Chapter 2: Motion in a Straight Line

Instantaneous velocity. The instantaneous velocity is the time derivative of position:

$$v = \dfrac{dx}{dt} = \dfrac{d}{dt}\left(a + bt^2\right) = 2bt.$$

With $$b = 2.5 \, \mathrm{m\,s^{-2}}$$:

At $$t = 0 \, \mathrm{s}$$: $$v = 2(2.5)(0) = 0 \, \mathrm{m\,s^{-1}}.$$

At $$t = 2.0 \, \mathrm{s}$$: $$v = 2(2.5)(2.0) = 10 \, \mathrm{m\,s^{-1}}.$$

Average velocity between $$t = 2.0 \, \mathrm{s}$$ and $$t = 4.0 \, \mathrm{s}$$. By definition,

$$\bar{v} = \dfrac{x(4.0) - x(2.0)}{4.0 - 2.0}.$$

$$x(4.0) = 8.5 + 2.5(4.0)^2 = 8.5 + 40 = 48.5 \, \mathrm{m},$$
$$x(2.0) = 8.5 + 2.5(2.0)^2 = 8.5 + 10 = 18.5 \, \mathrm{m}.$$

$$\bar{v} = \dfrac{48.5 - 18.5}{2.0} = \dfrac{30}{2.0} = 15 \, \mathrm{m\,s^{-1}}.$$

Let the object move with a constant acceleration $$a$$ along the $$x$$-axis. Let $$v_0$$ and $$x_0$$ denote the velocity and position at $$t = 0$$, and $$v$$ and $$x$$ their values at a later time $$t$$.

(i) Velocity–time relation. Acceleration is $$a = \dfrac{dv}{dt}$$, so

$$dv = a\,dt.$$

Integrating between the limits $$v_0$$ (at $$t = 0$$) and $$v$$ (at time $$t$$):

$$\int_{v_0}^{v} dv = \int_{0}^{t} a\,dt \quad\Longrightarrow\quad v - v_0 = at.$$

$$v = v_0 + at.$$

(ii) Position–time relation. Since $$v = \dfrac{dx}{dt}$$,

$$dx = v\,dt = (v_0 + at)\,dt.$$

Integrating between $$x_0$$ (at $$t = 0$$) and $$x$$ (at time $$t$$):

$$\int_{x_0}^{x} dx = \int_{0}^{t} (v_0 + at)\,dt \quad\Longrightarrow\quad x - x_0 = v_0 t + \tfrac{1}{2} a t^2.$$

$$x = x_0 + v_0 t + \tfrac{1}{2} a t^2.$$

(iii) Velocity–displacement relation. Using $$a = \dfrac{dv}{dt} = \dfrac{dv}{dx} \cdot \dfrac{dx}{dt} = v\,\dfrac{dv}{dx}$$, we get

$$v\,dv = a\,dx.$$

Integrating from $$v_0$$ to $$v$$ on the left and $$x_0$$ to $$x$$ on the right:

$$\int_{v_0}^{v} v\,dv = \int_{x_0}^{x} a\,dx \quad\Longrightarrow\quad \tfrac{1}{2}\left(v^2 - v_0^2\right) = a(x - x_0).$$

$$v^2 = v_0^2 + 2a(x - x_0).$$

These three relations are the standard kinematic equations for uniformly accelerated motion in one dimension.

Take the $$y$$-axis vertically upward with origin at the point of throw. Then $$a = -g = -10 \, \mathrm{m\,s^{-2}}$$ throughout the motion (treated by a single equation valid for both upward and downward parts).

(a) Maximum height. At the topmost point the velocity is zero. Using $$v^2 = v_0^2 + 2ay$$ with $$v = 0$$, $$v_0 = 20 \, \mathrm{m\,s^{-1}}$$, $$a = -10 \, \mathrm{m\,s^{-2}}$$:

$$0 = (20)^2 + 2(-10) y \quad\Longrightarrow\quad y = \dfrac{400}{20} = 20 \, \mathrm{m}.$$

The ball rises $$20 \, \mathrm{m}$$ above the point of throw, i.e. to a height of $$25.0 + 20 = 45 \, \mathrm{m}$$ above the ground.

