{55}{324} = \frac{55}{25} = \frac{11}{5}.$$

Thus, the required ratio of the times is $$\frac{11}{5}.$$

Hence, the correct answer is Option C.

We have the displacement of the particle as a function of time, $$x(t)=at+bt^2-ct^3,$$ where the symbols $$a,\;b$$ and $$c$$ are constants.

The velocity is defined as the first derivative of displacement with respect to time, that is $$v(t)=\dfrac{dx}{dt}.$$ Differentiating term-by-term,

$$\begin{aligned} v(t) &= \dfrac{d}{dt}\!\bigl(at\bigr) + \dfrac{d}{dt}\!\bigl(bt^2\bigr) - \dfrac{d}{dt}\!\bigl(ct^3\bigr) \\ &= a + 2bt - 3ct^2. \end{aligned}$$

Acceleration is the derivative of velocity with respect to time, $$a(t)=\dfrac{dv}{dt}.$$ Differentiating the expression for $$v(t)$$ gives

$$\begin{aligned} a(t) &= \dfrac{d}{dt}\!\bigl(a\bigr) + \dfrac{d}{dt}\!\bigl(2bt\bigr) - \dfrac{d}{dt}\!\bigl(3ct^2\bigr) \\ &= 0 + 2b - 6ct \\ &= 2b - 6ct. \end{aligned}$$

The problem asks for the velocity when the acceleration becomes zero. So we set $$a(t)=0$$:

$$2b - 6ct = 0.$$

Solving for $$t$$ yields

$$\begin{aligned} 2b &= 6ct \\ t &= \frac{2b}{6c} \\ t &= \frac{b}{3c}. \end{aligned}$$

Now we substitute this value of $$t$$ back into the velocity expression $$v(t)=a+2bt-3ct^2$$ to find the required velocity:

$$\begin{aligned} v &= a + 2b\left(\frac{b}{3c}\right) - 3c\left(\frac{b}{3c}\right)^2 \\ &= a + \frac{2b^2}{3c} - 3c\left(\frac{b^2}{9c^2}\right) \\ &= a + \frac{2b^2}{3c} - \frac{b^2}{3c} \\ &= a + \frac{2b^2 - b^2}{3c} \\ &= a + \frac{b^2}{3c}. \end{aligned}$$

Hence, the velocity of the particle at the instant when its acceleration is zero is $$a + \dfrac{b^2}{3c}.$$

Hence, the correct answer is Option A.

Let the total distance of the race be $$S$$. Both cars start from rest, so their initial velocities are zero.

For motion with uniform (constant) acceleration, the basic kinematic relations we shall use are

1. Distance covered from rest in time $$T$$: $$S=\dfrac12\,a\,T^{2}.$$

2. Final speed attained from rest in time $$T$$: $$v_{\text{final}}=a\,T.$$

Denote

$$T_B=$$ time taken by car $$B,$$

$$T_A=$$ time taken by car $$A.$$

The statement “car A takes a time $$t$$ less than car B” gives us

$$T_A=T_B-t. \quad -(1)$$

Since both cars run over the same total distance $$S$$, we write the distance formula for each:

For car $$A$$ (acceleration $$a_1$$):

$$S=\dfrac12\,a_1\,T_A^{2}.$$

For car $$B$$ (acceleration $$a_2$$):

$$S=\dfrac12\,a_2\,T_B^{2}.$$

Because the right-hand sides are both equal to the same $$S$$, we equate them:

$$\dfrac12\,a_1\,T_A^{2}=\dfrac12\,a_2\,T_B^{2}.$$

The factor $$\dfrac12$$ appears on both sides, so we cancel it, obtaining

$$a_1\,T_A^{2}=a_2\,T_B^{2}. \quad -(2)$$

Now substitute the relation (1), $$T_A=T_B-t,$$ into equation (2):

$$a_1\,(T_B-t)^{2}=a_2\,T_B^{2}.$$

We expand the square carefully:

$$a_1\left(T_B^{2}-2\,T_B\,t+t^{2}\right)=a_2\,T_B^{2}.$$

Multiplying term by term gives

$$a_1\,T_B^{2}-2\,a_1\,T_B\,t+a_1\,t^{2}=a_2\,T_B^{2}.$$

Next we bring every term to the left side to form a quadratic equation in $$T_B$$:

$$(a_1-a_2)\,T_B^{2}-2\,a_1\,T_B\,t+a_1\,t^{2}=0. \quad -(3)$$

Equation (3) is a quadratic of the standard form $$A\,T_B^{2}+B\,T_B+C=0$$ with

$$A=a_1-a_2,\quad B=-2\,a_1\,t,\quad C=a_1\,t^{2}.$$

We need $$T_B$$, so we apply the quadratic-formula statement

$$T_B=\dfrac{-B\pm\sqrt{B^{2}-4AC}}{2A}.$$

Substituting $$A,B,C$$ from above:

$$T_B=\dfrac{-(-2\,a_1\,t)\pm\sqrt{(-2\,a_1\,t)^{2}-4\,(a_1-a_2)\,(a_1\,t^{2})}}{2\,(a_1-a_2)}.$$

Simplifying step by step:

• The first numerator term: $$-(-2\,a_1\,t)=2\,a_1\,t.$$

• Inside the square root:

$$(-2\,a_1\,t)^{2}=4\,a_1^{2}\,t^{2},$$

$$4AC=4\,(a_1-a_2)\,(a_1\,t^{2})=4\,a_1\,(a_1-a_2)\,t^{2}.$$

Therefore

$$B^{2}-4AC=4\,a_1^{2}\,t^{2}-4\,a_1\,(a_1-a_2)\,t^{2} =4\,t^{2}\left(a_1^{2}-a_1(a_1-a_2)\right).$$

Inside the brackets we simplify the expression:

$$a_1^{2}-a_1(a_1-a_2)=a_1^{2}-a_1^{2}+a_1\,a_2=a_1\,a_2.$$

So the discriminant becomes

$$B^{2}-4AC=4\,t^{2}\,a_1\,a_2.$$

The square root therefore is

$$\sqrt{B^{2}-4AC}=2\,t\,\sqrt{a_1\,a_2}.$$

Putting everything back into the expression for $$T_B$$:

$$T_B=\dfrac{2\,a_1\,t\;\pm\;2\,t\,\sqrt{a_1\,a_2}}{2\,(a_1-a_2)} =\dfrac{a_1\,t\;\pm\;t\,\sqrt{a_1\,a_2}}{a_1-a_2}.$$

Only the “plus” or “minus” that keeps $$T_B$$ positive is physically acceptable; we shall keep both for the moment and the correct sign will reveal itself automatically when we calculate the speed difference.

