Kinematics JEE Notes PDF, Formulas, Practice Questions

Kinematics is the study of motion, meaning how objects move through space and time, without asking why they move (that comes later with Newton's laws). It is the second chapter of JEE Physics and the foundation for laws of motion, work and energy, and rotational motion. These Kinematics JEE notes build the language of motion step by step (position, distance, displacement, speed, velocity, acceleration) and then cover the SUVAT equations, free fall, relative motion, projectile motion and motion graphs, along with important JEE questions for practice and fast revision.

Kinematics JEE Notes: Important Concepts

Almost every trap in this chapter comes from confusing a scalar with its vector counterpart. Get these three pairs right and most of the chapter follows.

Distance and Displacement

Distance: the total length of the path actually travelled. It is a scalar, always positive or zero.

Displacement: the shortest straight-line distance from the starting point to the ending point, together with direction. It is a vector, and can be positive, negative or zero.

• $$\text{Distance} \geq |\text{Displacement}|$$, always.
• $$\text{Distance} = |\text{Displacement}|$$ only for straight-line motion without a change of direction.
• If an object returns to its starting point, $$\text{Displacement} = 0$$ but $$\text{Distance} \neq 0$$.

Worked example: a person walks 4 m east, then 3 m north.

• $$\text{Distance} = 4 + 3 = 7\,\text{m}$$ (total path length)
• $$\text{Displacement} = \sqrt{4^2 + 3^2} = \sqrt{25} = 5\,\text{m}$$, at $$\theta = \tan^{-1}\left(\frac{3}{4}\right) \approx 36.87^\circ$$ north of east
• Distance $$7\,\text{m}$$ is greater than displacement $$5\,\text{m}$$.

Speed and Velocity

Speed is the rate at which distance is covered, a scalar, always positive or zero. Velocity is the rate of change of displacement, a vector, which can be positive, negative or zero.

Worked example: a car travels 100 km in 2 hours, then returns 60 km in 1 hour.

• Total distance $$= 160\,\text{km}$$, total time $$= 3\,\text{h}$$ ⇒ $$\text{Average speed} = \frac{160}{3} \approx 53.3\,\text{km/h}$$
• Net displacement $$= 100 - 60 = 40\,\text{km}$$ ⇒ $$\text{Average velocity} = \frac{40}{3} \approx 13.3\,\text{km/h}$$ in the original direction

JEE tip: For equal distances at speeds $$v_1$$ and $$v_2$$, average speed is the harmonic mean, $$v_{\text{avg}} = \frac{2v_1v_2}{v_1+v_2}$$. For equal times, it is the arithmetic mean, $$v_{\text{avg}} = \frac{v_1+v_2}{2}$$. This distinction is tested very often.

Acceleration

Acceleration is the rate of change of velocity with time. It is a vector, with SI unit $$\text{m/s}^2$$. Constant velocity means zero acceleration; increasing velocity means acceleration along the motion; decreasing velocity means acceleration opposite to the motion, called deceleration or retardation.

• Average acceleration: $$a_{\text{avg}} = \frac{\Delta v}{\Delta t} = \frac{v-u}{t}$$
• Instantaneous acceleration: $$a = \frac{dv}{dt} = \frac{d^2x}{dt^2}$$

Worked example: a bike goes from $$10\,\text{m/s}$$ to $$30\,\text{m/s}$$ in $$5\,\text{s}$$.

$$a = \frac{30-10}{5} = 4\,\text{m/s}^2$$, so the velocity gains $$4\,\text{m/s}$$ every second.

Equations of Motion (SUVAT) and Free Fall

The Three Equations of Motion

For straight-line motion with constant acceleration, five variables describe everything: $$s$$ (displacement), $$u$$ (initial velocity), $$v$$ (final velocity), $$a$$ (acceleration) and $$t$$ (time). Any three fix the remaining two.

Important: these hold only for constant acceleration. If acceleration varies with time, use calculus instead: $$v = \int a\,dt$$ and $$s = \int v\,dt$$.

Worked example: a car starts from rest and accelerates uniformly at $$3\,\text{m/s}^2$$. Distance in $$10\,\text{s}$$?

$$s = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2}(3)(10^2) = 150\,\text{m}$$

Worked example: a train at $$20\,\text{m/s}$$ brakes at $$2\,\text{m/s}^2$$. Stopping distance?

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

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

$$0 = 400 - 4s$$

$$s = 100\,\text{m}$$

Displacement in the nth Second

JEE often asks for the displacement during the nth second, not the displacement in n seconds. For uniformly accelerated motion, the direct formula is:

$$s_n = u + \frac{a}{2}(2n-1)$$

Here, $$s_n$$ represents the displacement during the nth second. If the object does not reverse direction during that second, its magnitude is also the distance covered in the nth second.

Worked example: an object starts from rest with $$a = 4\,\text{m/s}^2$$. Distance in the 3rd second?

$$s_3 = 0 + \frac{4}{2}(2\times3-1) = 2(5) = 10\,\text{m}$$

Free Fall

When an object is dropped or thrown near Earth's surface with negligible air resistance, it moves under constant acceleration due to gravity, $$g \approx 9.8\,\text{m/s}^2$$, often approximated as $$10\,\text{m/s}^2$$ in JEE problems. Free fall is simply uniformly accelerated motion with acceleration due to gravity.