(b) Time to hit the ground. Let $$t$$ be the total time from the throw to the moment the ball reaches the ground. Taking upward positive and the point of throw as origin, the ground is at $$y = -25 \, \mathrm{m}$$.

$$y = v_0 t + \tfrac{1}{2} a t^2 \quad\Longrightarrow\quad -25 = 20 t + \tfrac{1}{2}(-10) t^2.$$

$$-25 = 20 t - 5 t^2 \quad\Longrightarrow\quad t^2 - 4t - 5 = 0.$$

$$t = \dfrac{4 \pm \sqrt{16 + 20}}{2} = \dfrac{4 \pm 6}{2}.$$

The positive root gives $$t = 5 \, \mathrm{s}$$ (the negative root $$t = -1\,\mathrm{s}$$ is unphysical).

When air resistance is neglected the only force on a falling object is gravity, which produces a constant downward acceleration $$g \approx 9.8 \, \mathrm{m\,s^{-2}}$$ near the Earth's surface. The motion is therefore uniformly accelerated motion in one dimension.

Choose the $$y$$-axis vertically downward with origin at the point of release, so that $$a = +g$$. Assume the body is released from rest, i.e. $$v_0 = 0$$ at $$t = 0$$.

The kinematic equations become

$$v = g t,$$
$$y = \tfrac{1}{2} g t^2,$$
$$v^2 = 2 g y.$$

Thus the speed grows linearly with time, while the distance fallen grows quadratically with time. With $$g = 9.8 \, \mathrm{m\,s^{-2}}$$, in successive seconds the body falls $$4.9 \, \mathrm{m}$$, $$14.7 \, \mathrm{m}$$, $$24.5 \, \mathrm{m}, \ldots$$ — odd-number multiples of $$4.9 \, \mathrm{m}$$ (Galileo's law).

Graphically:

• The acceleration–time graph is a horizontal line at $$a = g$$.
• The velocity–time graph is a straight line through the origin with slope $$g$$.
• The position–time (distance fallen versus time) graph is the upward-opening parabola $$y = \tfrac{1}{2} g t^2$$.

Because every body, irrespective of mass, falls with the same acceleration $$g$$ in vacuum, a feather and a stone released together strike the ground simultaneously (Galileo's experiment).

Let the body start from rest at $$t = 0$$ and choose the downward direction as positive. The acceleration is $$g$$ and the distance fallen in time $$t$$ is

$$y(t) = \tfrac{1}{2} g t^2.$$

Divide the total time into equal intervals of length $$\tau$$. The distance fallen up to the end of the $$n^{\text{th}}$$ interval is

$$y_n = \tfrac{1}{2} g (n\tau)^2 = \tfrac{1}{2} g \tau^2 \, n^2.$$

The distance traversed during the $$n^{\text{th}}$$ interval alone is

$$D_n = y_n - y_{n-1} = \tfrac{1}{2} g \tau^2 \left[n^2 - (n-1)^2\right] = \tfrac{1}{2} g \tau^2 (2n - 1).$$

Hence, putting $$n = 1, 2, 3, 4, \ldots$$

$$D_1 : D_2 : D_3 : D_4 : \ldots = 1 : 3 : 5 : 7 : \ldots$$

which is Galileo's law of odd numbers. Q.E.D.

Let $$d_s$$ be the stopping distance. The vehicle has initial velocity $$v_0$$, final velocity $$0$$ (it stops), and constant acceleration $$-a$$ (deceleration $$a$$). Using the kinematic equation $$v^2 = v_0^2 + 2(\text{acc.})\,d_s$$,

$$0 = v_0^2 + 2(-a)\,d_s.$$

Solving for $$d_s$$:

$$d_s = \dfrac{v_0^2}{2a}.$$

The stopping distance is proportional to the square of the initial velocity. So doubling the speed quadruples the stopping distance — a key reason highway speed limits matter. For two vehicles with the same braking ability $$a$$, the heavier vehicle and the lighter vehicle have the same stopping distance (it does not depend on mass in this idealisation).