Now we focus on the required quantity, namely the speed difference at the finish. Using the second kinematic relation $$v_{\text{final}}=a\,T$$, the individual finish speeds are

For car $$A$$: $$v_A=a_1\,T_A=a_1\,(T_B-t).$$

For car $$B$$: $$v_B=a_2\,T_B.$$

The problem defines $$v$$ as “the speed of car A at the finish is $$v$$ more than that of car B,” which translates to

$$v=v_A-v_B.$$

Substituting the expressions of $$v_A$$ and $$v_B$$, we have

$$v=a_1\,(T_B-t)-a_2\,T_B =a_1\,T_B-a_1\,t-a_2\,T_B =(a_1-a_2)\,T_B-a_1\,t. \quad -(4)$$

Equation (4) already contains the factor $$(a_1-a_2)\,T_B,$$ which suggests inserting the value of $$T_B$$ that we just derived. So we substitute $$T_B=\dfrac{a_1\,t\;\pm\;t\,\sqrt{a_1\,a_2}}{a_1-a_2}$$ into (4):

$$v=(a_1-a_2)\left(\dfrac{a_1\,t\;\pm\;t\,\sqrt{a_1\,a_2}}{a_1-a_2}\right)-a_1\,t.$$

The factor $$(a_1-a_2)$$ cancels immediately, leaving

$$v=a_1\,t\;\pm\;t\,\sqrt{a_1\,a_2}-a_1\,t=t\,\sqrt{a_1\,a_2},$$

because the $$a_1\,t$$ terms subtract out. The sign that survives is the positive one; speed difference must be positive. Thus we finally obtain

$$v=t\,\sqrt{a_1\,a_2}.$$

Looking at the given options, this matches exactly with Option C: $$\sqrt{a_1\,a_2}\;t.$$

Hence, the correct answer is Option C.

Given that the particle starts from rest ($$u = 0$$) with uniform acceleration ($$a = \text{constant}$$):

At time $$t$$:

$$v = u + at \implies v = at$$

Since $$a$$ is a positive constant, the velocity-time ($$v-t$$) graph is a straight line passing through the origin with a positive slope.

For displacement:

$$x = ut + \frac{1}{2}at^2 \implies x = \frac{1}{2}at^2$$

The displacement-time ($$x-t$$) graph is a parabola opening upwards.

For acceleration:

$$a = \text{constant}$$

The acceleration-time ($$a-t$$) graph is a horizontal line.

Based on these relations:

• Graph (A): Correct (represents $$x \propto t^2$$)
• Graph (B): Correct (represents $$v \propto t$$)
• Graph (D): Correct (represents constant $$a$$)

Hence, graphs (A), (B), and (D) are correct.

We begin by recalling the equation of motion for uniformly retarded (decelerated) motion:

$$v^{2}=u^{2}+2as$$

Here, $$u$$ is the initial speed, $$v$$ is the final speed, $$a$$ is the uniform acceleration (negative for braking), and $$s$$ is the displacement during the braking period. For a vehicle that comes to rest, the final speed is zero, that is, $$v=0$$. Substituting $$v=0$$, we obtain

$$0=u^{2}+2as \implies -u^{2}=2as \implies a=-\dfrac{u^{2}}{2s}.$$

The quantity $$|a|=\dfrac{u^{2}}{2s}$$ is the magnitude of the retardation provided by the brakes. Because the same brakes are used in both situations, the magnitude of this retardation remains unchanged.

First we analyse the given data for the initial situation.

Initial speed: $$40\text{ km h}^{-1}.$$

We convert kilometres per hour to metres per second using the relation $$1\text{ km h}^{-1}=\dfrac{5}{18}\text{ m s}^{-1}$$:

$$u_{1}=40\times\dfrac{5}{18}=\dfrac{200}{18}=\dfrac{100}{9}\text{ m s}^{-1}.$$

Stopping distance in this case: $$s_{1}=40\text{ m}.$$

Substituting these values in $$|a|=\dfrac{u^{2}}{2s}$$ gives the magnitude of the retardation:

$$|a|=\dfrac{u_{1}^{2}}{2s_{1}}=\dfrac{\left(\dfrac{100}{9}\right)^{2}}{2\times40} =\dfrac{\dfrac{10000}{81}}{80} =\dfrac{10000}{81\times80} =\dfrac{10000}{6480} =\dfrac{125}{81}\text{ m s}^{-2}.$$

We now consider the second situation, where the automobile’s speed is doubled.

New speed: $$80\text{ km h}^{-1}.$$

Converting again to metres per second:

$$u_{2}=80\times\dfrac{5}{18}=\dfrac{400}{18}=\dfrac{200}{9}\text{ m s}^{-1}.$$

The same brakes give the same magnitude of retardation, $$|a|=\dfrac{125}{81}\text{ m s}^{-2}.$$ Using the formula $$|a|=\dfrac{u^{2}}{2s},$$ but now for the second set of values, we write

$$|a|=\dfrac{u_{2}^{2}}{2s_{2}}.$$

Solving for $$s_{2}$$ gives

$$s_{2}=\dfrac{u_{2}^{2}}{2|a|}.$$

Because $$|a|$$ is the same for both cases, we can avoid recalculation by forming a ratio:

$$\dfrac{s_{2}}{s_{1}}=\dfrac{u_{2}^{2}}{u_{1}^{2}}.$$

Substituting the known speeds,

$$\dfrac{s_{2}}{40}=\dfrac{\left(\dfrac{200}{9}\right)^{2}}{\left(\dfrac{100}{9}\right)^{2}} =\dfrac{200^{2}}{100^{2}} =\dfrac{40000}{10000}=4.$$

Hence

$$s_{2}=4\times40=160\text{ m}.$$

So the automobile will require a minimum stopping distance of $$160\text{ m}$$ when travelling at $$80\text{ km h}^{-1}.$$

Hence, the correct answer is Option B.