Object dropped from rest ($$u=0$$, downward positive):

• $$v = gt$$
• $$h = \frac{1}{2}gt^2$$
• $$v^2 = 2gh$$

Object thrown upward with initial velocity $$u$$ (upward positive, so $$a=-g$$):

• $$v = u-gt$$
• $$h = ut-\frac{1}{2}gt^2$$
• $$v^2 = u^2-2gh$$
• Time to maximum height: $$t_{\text{up}} = \frac{u}{g}$$
• Maximum height: $$H = \frac{u^2}{2g}$$
• Total time of flight back to the same level: $$T = \frac{2u}{g}$$

Worked example: a ball is thrown vertically upward at $$20\,\text{m/s}$$, with $$g = 10\,\text{m/s}^2$$.

• $$H = \frac{u^2}{2g} = \frac{400}{20} = 20\,\text{m}$$
• $$T = \frac{2u}{g} = \frac{40}{10} = 4\,\text{s}$$
• It rises for $$2\,\text{s}$$, reaches $$20\,\text{m}$$, and falls back in another $$2\,\text{s}$$.

JEE tip: For a body thrown upward, the speed at any height during ascent equals the speed at that same height during descent, and the time of ascent equals the time of descent when it returns to the same level.

Projectile Motion in JEE Physics

A projectile is launched into the air and moves under gravity alone, tracing a curved path. The whole topic rests on one idea: the horizontal and vertical motions are independent.

• Horizontal: no acceleration, so velocity stays constant, since gravity acts only vertically.
• Vertical: constant acceleration $$g$$ downward, exactly like free fall.

Projectile Launched at an Angle θ

For launch speed $$u$$ at angle $$\theta$$ above the horizontal from ground level:

Initial velocity components

• Horizontal: $$u_x = u\cos\theta$$
• Vertical: $$u_y = u\sin\theta$$

At any time $$t$$

Key results

Equation of trajectory (eliminate $$t$$ between $$x$$ and $$y$$):

$$y = x\tan\theta-\frac{gx^2}{2u^2\cos^2\theta}$$

This is a parabola opening downward.

Worked example: a ball is thrown at $$40\,\text{m/s}$$ at $$30^\circ$$ above the horizontal, with $$g=10\,\text{m/s}^2$$.

• $$T = \frac{2(40)\sin30^\circ}{10} = \frac{2(40)(0.5)}{10} = 4\,\text{s}$$
• $$H = \frac{40^2\sin^2 30^\circ}{2(10)} = \frac{1600(0.5)^2}{20} = 20\,\text{m}$$
• $$R = \frac{40^2\sin60^\circ}{10} = \frac{1600(\sqrt{3}/2)}{10} = 80\sqrt{3} \approx 138.6\,\text{m}$$

Note: Range is maximum at $$\theta=45^\circ$$, giving $$R_{\max}=\frac{u^2}{g}$$. Two complementary angles, $$\theta$$ and $$90^\circ-\theta$$, give the same range but different maximum heights.

Horizontal Projectile

An object thrown horizontally from a height, such as a ball rolling off a table, is the special case $$\theta=0^\circ$$.

• $$u_x=u,\qquad u_y=0$$
• $$x=ut,\qquad y=\frac{1}{2}gt^2$$
• Time to reach the ground from height $$h$$: $$t=\sqrt{\frac{2h}{g}}$$
• Range: $$R=u\sqrt{\frac{2h}{g}}$$
• Speed on hitting the ground: $$v=\sqrt{u^2+2gh}$$

Worked example: a stone is thrown horizontally at $$10\,\text{m/s}$$ from a $$45\,\text{m}$$ tower, with $$g=10\,\text{m/s}^2$$.

• $$t=\sqrt{\frac{2(45)}{10}}=\sqrt{9}=3\,\text{s}$$
• $$x=10(3)=30\,\text{m}$$ from the base
• $$v=\sqrt{10^2+2(10)(45)}=\sqrt{1000}=10\sqrt{10}\approx31.6\,\text{m/s}$$

Relative Motion and Motion Graphs

Relative Velocity

All motion is measured against a frame of reference. When two objects move, the velocity of A as seen from B differs from the velocity of A seen by a stationary observer.

$$\vec{v}_{AB}=\vec{v}_A-\vec{v}_B$$

Special cases in one dimension:

• Same direction: $$v_{\text{rel}}=|v_A-v_B|$$
• Opposite directions: $$v_{\text{rel}}=|v_A|+|v_B|$$

Worked example: trains A and B run on parallel tracks, A at $$60\,\text{km/h}$$ east and B at $$40\,\text{km/h}$$ east.

• $$v_{AB}=60-40=20\,\text{km/h}$$ east, so a passenger in B sees A pulling ahead at $$20\,\text{km/h}$$.
• If B instead moves west at $$40\,\text{km/h}$$: $$v_{AB}=60-(-40)=100\,\text{km/h}$$, the magnitude of their relative velocity.