During the reaction time $$t_r$$ the ruler is in free fall (released from rest) under gravity. With downward positive, $$v_0 = 0$$ and $$a = g$$, the distance fallen is

$$d = \tfrac{1}{2} g t_r^2 \quad\Longrightarrow\quad t_r = \sqrt{\dfrac{2 d}{g}}.$$

Substituting $$d = 21.0 \, \mathrm{cm} = 0.210 \, \mathrm{m}$$ and $$g = 9.8 \, \mathrm{m\,s^{-2}}$$,

$$t_r = \sqrt{\dfrac{2 \times 0.210}{9.8}} = \sqrt{0.0429} \approx 0.21 \, \mathrm{s}.$$

So the reaction time is about $$0.2 \, \mathrm{s}$$, which is typical for a healthy human.

Choose the home as origin ($$x = 0$$) and take $$t = 0$$ at 9.00 am. The motion has three phases.

Phase 1 — walking to office (9.00 am to 9.30 am).

Time taken $$= \dfrac{2.5}{5} = 0.5 \, \mathrm{h} = 30 \, \mathrm{min}$$.

So during $$t = 0$$ to $$t = 0.5 \, \mathrm{h}$$, position rises linearly from $$x = 0$$ to $$x = 2.5 \, \mathrm{km}$$ (slope $$= 5 \, \mathrm{km\,h^{-1}}$$).

Phase 2 — at office (9.30 am to 5.00 pm).

Position stays constant at $$x = 2.5 \, \mathrm{km}$$ from $$t = 0.5 \, \mathrm{h}$$ to $$t = 8.0 \, \mathrm{h}$$ (a horizontal segment).

Phase 3 — return by auto (5.00 pm onwards).

Time taken $$= \dfrac{2.5}{25} = 0.1 \, \mathrm{h} = 6 \, \mathrm{min}$$.

During $$t = 8.0 \, \mathrm{h}$$ to $$t = 8.1 \, \mathrm{h}$$, position falls linearly from $$x = 2.5 \, \mathrm{km}$$ to $$x = 0$$ (slope $$= -25 \, \mathrm{km\,h^{-1}}$$).

Plot. Take the $$t$$-axis with the scale $$1 \text{ unit} = 1 \, \mathrm{h}$$ and the $$x$$-axis with the scale $$1 \text{ unit} = 0.5 \, \mathrm{km}$$. Plot:

• A straight line from $$(0, 0)$$ to $$(0.5, 2.5)$$ — going to office.
• A horizontal line from $$(0.5, 2.5)$$ to $$(8.0, 2.5)$$ — at office.
• A steeper straight line from $$(8.0, 2.5)$$ to $$(8.1, 0)$$ — returning home.

The first segment has slope $$+5\,\mathrm{km\,h^{-1}}$$, the middle segment slope $$0$$, and the last segment slope $$-25\,\mathrm{km\,h^{-1}}$$.

One full cycle consists of $$5$$ forward steps and $$3$$ backward steps, i.e. $$8$$ steps taking $$8 \, \mathrm{s}$$. Net displacement per cycle $$= +5 - 3 = +2 \, \mathrm{m}$$.

After each complete cycle the position advances by $$2 \, \mathrm{m}$$:

At $$t = 8 \, \mathrm{s}$$: $$x = 2 \, \mathrm{m}$$.
At $$t = 16 \, \mathrm{s}$$: $$x = 4 \, \mathrm{m}$$.
At $$t = 24 \, \mathrm{s}$$: $$x = 6 \, \mathrm{m}$$.
At $$t = 32 \, \mathrm{s}$$: $$x = 8 \, \mathrm{m}$$.

However, the maximum position reached within a cycle (i.e. at the end of the $$5$$ forward steps) is $$5 \, \mathrm{m}$$ more than the position at the start of the cycle. So within the cycle the position reaches $$x = 2k + 5$$ at $$t = 8k + 5 \, \mathrm{s}$$, before falling back by $$3 \, \mathrm{m}$$.

The pit is at $$x = 13 \, \mathrm{m}$$. Look at when the drunkard's position first reaches $$13 \, \mathrm{m}$$.