Velocity is positive during the upwards journey and negative during the downwards journey. The magnitude of velocity decreases during the upwards journey, is zero at the highest point, and then increases during the downwards journey.

We are given that, at the initial instant, the bus and the car are both at rest and the car is 200 m behind the bus. Hence, the initial separation between them is $$200\;\text{m}$$.

Both vehicles begin to move simultaneously along the same straight road, but with different forward accelerations. The acceleration of the bus is

$$a_b = 2\;\text{m s}^{-2},$$

while the acceleration of the car is

$$a_c = 4\;\text{m s}^{-2}.$$

Because they start from rest, their initial velocities are both

$$u_b = 0\;\text{m s}^{-1}, \qquad u_c = 0\;\text{m s}^{-1}.$$

For motion with uniform acceleration we use the kinematic relation

$$s = ut + \frac{1}{2} a t^2,$$

where $$s$$ is the displacement in time $$t$$, $$u$$ is the initial velocity, and $$a$$ is the constant acceleration.

Let the time at which the car catches up with the bus be $$t$$ seconds. At that moment the car must cover the initial 200 m gap in addition to whatever distance the bus itself has travelled. An elegant way to handle this is to look at the relative motion between the two vehicles.

The relative (car with respect to bus) initial velocity is

$$u_{\text{rel}} = u_c - u_b = 0 - 0 = 0\;\text{m s}^{-1}.$$

The relative acceleration is

$$a_{\text{rel}} = a_c - a_b = 4 - 2 = 2\;\text{m s}^{-2}.$$

In the relative frame, the car has to cover the entire initial separation of 200 m. Therefore, the relative displacement to be achieved is

$$s_{\text{rel}} = 200\;\text{m}.$$

Applying the same kinematic equation to the relative motion, we write

$$s_{\text{rel}} = u_{\text{rel}}\,t + \frac{1}{2}\,a_{\text{rel}}\,t^2.$$

Substituting the known values, we have

$$200 = 0 \cdot t + \frac{1}{2}\,(2)\,t^2.$$

The term with the initial velocity drops out, leaving

$$200 = 1 \cdot t^2.$$

So

$$t^2 = 200.$$

Taking the positive square root (time must be positive), we get

$$t = \sqrt{200} = \sqrt{100 \times 2} = 10\sqrt{2}\;\text{s}.$$

Hence, the correct answer is Option D.

We want a motion with:

• positive velocity → graph stays above x-axis
• constant negative acceleration → velocity decreases uniformly

Using

$$v^2=u^2+2as$$

with a<0 velocity decreases with distance in a nonlinear (curved) way.

So the correct graph must:

• start at a positive velocity
• decrease continuously
• be curved (not a straight line)
• eventually approach zero

Graph (3) matches this:

• velocity is always positive
• it decreases with distance
• the curve bends downward (consistent with negative acceleration)

Why others are wrong:

• straight line → constant slope → would mean constant acceleration in v vs t, not v vs x
• increasing curve → implies positive acceleration

Final answer: Graph (3)

We are given the force $$ F = \frac{R}{t^2} v(t) $$ acting on a particle of mass $$ m $$. According to Newton's second law, force is also equal to mass times acceleration, so $$ F = m \frac{dv}{dt} $$. Setting these equal, we get:

$$ m \frac{dv}{dt} = \frac{R}{t^2} v(t) $$

To solve this differential equation, we separate the variables. Divide both sides by $$ v $$ (assuming $$ v \neq 0 $$) and multiply by $$ dt $$:

$$ \frac{m}{v} dv = \frac{R}{t^2} dt $$

Now, integrate both sides. The left side with respect to $$ v $$ and the right side with respect to $$ t $$:

$$ \int \frac{m}{v} dv = \int \frac{R}{t^2} dt $$

Integrating the left side: $$ m \int \frac{1}{v} dv = m \ln|v| $$. Integrating the right side: $$ R \int t^{-2} dt = R \left( \frac{t^{-1}}{-1} \right) = -\frac{R}{t} $$. So we have:

$$ m \ln|v| = -\frac{R}{t} + C $$

where $$ C $$ is the constant of integration. Since the particle starts from rest, as $$ t \to 0^+ $$, $$ v \to 0 $$. As $$ t \to 0^+ $$, $$ -\frac{R}{t} \to -\infty $$, so $$ \ln|v| \to -\infty $$, which implies $$ v \to 0 $$, satisfying the initial condition.

For $$ t > 0 $$ and $$ v > 0 $$, we can drop the absolute value:

$$ m \ln v = -\frac{R}{t} + C $$

Solve for $$ \ln v $$ by dividing both sides by $$ m $$:

$$ \ln v = -\frac{R}{m t} + \frac{C}{m} $$

Let $$ K = \frac{C}{m} $$, so:

$$ \ln v = -\frac{R}{m t} + K $$

Exponentiate both sides to solve for $$ v $$:

$$ v = e^K e^{-\frac{R}{m t}} $$

Let $$ A = e^K $$, so:

$$ v = A e^{-\frac{R}{m t}} $$

To test this law experimentally, we need a linear relationship that can be verified with a straight-line plot. Take the natural logarithm of both sides:

$$ \ln v = \ln A - \frac{R}{m t} $$

This equation is of the form $$ y = mx + b $$, where $$ y = \ln v $$ and $$ x = \frac{1}{t} $$. The slope is $$ -\frac{R}{m} $$ and the intercept is $$ \ln A $$. Thus, plotting $$ \ln v $$ against $$ \frac{1}{t} $$ should yield a straight line.