Reading Motion Graphs

JEE tests graph interpretation regularly, and the whole skill reduces to knowing what slope and area mean on each graph.

Position-time (x-t) graph

• Straight line ⇒ constant velocity (uniform motion)
• Horizontal line ⇒ object at rest
• Greater magnitude of slope ⇒ greater speed
• Slope at any point gives instantaneous velocity

Velocity-time (v-t) graph

• Horizontal line ⇒ constant velocity, zero acceleration
• Positive slope ⇒ positive acceleration; negative slope ⇒ negative acceleration
• Area above the time axis gives positive displacement; area below gives negative displacement

Acceleration-time (a-t) graph

• $$a=0$$ ⇒ constant velocity
• $$a=\text{constant}$$ ⇒ uniformly accelerated motion
• Area under the graph gives $$\Delta v$$

JEE tip: To get distance rather than displacement from a v-t graph, take the absolute value of each area segment and add them. This matters whenever the velocity becomes negative and the object reverses direction.

Worked example: a v-t graph rises in a straight line from $$v=0$$ at $$t=0$$ to $$v=20\,\text{m/s}$$ at $$t=5\,\text{s}$$, then stays horizontal at $$20\,\text{m/s}$$ until $$t=10\,\text{s}$$.

• Phase 1 (triangle): $$\text{Area}=\frac{1}{2}(5)(20)=50\,\text{m}$$, $$a=\frac{20-0}{5-0}=4\,\text{m/s}^2$$
• Phase 2 (rectangle): $$\text{Area}=20(5)=100\,\text{m}$$, $$a=0$$
• Total displacement: $$50+100=150\,\text{m}$$

Kinematics Formula Sheet at a Glance

JEE Important Points, Common Mistakes and Quick Revision

Points JEE Repeatedly Tests

• The scalar and vector pairs: distance against displacement, speed against velocity. A question asking for average velocity when you computed average speed is the classic trap.
• Equal distances need the harmonic mean, $$v_{\text{avg}}=\frac{2v_1v_2}{v_1+v_2}$$, while equal times need the arithmetic mean, $$v_{\text{avg}}=\frac{v_1+v_2}{2}$$.
• SUVAT applies only to constant acceleration. Variable acceleration generally requires calculus.
• Complementary launch angles, $$\theta$$ and $$90^\circ-\theta$$, give equal ranges.
• On a v-t graph, signed area is displacement, not total distance.
• For vertical motion returning to the same height under constant gravity, ascent and descent times are equal and speeds at equal heights have equal magnitudes.

Common Mistakes to Avoid

1. Sign errors in deceleration and free fall. Fix a positive direction at the start and keep it for the whole problem.
2. Using SUVAT with variable acceleration. If acceleration depends on time or position, use the appropriate calculus relation.
3. Confusing $$s_n$$ with displacement in $$n$$ seconds. The nth-second formula gives displacement during that one-second interval.
4. Averaging speeds directly when the two legs cover equal distances.
5. Applying vertical equations to the horizontal component of an ideal projectile. Horizontal velocity remains constant.
6. Forgetting that vertical velocity is zero at maximum height, while horizontal velocity is generally not zero.
7. Adding relative velocities without signs. Assign directions first, then use $$\vec{v}_{AB}=\vec{v}_A-\vec{v}_B$$.
8. Reading area under a v-t graph as distance when part of the graph lies below the time axis.

Quick Revision Notes for Kinematics

• $$\text{Distance}\geq|\text{Displacement}|$$; equality occurs for straight-line motion without reversal.
• $$v_{\text{avg}}=\frac{\text{Displacement}}{\text{Total time}}$$, while $$\text{Average speed}=\frac{\text{Distance}}{\text{Total time}}$$.
• Three SUVAT equations: $$v=u+at$$, $$s=ut+\frac{1}{2}at^2$$, $$v^2=u^2+2as$$, valid for constant acceleration only.
• nth-second displacement: $$s_n=u+\frac{a}{2}(2n-1)$$.
• Free fall or vertical throw: $$H=\frac{u^2}{2g}$$, $$t_{\text{up}}=\frac{u}{g}$$, $$T=\frac{2u}{g}$$.
• Projectile: $$T=\frac{2u\sin\theta}{g}$$, $$H=\frac{u^2\sin^2\theta}{2g}$$, $$R=\frac{u^2\sin2\theta}{g}$$; the trajectory is a downward-opening parabola.
• Horizontal projectile: $$t=\sqrt{\frac{2h}{g}}$$, $$R=u\sqrt{\frac{2h}{g}}$$, $$v=\sqrt{u^2+2gh}$$.
• Relative velocity: $$\vec{v}_{AB}=\vec{v}_A-\vec{v}_B$$.
• Graphs: $$\frac{dx}{dt}=v$$, $$\frac{dv}{dt}=a$$, area under a v-t graph gives displacement, and area under an a-t graph gives $$\Delta v$$.

Problem-solving routine: Use the JEE formula sheet to quickly recall the relevant equations, then list the known quantities u, v, a, s, t, identify which quantities are known, and choose the SUVAT equation containing those quantities and the required unknown. Choosing the right equation first removes most of the algebra