After $$4$$ complete cycles ($$t = 32 \, \mathrm{s}$$) the drunkard is at $$x = 8 \, \mathrm{m}$$. He has just begun his $$5^{\text{th}}$$ cycle. Now he takes $$5$$ forward steps:

• $$t = 33 \, \mathrm{s}$$: $$x = 9 \, \mathrm{m}$$.
• $$t = 34 \, \mathrm{s}$$: $$x = 10 \, \mathrm{m}$$.
• $$t = 35 \, \mathrm{s}$$: $$x = 11 \, \mathrm{m}$$.
• $$t = 36 \, \mathrm{s}$$: $$x = 12 \, \mathrm{m}$$.
• $$t = 37 \, \mathrm{s}$$: $$x = 13 \, \mathrm{m}$$ — the drunkard falls into the pit.

Plot. The $$x$$-$$t$$ graph is a zig-zag staircase: from $$(0,0)$$ a straight line of slope $$+1\,\mathrm{m\,s^{-1}}$$ up to $$(5, 5)$$, then a line of slope $$-1\,\mathrm{m\,s^{-1}}$$ down to $$(8, 2)$$, then up to $$(13, 7)$$, down to $$(16, 4)$$, up to $$(21, 9)$$, down to $$(24, 6)$$, up to $$(29, 11)$$, down to $$(32, 8)$$, and finally up to $$(37, 13)$$ where the motion ends.

Convert the initial speed to SI units:

$$u = 126 \, \mathrm{km\,h^{-1}} = 126 \times \dfrac{1000}{3600} \, \mathrm{m\,s^{-1}} = 35 \, \mathrm{m\,s^{-1}}.$$

Final speed $$v = 0$$ and distance $$s = 200 \, \mathrm{m}$$. Using $$v^2 = u^2 + 2 a s$$:

$$0 = (35)^2 + 2 a (200) \quad\Longrightarrow\quad a = -\dfrac{1225}{400} = -3.0625 \, \mathrm{m\,s^{-2}}.$$

So the retardation (magnitude of the deceleration) is

$$|a| \approx 3.06 \, \mathrm{m\,s^{-2}}.$$

Time to stop. Using $$v = u + at$$:

$$0 = 35 + (-3.0625) t \quad\Longrightarrow\quad t = \dfrac{35}{3.0625} \approx 11.43 \, \mathrm{s}.$$

Take $$g = 9.8 \, \mathrm{m\,s^{-2}}$$ and the ball starting from rest at $$h = 90 \, \mathrm{m}$$.

First fall (0 to first collision). Using $$h = \tfrac{1}{2} g t^2$$:

$$t_1 = \sqrt{\dfrac{2 h}{g}} = \sqrt{\dfrac{2 \times 90}{9.8}} = \sqrt{18.37} \approx 4.29 \, \mathrm{s}.$$

Speed just before first impact:

$$v_1 = g t_1 = 9.8 \times 4.29 \approx 42 \, \mathrm{m\,s^{-1}}.$$

After the collision the ball loses one tenth of its speed, retaining $$9/10$$, so

$$v_1' = 0.9 \times 42 = 37.8 \, \mathrm{m\,s^{-1}}.$$

First bounce. The ball moves up at $$37.8 \, \mathrm{m\,s^{-1}}$$, decelerating at $$g$$. Time to reach the top: $$37.8/9.8 \approx 3.86 \, \mathrm{s}$$. By symmetry, total time for the up-and-down bounce is

$$\Delta t = \dfrac{2 \times 37.8}{9.8} \approx 7.71 \, \mathrm{s}.$$

So the next (second) collision occurs at $$t \approx 4.29 + 7.71 = 12.0 \, \mathrm{s}$$, with speed again $$37.8 \, \mathrm{m\,s^{-1}}$$ just before impact.

Speed–time graph for $$0 \le t \le 12 \, \mathrm{s}$$.

• $$0 \le t \le 4.29 \, \mathrm{s}$$: speed rises linearly from $$0$$ to $$42 \, \mathrm{m\,s^{-1}}$$ with slope $$+g$$.
• At $$t = 4.29 \, \mathrm{s}$$ there is a sudden vertical drop from $$42$$ to $$37.8 \, \mathrm{m\,s^{-1}}$$ (the collision).
• $$4.29 \le t \le 8.14 \, \mathrm{s}$$: ball moves upward; speed decreases linearly from $$37.8$$ to $$0$$ (slope $$-g$$).
• $$8.14 \le t \le 12.0 \, \mathrm{s}$$: ball falls back; speed increases linearly from $$0$$ to $$37.8 \, \mathrm{m\,s^{-1}}$$ (slope $$+g$$).
• At $$t = 12.0 \, \mathrm{s}$$ there is again a vertical drop, this time from $$37.8$$ to $$34.02 \, \mathrm{m\,s^{-1}}$$.