In experimental physics, "log" often means base 10 logarithm. Using $$ \log_{10} v $$:

$$ \log_{10} v = \frac{\ln v}{\ln 10} = \frac{1}{\ln 10} \left( \ln A - \frac{R}{m t} \right) = -\frac{R}{m \ln 10} \cdot \frac{1}{t} + \frac{\ln A}{\ln 10} $$

This is still linear in $$ \frac{1}{t} $$, with slope $$ -\frac{R}{m \ln 10} $$ and intercept $$ \frac{\ln A}{\ln 10} $$. Therefore, plotting $$ \log v(t) $$ against $$ \frac{1}{t} $$ will give a straight line.

Now, evaluate the options:

• Option A: $$ \log v(t) $$ against $$ \frac{1}{t} $$ - This matches our derived linear relationship.
• Option B: $$ v(t) $$ against $$ t^2 $$ - From $$ v = A e^{-\frac{R}{m t}} $$, this is not linear in $$ t^2 $$.
• Option C: $$ \log v(t) $$ against $$ \frac{1}{t^2} $$ - Our equation has $$ \log v $$ linear in $$ \frac{1}{t} $$, not $$ \frac{1}{t^2} $$. Plotting against $$ \frac{1}{t^2} $$ would not yield a straight line.
• Option D: $$ \log v(t) $$ against $$ t $$ - Our equation shows $$ \log v $$ depends on $$ \frac{1}{t} $$, not directly on $$ t $$. This would not be linear.

Hence, the best way to test the law experimentally is to plot $$ \log v(t) $$ against $$ \frac{1}{t} $$, which is Option A.

Hence, the correct answer is Option A.

A block of mass $$ m = 10 $$ kg rests on a horizontal table with coefficient of friction $$ \mu = 0.05 $$. A bullet of mass $$ 50 $$ g, which is $$ 0.05 $$ kg, moving with speed $$ v $$, embeds itself in the block. After the collision, the combined system moves and stops after traveling $$ 2 $$ m. We need to find the height $$ H $$ from which a freely falling object would acquire speed $$ \frac{v}{10} $$, neglecting energy losses and using $$ g = 10 $$ m/s².

First, consider the inelastic collision between the bullet and the block. Conservation of momentum applies:

Initial momentum = Momentum of bullet + Momentum of block = $$ (0.05) \times v + (10) \times 0 = 0.05v $$ kg·m/s.

After collision, the combined mass is $$ 10 + 0.05 = 10.05 $$ kg. Let the common velocity be $$ V $$.

Final momentum = $$ 10.05 \times V $$ kg·m/s.

Equating initial and final momentum:

$$ 0.05v = 10.05V $$

Solving for $$ V $$:

$$ V = \frac{0.05v}{10.05} = \frac{5v}{1005} = \frac{v}{201} $$

So, $$ V = \frac{v}{201} $$ m/s.

After the collision, the block with embedded bullet slides on the table with initial velocity $$ V $$ and stops after $$ s = 2 $$ m due to friction. The friction force causes deceleration.

The normal force $$ N $$ equals the weight of the combined system:

$$ N = (10.05) \times 10 = 100.5 \text{ N} $$

Friction force $$ f = \mu N = 0.05 \times 100.5 = 5.025 $$ N.

Deceleration $$ a = \frac{f}{\text{total mass}} = \frac{5.025}{10.05} = 0.5 $$ m/s². Alternatively, $$ a = \mu g = 0.05 \times 10 = 0.5 $$ m/s².

Using the equation of motion: final velocity² = initial velocity² + 2 × acceleration × distance

$$ 0^2 = V^2 + 2 \times (-0.5) \times 2 $$

$$ 0 = V^2 - 2 $$

$$ V^2 = 2 $$

$$ V = \sqrt{2} \text{ m/s} \quad (\text{taking positive value}) $$

But $$ V = \frac{v}{201} $$, so:

$$ \frac{v}{201} = \sqrt{2} $$

$$ v = 201 \sqrt{2} \text{ m/s} $$

Now, for the freely falling object: it starts from height $$ H $$ with initial velocity 0 and acquires speed $$ \frac{v}{10} $$ at the bottom. Using conservation of energy:

Potential energy loss = Kinetic energy gain

$$ m g H = \frac{1}{2} m \left( \frac{v}{10} \right)^2 $$

Cancel $$ m $$ from both sides:

$$ g H = \frac{1}{2} \times \frac{v^2}{100} $$

$$ 10 H = \frac{v^2}{200} $$

$$ H = \frac{v^2}{2000} $$

Substitute $$ v = 201 \sqrt{2} $$:

$$ H = \frac{(201 \sqrt{2})^2}{2000} = \frac{201^2 \times 2}{2000} $$

Calculate $$ 201^2 $$:

$$ 201^2 = (200 + 1)^2 = 200^2 + 2 \times 200 \times 1 + 1^2 = 40000 + 400 + 1 = 40401 $$

So,

$$ H = \frac{40401 \times 2}{2000} = \frac{80802}{2000} = 40.401 \text{ m} $$

Convert meters to kilometers (1 km = 1000 m):

$$ H = \frac{40.401}{1000} = 0.040401 \text{ km} $$

The options are:

A. 0.2 km

B. 0.5 km

C. 0.4 km

D. None of these

Comparing $$ H \approx 0.0404 $$ km with the options:

0.0404 km is not close to 0.2 km, 0.5 km, or 0.4 km (which are 200 m, 500 m, and 400 m respectively, while 0.0404 km is 40.4 m).

Hence, the correct answer is Option D.

1. When both stones are in air:

Position of stone 1: $$y_1 = u_1 t - \frac{1}{2}gt^2 = 10t - 5t^2$$

Position of stone 2: $$y_2 = u_2 t - \frac{1}{2}gt^2 = 40t - 5t^2$$

Relative position, $$(y_2 - y_1) = (40t - 5t^2) - (10t - 5t^2) = 30t$$

This is a straight line passing through the origin with a positive slope of $$30$$. This phase ends when the first stone hits the ground.

Time of impact:

For Stone 1: $$-240 = 10t - 5t^2 \implies t^2 - 2t - 48 = 0 \implies (t-8)(t+6)=0$$. Thus, $$t_1 = 8\text{ seconds}$$.

For Stone 2: $$-240 = 40t - 5t^2 \implies t^2 - 8t - 48 = 0 \implies (t-12)(t+4)=0$$.

Thus, $$t_2 = 12\text{ seconds}$$.

2. Stone 1 is on the ground, Stone 2 is in the air ($$8 < t \le 12\text{ s}$$)

Stone 1 is stationary: $$y_1 = -240\text{ m}$$.

Stone 2 continues moving: $$y_2 = 40t - 5t^2$$.

Relative position, $$(y_2 - y_1) = (40t - 5t^2) - (-240) = -5t^2 + 40t + 240$$

This is a downward-opening parabola.

At $$t = 8\text{ s}$$, $$y_2 - y_1 = 240\text{ m}$$.

At $$t = 12\text{ s}$$, $$y_2 - y_1 = 0\text{ m}$$.

Graph D correctly depicts the linear increase followed by a downward parabolic decrease.

We are given that the stone is projected vertically upward from the top of a tower of height $$64\ \text{m}$$ with an initial velocity $$u = 48\ \text{m s}^{-1}$$. The acceleration due to gravity acts downward with magnitude $$g = 32\ \text{m s}^{-2}$$. We wish to find the greatest height that the stone reaches above the ground.

First we determine how high the stone rises above the point of projection (the tower-top). At the highest point, the vertical velocity becomes zero. Using the first equation of motion, which in general form is written as

$$v = u - g t,$$

we set the final velocity $$v = 0$$ (because at the greatest height the stone momentarily comes to rest). Substituting $$u = 48\ \text{m s}^{-1}$$ and $$g = 32\ \text{m s}^{-2}$$, we get

$$0 = 48 - 32 t.$$

Solving for the time $$t$$ it takes to reach the maximum point:

$$32 t = 48 \quad\Longrightarrow\quad t = \frac{48}{32} = \frac{3}{2} = 1.5\ \text{s}.$$

Now we calculate the vertical displacement $$s$$ in this time interval. For vertical motion with constant acceleration, the displacement is obtained from the second equation of motion, stated as

$$s = u t - \frac{1}{2} g t^2.$$

Substituting the known values $$u = 48\ \text{m s}^{-1},\ t = 1.5\ \text{s},\ g = 32\ \text{m s}^{-2},$$ we have

$$\begin{aligned} s &= 48 \times 1.5 \;-\; \frac12 \times 32 \times (1.5)^2 \\ &= 72 \;-\; 16 \times 2.25 \\ &= 72 \;-\; 36 \\ &= 36\ \text{m}. \end{aligned}$$

Thus the stone rises $$36\ \text{m}$$ above the top of the tower.

The tower itself is $$64\ \text{m}$$ high, so the greatest height above ground level reached by the stone is

$$\text{Maximum height} = 64\ \text{m} + 36\ \text{m} = 100\ \text{m}.$$

Hence, the correct answer is Option D.

A bullet loses $$\left(\frac{1}{n}\right)^{\text{th}}$$ of its velocity when passing through one plank. This means if the initial velocity is $$u$$, after passing through one plank, the velocity becomes $$u - \frac{u}{n} = u \left(1 - \frac{1}{n}\right) = u \left(\frac{n-1}{n}\right)$$. The retardation is uniform, so we can use the equation of motion $$v^2 = u^2 + 2as$$, where $$v$$ is the final velocity, $$u$$ is the initial velocity, $$a$$ is the acceleration (which is negative for retardation), and $$s$$ is the distance traveled (thickness of one plank).

For one plank, initial velocity $$u$$, final velocity $$v = u \frac{n-1}{n}$$, acceleration $$a = -a$$ (where $$a > 0$$ is the magnitude of retardation), and distance $$s$$. Substituting into the equation:

$$\left(u \frac{n-1}{n}\right)^2 = u^2 + 2(-a)s$$

$$u^2 \frac{(n-1)^2}{n^2} = u^2 - 2as$$

Bring all terms to one side:

$$u^2 \frac{(n-1)^2}{n^2} - u^2 = -2as$$

$$u^2 \left( \frac{(n-1)^2}{n^2} - 1 \right) = -2as$$

Simplify the expression inside the parentheses:

$$\frac{(n-1)^2}{n^2} - 1 = \frac{(n-1)^2 - n^2}{n^2} = \frac{n^2 - 2n + 1 - n^2}{n^2} = \frac{-2n + 1}{n^2}$$

So:

$$u^2 \left( \frac{-2n + 1}{n^2} \right) = -2as$$

Multiply both sides by $$-1$$:

$$u^2 \left( \frac{2n - 1}{n^2} \right) = 2as$$

Solve for $$s$$:

$$s = \frac{u^2 (2n - 1)}{2a n^2}$$

Now, to stop the bullet, the final velocity must be zero. Let $$N$$ be the number of planks required. The total distance covered is $$N s$$. Using the equation of motion again, with initial velocity $$u$$, final velocity $$0$$, acceleration $$-a$$, and distance $$N s$$:

$$0^2 = u^2 + 2(-a)(N s)$$

$$0 = u^2 - 2a N s$$

$$2a N s = u^2$$

$$N = \frac{u^2}{2a s}$$

Substitute the expression for $$s$$:

$$N = \frac{u^2}{2a \left( \frac{u^2 (2n - 1)}{2a n^2} \right)} = \frac{u^2 \cdot 2a n^2}{2a u^2 (2n - 1)} = \frac{n^2}{2n - 1}$$

Thus, the number of planks required to stop the bullet is $$\frac{n^2}{2n - 1}$$. Comparing with the options:

A. Infinite
B. $$n$$
C. $$\frac{n^2}{2n-1}$$
D. $$\frac{2n^2}{n-1}$$

The expression $$\frac{n^2}{2n-1}$$ matches option C. Therefore, the correct answer is option C.