The graph is a saw-tooth shape whose peaks decay (each peak is $$0.9$$ times the previous one) because of the energy loss at each collision.

Instantaneous velocity is

$$\vec{v} = \lim_{\Delta t \to 0} \dfrac{\Delta \vec{r}}{\Delta t},$$

so its magnitude is $$|\vec{v}| = \lim_{\Delta t \to 0} \dfrac{|\Delta \vec{r}|}{\Delta t}$$.

Instantaneous speed is

$$s = \lim_{\Delta t \to 0} \dfrac{\Delta L}{\Delta t},$$

where $$\Delta L$$ is the path length traversed in the time interval $$\Delta t$$. In the infinitesimal limit $$\Delta t \to 0$$, the particle cannot reverse direction during the interval; the path traversed is a single infinitesimal segment, hence $$\Delta L = |\Delta \vec{r}|$$. Therefore

$$s = |\vec{v}|.$$

The discrepancy between average speed and the magnitude of average velocity arose precisely from changes in direction during a finite interval; over an infinitesimal interval there is no scope for such a change, so the two quantities coincide.

None of the four graphs in Fig. 2.10 can represent one-dimensional motion of a particle.

(a) The graph is an $$x$$-$$t$$ plot in which a single instant $$t$$ has two different values of $$x$$. A particle cannot be in two positions at the same instant, so this is impossible.

(b) The graph is a $$v$$-$$t$$ plot showing two values of velocity at the same instant $$t$$. A particle cannot have two velocities simultaneously, so this is impossible.

(c) The graph plots speed against time and shows speed taking negative values. Speed is by definition the magnitude of velocity and cannot be negative, so this is impossible.

(d) The graph plots total path length against time and shows path length decreasing. Path length is monotonically non-decreasing in time (the particle keeps adding distance traversed, even if it reverses direction), so this is impossible.

No, it is not correct. Figure 2.11 is an $$x$$-$$t$$ (position–time) graph; it is not the trajectory or path of the particle in space. A particle in one-dimensional motion is constrained to move along the $$x$$-axis, so its actual path is always a straight line — both for $$t < 0$$ and for $$t > 0$$. The shape of the $$x$$-$$t$$ curve tells us only how the position changes with time, not the geometric path traced out.

A suitable physical context. For $$t < 0$$ the graph is a horizontal line, so the position does not change with time — the particle is simply at rest. For $$t > 0$$ the $$x$$-$$t$$ graph is a parabola ($$x \propto t^2$$), which represents uniformly accelerated motion. A suitable situation is therefore a ball held at rest in the hand up to $$t = 0$$, and then released at $$t = 0$$ so that for $$t > 0$$ it falls freely under gravity with constant acceleration, $$x = \tfrac{1}{2} g t^2$$.

The damage to the thief's car depends on the speed of the bullet relative to the car, not on its ground speed. Convert all speeds to $$\mathrm{m\,s^{-1}}$$:

$$v_{\text{van}} = 30 \, \mathrm{km\,h^{-1}} = 30 \times \dfrac{5}{18} = \dfrac{25}{3} \, \mathrm{m\,s^{-1}}.$$

$$v_{\text{thief}} = 192 \, \mathrm{km\,h^{-1}} = 192 \times \dfrac{5}{18} = \dfrac{160}{3} \, \mathrm{m\,s^{-1}}.$$

The muzzle speed of the bullet ($$150 \, \mathrm{m\,s^{-1}}$$) is relative to the van. So the bullet's speed in the ground frame is

$$v_{\text{bullet}} = 150 + \dfrac{25}{3} = \dfrac{475}{3} \, \mathrm{m\,s^{-1}}.$$

The speed of the bullet relative to the thief's car (both moving in the same direction) is

$$v_{\text{bullet/thief}} = v_{\text{bullet}} - v_{\text{thief}} = \dfrac{475}{3} - \dfrac{160}{3} = \dfrac{315}{3} = 105 \, \mathrm{m\,s^{-1}}.$$

(a) The $$x$$-$$t$$ graph that rises and then suddenly drops back to zero.