Hence, the correct answer is Option C.

Let the length of the escalator be $$ d $$. We need to find the time taken by the person to walk up the moving escalator.

First, when the escalator is stalled (not moving), the person climbs the entire length $$ d $$ in 60 seconds. So, the speed of the person relative to the escalator is $$ v_p = \frac{d}{60} $$.

Second, when the person stands still on the moving escalator, it takes 40 seconds to cover the same distance $$ d $$. So, the speed of the escalator is $$ v_e = \frac{d}{40} $$.

Now, when the person walks up the moving escalator, both the person and the escalator are moving in the same upward direction. Therefore, the effective speed of the person relative to the ground is the sum of their individual speeds: $$ v_p + v_e $$.

The time $$ t $$ taken to cover distance $$ d $$ at this combined speed is:

$$ t = \frac{d}{v_p + v_e} $$

Substitute the expressions for $$ v_p $$ and $$ v_e $$:

$$ t = \frac{d}{\frac{d}{60} + \frac{d}{40}} $$

Factor $$ d $$ out of the denominator:

$$ t = \frac{d}{d \left( \frac{1}{60} + \frac{1}{40} \right)} $$

Cancel $$ d $$ from numerator and denominator (since $$ d \neq 0 $$):

$$ t = \frac{1}{\frac{1}{60} + \frac{1}{40}} $$

Now, compute the denominator. Find the sum $$ \frac{1}{60} + \frac{1}{40} $$. The least common multiple of 60 and 40 is 120. Rewrite each fraction:

$$ \frac{1}{60} = \frac{2}{120}, \quad \frac{1}{40} = \frac{3}{120} $$

Add them:

$$ \frac{2}{120} + \frac{3}{120} = \frac{5}{120} = \frac{1}{24} $$

So,

$$ t = \frac{1}{\frac{1}{24}} = 24 \text{ seconds} $$

Hence, the time taken by the person to walk up the moving escalator is 24 seconds. Comparing with the options, Option C is 24 s.

So, the answer is 24 seconds.

We have a boy of mass 20 kg standing on a cart of mass 80 kg. The cart is free to move and there is negligible friction between the cart and the ground. Initially, the boy is 25 m from a wall. The boy walks 10 m towards the wall on the cart. We need to find the final distance of the boy from the wall.

Since there is no external force acting on the system (friction is negligible), the center of mass of the system (boy + cart) remains stationary. This is a key principle we will use.

Set up a coordinate system with the wall at position $$x = 0$$. Initially, the boy is at $$x_{\text{boy, initial}} = 25$$ m. Let the initial position of the center of mass of the cart be $$X$$. The center of mass of the entire system ($$x_{\text{cm}}$$) is given by:

$$ x_{\text{cm}} = \frac{m_{\text{boy}} \cdot x_{\text{boy, initial}} + m_{\text{cart}} \cdot X}{m_{\text{boy}} + m_{\text{cart}}} = \frac{20 \cdot 25 + 80 \cdot X}{100} = \frac{500 + 80X}{100} = 5 + 0.8X $$

Now, the boy walks 10 m towards the wall relative to the cart. Since the wall is to the left (negative x-direction), the boy moves $$-10$$ m relative to the cart. However, as the boy moves, the cart may also move. Let the displacement of the cart's center of mass be $$D$$ (positive if to the right). The displacement of the boy relative to the ground is then the displacement relative to the cart plus the displacement of the cart:

Displacement of boy relative to ground $$ = -10 + D $$

Thus, the final position of the boy is:

$$ x_{\text{boy, final}} = x_{\text{boy, initial}} + (-10 + D) = 25 - 10 + D = 15 + D $$

The final position of the cart's center of mass is:

$$ x_{\text{cart, final}} = X + D $$

The center of mass remains unchanged, so:

$$ x_{\text{cm}} = \frac{m_{\text{boy}} \cdot x_{\text{boy, final}} + m_{\text{cart}} \cdot x_{\text{cart, final}}}{m_{\text{boy}} + m_{\text{cart}}} = \frac{20 \cdot (15 + D) + 80 \cdot (X + D)}{100} $$

Set this equal to the initial center of mass:

$$ \frac{20(15 + D) + 80(X + D)}{100} = 5 + 0.8X $$

Multiply both sides by 100 to clear the denominator:

$$ 20(15 + D) + 80(X + D) = 100(5 + 0.8X) $$

Expand both sides:

Left side: $$20 \cdot 15 + 20 \cdot D + 80 \cdot X + 80 \cdot D = 300 + 20D + 80X + 80D = 300 + 100D + 80X$$

Right side: $$100 \cdot 5 + 100 \cdot 0.8X = 500 + 80X$$

So:

$$ 300 + 100D + 80X = 500 + 80X $$

Subtract $$80X$$ from both sides:

$$ 300 + 100D = 500 $$

Subtract 300 from both sides:

$$ 100D = 200 $$

Divide by 100:

$$ D = 2 $$

So the cart moves 2 m to the right. The final position of the boy is:

$$ x_{\text{boy, final}} = 15 + D = 15 + 2 = 17 \text{ m} $$

Therefore, the final distance of the boy from the wall is 17 m.

Hence, the correct answer is Option D.

For motion under uniform gravity (air resistance neglected) we can use the constant-acceleration relation

$$v^{2}=u^{2}+2g\,(h_{\text{initial}}-h)\qquad -(1)$$

Here $$h$$ is the vertical coordinate measured from the ground (origin at the ground, positive upward), $$u$$ is the speed when the particle is at height $$h_{\text{initial}}$$ and $$g$$ is the magnitude of acceleration due to gravity.

Case 1: Descent from the release point ($$h=H$$) to the ground ($$h=0$$)
The ball is dropped from rest, so $$u=0$$ at $$h=H$$. Putting $$h_{\text{initial}}=H$$ and $$u=0$$ in (1):

$$v^{2}=2g\,(H-h)\qquad (0\le h\le H)$$

Thus $$v=\sqrt{2g\,(H-h)}$$, which is the upper half of a parabola on a $$v$$ vs $$h$$ plot. At the starting point $$h=H$$, $$v=0$$; just before impact $$h=0$$, $$v=\sqrt{2gH}$$ (maximum speed).

Case 2: Ascent after the inelastic bounce
Let the coefficient of restitution with the ground be $$e\,(0\lt e\lt 1)$$. Speed just after the bounce (directed upward) is therefore $$e\sqrt{2gH}$$.

During upward motion the acceleration is still downward $$(-g)$$, so we again use (1) with $$u=e\sqrt{2gH}$$ at $$h=0$$:

$$v^{2}=e^{2}\,2gH-2g\,h\qquad (0\le h\le h_{\max})$$

where $$h_{\max}=e^{2}H$$ (height reached after bounce). Hence $$v=\sqrt{e^{2}2gH-2g\,h}\,,$$ another upper-parabolic branch but this time terminating at $$h=e^{2}H$$ where $$v=0$$.

Shape of the complete $$v$$-$$h$$ graph
1. From $$h=H$$ to $$h=0$$: a parabolic arc opening toward the $$h$$-axis, starting at $$v=0$$ and ending at $$v=\sqrt{2gH}$$.
2. Immediately after impact: vertical jump to the point $$h=0,\;v=e\sqrt{2gH}$$ (because direction has reversed but we plot speed, a positive quantity).
3. From $$h=0$$ to $$h=e^{2}H$$: a second, shorter parabolic arc (same orientation) ending at $$v=0$$.

Among the given choices only Option D displays (i) the first long parabolic branch, (ii) a discontinuity in speed at the ground, and (iii) a second shorter parabolic branch terminating below the initial height. Therefore,

Option D which is: the only correct $$v$$-$$h$$ graph.

Solution & Explanation

1. Define System Parameters and Set Up Equations

Let the total length of the train be $$L$$. Therefore, the middle wagon sits exactly at a distance of $$\frac{L}{2}$$ from both the engine (front) and the guard's room (back). The train is moving with a constant uniform acceleration $$a$$.

Using the third equation of motion ($$v^2 = u^2 + 2as$$) for the entire length of the train as it passes the pole:

• Initial velocity (when engine passes) = $$u$$
• Final velocity (when guard's room passes) = $$v$$
• Total displacement = $$L$$

$$v^2 = u^2 + 2aL \implies 2aL = v^2 - u^2 \quad \text{--- (Eq. 1)}$$

2. Analyze the Motion of the Middle Wagon

Let $$v_m$$ be the instantaneous velocity of the train at the exact moment its middle wagon passes the pole. Over this interval, the train has covered a displacement of exactly $$s = \frac{L}{2}$$ past the pole.

Applying the third equation of motion over this half-length interval:

$$v_m^2 = u^2 + 2a\left(\frac{L}{2}\right)$$

$$v_m^2 = u^2 + aL \quad \text{--- (Eq. 2)}$$

3. Substitute and Solve for $v_m$

From Equation 1, we can isolate the term $$aL$$:

$$aL = \frac{v^2 - u^2}{2}$$

Substitute this expression for $$aL$$ directly back into Equation 2:

$$v_m^2 = u^2 + \frac{v^2 - u^2}{2}$$

$$v_m^2 = \frac{2u^2 + v^2 - u^2}{2}$$

$$v_m^2 = \frac{u^2 + v^2}{2}$$

Taking the positive square root to find the actual velocity value:

$$v_m = \sqrt{\frac{u^2 + v^2}{2}}$$

Concept Check: Because kinetic energy scales with the square of velocity ($$v^2$$), the velocity at the spatial midpoint under constant acceleration matches the root-mean-square (RMS) value of the initial and final speeds rather than a simple arithmetic average.

Correct Option Key: Option D ($$\sqrt{\frac{u^2 + v^2}{2}}$$)

The graph provided is a displacement-time ($$x$$ vs $$t$$) plot. For such a graph, the instantaneous velocity at any point equals the slope of the tangent at that point:

$$v = \frac{dx}{dt} = \text{slope of the } x\text{-}t \text{ graph}$$

Case 1: Point $$A$$ lies on a straight-line segment of the curve. Two easily readable neighbouring points on that segment are, for example, $$(2\;\text{s},\,1\;\text{m})$$ and $$(4\;\text{s},\,2\;\text{m})$$. The slope (and hence the velocity at $$A$$) is

$$v_A \;=\; \frac{2\;\text{m} - 1\;\text{m}}{4\;\text{s} - 2\;\text{s}} \;=\; \frac{1}{2}\;\text{m s}^{-1} \;=\; 0.5\;\text{m s}^{-1}$$

Case 2: Point $$B$$ also falls on a straight portion of the graph. Picking two nearby points on that segment—for instance $$(10\;\text{s},\,3\;\text{m})$$ and $$(14\;\text{s},\,5\;\text{m})$$—gives

$$v_B \;=\; \frac{5\;\text{m} - 3\;\text{m}}{14\;\text{s} - 10\;\text{s}} \;=\; \frac{2}{4}\;\text{m s}^{-1} \;=\; 0.5\;\text{m s}^{-1}$$

Because both points lie on straight segments having the same slope, the instantaneous velocities are identical:

$$v_A = v_B = 0.5\;\text{m s}^{-1}$$

Therefore, the correct choice is
Option A which is: $$v_A = v_B = 0.5\ \text{m/s}$$

The first body starts from rest with uniform acceleration $$a$$, so after time $$t$$ its displacement from the origin is

$$x_1(t)=\tfrac12 a t^{2}\qquad -(1)$$

The second body crosses the origin at $$t=0$$ with a constant speed $$v$$, hence

$$x_2(t)=v\,t\qquad -(2)$$

The required function is the difference of the two displacements:

$$x_1-x_2=\tfrac12 a t^{2}-v t=t\bigl(\tfrac12 a t-v\bigr)\qquad -(3)$$

Equation (3) is a quadratic in $$t$$ whose leading coefficient $$\tfrac12 a$$ is positive, so its graph is an upward-opening parabola.