A ball is thrown vertically upward by a boy from the ground. It rises (position $$x$$ increases with time), reaches the highest point and falls back; on hitting the ground at the same horizontal location the ball is suddenly stopped (caught by the boy or by the ground). After that the ball is at rest.

(b) The $$v$$-$$t$$ graph showing alternating positive and (smaller-magnitude) negative spikes that gradually die out.

A ball is dropped from a height onto a hard floor. It hits the floor (velocity reverses direction abruptly — vertical jump in the graph), rebounds with a smaller speed, falls back, hits the floor again, and so on. Each successive rebound has a smaller magnitude because energy is lost in each collision.

(c) The $$a$$-$$t$$ graph that is zero except for a short pulse.

A cricket ball is moving with uniform velocity and is struck by a batsman. During the brief contact between the bat and the ball, a large acceleration acts on the ball (the short non-zero pulse); before and after the impact the ball moves uniformly (acceleration zero).

For simple harmonic motion the displacement can be written as

$$x(t) = -A \sin\!\left(\dfrac{2\pi t}{T}\right)$$

with the period $$T = 2 \, \mathrm{s}$$, consistent with Fig. 2.13. Then

$$v(t) = \dfrac{dx}{dt} = -A \dfrac{2\pi}{T} \cos\!\left(\dfrac{2\pi t}{T}\right), \qquad a(t) = -\left(\dfrac{2\pi}{T}\right)^2 x(t).$$

Crucially, $$a$$ always has the sign opposite to $$x$$.

At $$t = 0.3 \, \mathrm{s}$$. Read the graph: $$x$$ is negative. Since the curve is still moving towards its negative extreme, $$v$$ is negative (slope $$dx/dt$$ is negative). $$a = -\omega^2 x$$ is positive.

At $$t = 1.2 \, \mathrm{s}$$. From the graph: $$x$$ is positive (the curve has swung past zero into positive territory). Its slope is positive ($$v > 0$$). $$a = -\omega^2 x$$ is negative.

At $$t = -1.2 \, \mathrm{s}$$. $$x$$ is negative (graph below the $$t$$-axis). The slope is positive (curve rising towards zero from the negative trough), so $$v > 0$$. $$a = -\omega^2 x$$ is positive.

On an $$x$$-$$t$$ graph the slope of the chord drawn between the two endpoints of an interval gives the average velocity; its magnitude is the average speed (the path length equals the magnitude of displacement here, since the $$x$$-$$t$$ curve in Fig. 2.14 is monotonic within each interval).

Average speed. Visually compare the steepness of the chords joining the endpoints of the three equal-width intervals. The chord is steepest in interval 3 and almost flat (horizontal) in interval 2.

Hence the average speed is greatest in interval 3 and least in interval 2.

Sign of average velocity. The slope of the chord is positive in intervals 1 and 2 (the curve is rising) and negative in interval 3 (the curve is falling).

So the average velocity is positive in intervals 1 and 2 and negative in interval 3.

Since the motion is along a constant direction taken as positive, the speed equals the velocity: $$v$$ is positive throughout. The slope of the speed-time curve gives the instantaneous acceleration.

Average acceleration over each interval. Average acceleration $$=$$ (change in speed)/(time interval). The change in speed is largest in interval 2, where the curve falls steeply. So the magnitude of the average acceleration is greatest in interval 2.

Average speed. The area under the speed-time curve (over an interval) divided by the duration gives the average speed. In interval 3 the curve is uniformly high (the highest values of speed are attained there), so the average speed is greatest in interval 3.

Signs of $$v$$ and $$a$$.

Accelerations at A, B, C and D. The points A, B, C and D lie at the local maxima and minima of the speed-time curve. At a local extremum the tangent to the curve is horizontal, so the slope (and hence the acceleration) is zero. Thus

$$a_A = a_B = a_C = a_D = 0.$$