Key features of the parabola:

• At $$t=0$$, $$x_1-x_2=0$$ (the two bodies are together).
• The initial slope is $$\left.\frac{d}{dt}(x_1-x_2)\right|_{t=0}= -v$$, i.e. the curve leaves the origin with a negative slope.
• The vertex occurs when $$\dfrac{d}{dt}(x_1-x_2)=a t-v=0\;\Longrightarrow\;t=\dfrac{v}{a}$$. The minimum value there is $$-\dfrac{v^{2}}{2a}$$, so the curve lies below the $$t$$-axis between $$t=0$$ and $$t=\dfrac{2v}{a}$$.
• It crosses the $$t$$-axis again when $$x_1-x_2=0\Rightarrow t=0$$ or $$t=\dfrac{2v}{a}$$ and thereafter remains positive.

Therefore the correct plot begins at the origin, dips below the $$t$$-axis, attains a minimum at $$t=\dfrac{v}{a}$$, emerges to cut the axis at $$t=\dfrac{2v}{a}$$, and then rises indefinitely — exactly the shape shown in Option B.

Option B which is: the upward-opening parabola starting at the origin with an initial negative slope and crossing the axis once more at $$t=\dfrac{2v}{a}$$

Solution & Explanation

1. Express Velocity as a Differential Equation

Velocity ($$v$$) is defined as the instantaneous rate of change of displacement ($$x$$) with respect to time ($$t$$):

$$v = \frac{dx}{dt}$$

We substitute the given velocity expression into this definition to form our differential equation:

$$\frac{dx}{dt} = v_0 + g \cdot t + f \cdot t^2$$

Separating the variables to prepare for integration gives:

$$dx = (v_0 + g \cdot t + f \cdot t^2) \cdot dt$$

2. Perform Definite Integration with Limits

We now integrate both sides using the boundary conditions specified in the problem statement:

• At the initial boundary condition, the time is $$t = 0$$ and the position is $$x = 0$$.
• At the final boundary condition, the time reaches unit time $$t = 1$$ and the position reaches $$x = x$$.

Setting up the integration limits:

$$\int_{0}^{x} dx = \int_{0}^{1} (v_0 + g \cdot t + f \cdot t^2) \cdot dt$$

Applying the standard power rule for calculus integration ($$\int t^n \cdot dt = \frac{t^{n+1}}{n+1}$$):

$$[x]_{0}^{x} = \left[ v_0 \cdot t + \frac{g \cdot t^2}{2} + \frac{f \cdot t^3}{3} \right]_{0}^{1}$$

3. Substitute Limits to Find the Final Displacement

Evaluating the limits by substituting the upper value ($$1$$) and subtracting the lower value ($$0$$):

$$x - 0 = \left( v_0 \cdot (1) + \frac{g \cdot (1)^2}{2} + \frac{f \cdot (1)^3}{3} \right) - (0)$$

$$x = v_0 + \frac{g}{2} + \frac{f}{3}$$

Concept Check: Because acceleration is not constant (due to the time-dependent terms $$g \cdot t$$ and $$f \cdot t^2$$), standard algebraic equations of motion cannot be used. Continuous calculus integration over the entire time interval correctly accounts for the changing acceleration components.

Correct Option Key: Option B ($$v_0 + \frac{g}{2} + \frac{f}{3}$$)

Let the common speed finally acquired by both points be $$v$$.

For point $$A$$ (acceleration $$f$$, starting from rest):
  Final speed  $$v = f\,t_A$$  $$\Rightarrow \; t_A = \frac{v}{f}$$

For point $$B$$ (acceleration $$f'$$, starting from rest):
  Final speed  $$v = f'\,t_B$$  $$\Rightarrow \; t_B = \frac{v}{f'}$$

Given that point $$A$$ takes $$m$$ seconds more than point $$B$$ to reach this speed,

$$t_A - t_B = m$$ $$\frac{v}{f} - \frac{v}{f'} = m$$ $$v\left(\frac{1}{f} - \frac{1}{f'}\right) = m$$  $$-(1)$$

Distances travelled while accelerating from rest:

For $$A$$: $$s_A = \frac12 f\,t_A^2 = \frac12 f\left(\frac{v}{f}\right)^2 = \frac{v^2}{2f}$$
For $$B$$: $$s_B = \frac12 f'\,t_B^2 = \frac12 f'\left(\frac{v}{f'}\right)^2 = \frac{v^2}{2f'}$$

Given that $$A$$ covers $$n$$ units more than $$B$$,

$$s_A - s_B = n$$ $$\frac{v^2}{2}\left(\frac{1}{f} - \frac{1}{f'}\right) = n$$  $$-(2)$$

Let $$\Delta = \frac{1}{f} - \frac{1}{f'}$$. From $$(1):\; v\Delta = m \;\Rightarrow\; v = \frac{m}{\Delta}$$

Substituting $$v$$ in $$(2):$$

$$\frac{1}{2}\left(\frac{m}{\Delta}\right)^2 \Delta = n$$ $$\frac{m^2}{2\Delta} = n$$ $$\Delta = \frac{m^2}{2n}$$  $$-(3)$$

But $$\Delta = \frac{1}{f} - \frac{1}{f'} = \frac{f' - f}{ff'}$$. Equating this with $$(3):$$

$$\frac{f' - f}{ff'} = \frac{m^2}{2n}$$ $$\bigl(f' - f\bigr)\,n = \frac12\,f\,f'\,m^2$$

Thus, the required relation is
$$\boxed{(f' - f)\,n = \dfrac12\,f\,f'\,m^2}$$

Option D which is: $$(f' - f)n = \frac{1}{2}ff'm^2$$