Chapter 2: Motion in a Straight Line (Kinematics)

A body can be treated as a point object when its size is negligible compared with the distance it moves.

(a) The carriage is small compared with the distance between stations: point object.

(b) The monkey is small compared with the length of the circular track: point object.

(c) and (d) The ball and the beaker rotate and move only a short distance, comparable to their own size, so their size cannot be ignored.

(a) and (b)

(a) OP < OQ, so A lives closer to the school than B.

(b) A's graph starts at t = 0 while B's starts later, so A starts earlier than B.

(c) B's line is steeper (larger slope = larger speed), so B walks faster than A.

(d) Both graphs become horizontal (child at home) at the same instant: the same time.

(e) The lines cross once; after the crossing B is ahead, so B overtakes A once.

(a) A, B (b) A, B (c) B, A (d) same (e) B, A, once

Walking time $=\frac{2.5}{5}=0.5$ h, so she reaches the office at 9.30 am (x = 2.5 km).

She stays at x = 2.5 km from 9.30 am to 5.00 pm (horizontal line).

Auto time $=\frac{2.5}{25}=0.1$ h $=6$ min, so she is back home (x = 0) at 5.06 pm. The return line is much steeper (higher speed).

Scale: on the t-axis 1 cm = 1 h; on the x-axis 1 cm = 0.5 km.

Rising line 9.00–9.30 am to 2.5 km, flat till 5.00 pm, steep line back to 0 at 5.06 pm.

One cycle (5 forward + 3 back) takes 8 s and gives a net displacement of 2 m.

After 4 cycles (32 s) he is at 8 m. In the next 5 forward steps he reaches 8 + 5 = 13 m, the pit. Time $=32+5=37$ s.

On the graph the zigzag first touches x = 13 m at t = 37 s.

37 s

$v_0=126\times \frac{5}{18}=35{\text{m s}}^{-1}$, $v=0$, $x=200$ m.

$v^2=v_0^2+2ax\Rightarrow 0=35^2+2a(200)\Rightarrow a=-\frac{1225}{400}=-3.06{\text{m s}}^{-2}$.

$t=\frac{v-v_0}{a}=\frac{0-35}{-3.06}=11.4$ s.

Retardation 3.06 m s⁻²; time 11.4 s

(a) Vertically downwards (acceleration due to gravity).

(b) At the top the velocity is zero; the acceleration is still 9.8 m s⁻² downwards.

(c) With x = 0 at the top and downward positive, the ball is always below the top, so x > 0. Acceleration g is downward: a > 0 throughout. Going up, the ball moves upward: v < 0; coming down: v > 0.

(d) Height $H=\frac{v_0^2}{2g}=\frac{29.4^2}{2\times 9.8}=44.1$ m. Time up $=\frac{29.4}{9.8}=3$ s; time down is the same, so it returns after 6 s.

(a) Downwards (b) v = 0, a = 9.8 m s⁻² downwards (c) Up: x > 0, v < 0, a > 0; down: x > 0, v > 0, a > 0 (d) 44.1 m; 6 s

(a) True. A ball thrown up has zero speed at the top, but its acceleration is g.

(b) False. Speed is the magnitude of velocity, so zero speed means zero velocity.

(c) True. In one dimension, constant speed means the particle keeps the same velocity (a sudden reversal at the same speed would need infinite acceleration, which is unphysical).

(d) False. It speeds up only when the positive direction is the direction of motion. A body moving along −x with positive acceleration slows down.

(a) True (b) False (c) True (d) False

Fall: $t_1=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2\times 90}{9.8}}=4.29$ s; speed at impact $=9.8\times 4.29=42.0{\text{m s}}^{-1}$.

After the bounce: speed $=\frac{9}{10}\times 42.0=37.8{\text{m s}}^{-1}$. Time to rise to rest $=\frac{37.8}{9.8}=3.86$ s (at t = 8.14 s); the same time to fall back, so the next impact is at $4.29+7.71=12.0$ s with speed 37.8 m s⁻¹.

Speed rises linearly to 42.0 m s⁻¹ at 4.29 s, drops to 37.8 m s⁻¹, falls linearly to 0 at 8.14 s, rises to 37.8 m s⁻¹ at 12 s.

(a) Displacement is the change in position (from start to end); path length is the actual distance travelled. If a particle goes 5 m forward and 3 m back, the path length is 8 m but the magnitude of displacement is 2 m. Since the straight line between two points is the shortest, path length ≥ |displacement|.

(b) |Average velocity| $=\frac{|\text{displacement}|}{\text{time}}$, average speed $=\frac{\text{path length}}{\text{time}}$. In the example above, over 8 s: 0.25 m s⁻¹ and 1 m s⁻¹. Dividing the inequality in (a) by the same time, average speed ≥ |average velocity|.

Equality holds in both when the particle moves in one direction only, without turning back.

Path length ≥ |displacement| and average speed ≥ |average velocity|; equal when the motion is in one direction without reversal.

Going takes $\frac{2.5}{5}=0.5$ h = 30 min; returning takes $\frac{2.5}{7.5}=\frac{1}{3}$ h = 20 min.

(i) 0–30 min: displacement = path = 2.5 km in 0.5 h. (a) 5 km h⁻¹ (b) 5 km h⁻¹.

(ii) 0–50 min: back home, displacement 0, path 5 km in $\frac{5}{6}$ h. (a) 0 (b) $\frac{5}{5/6}=6$ km h⁻¹.

(iii) 0–40 min: in the 10 min of return he covers $7.5\times \frac{1}{6}=1.25$ km. Displacement $=2.5-1.25=1.25$ km, path $=3.75$ km, time $=\frac{2}{3}$ h. (a) $\frac{1.25}{2/3}=\frac{15}{8}=1.875$ km h⁻¹ (b) $\frac{3.75}{2/3}=\frac{45}{8}=5.625$ km h⁻¹.

(i) 5, 5 km h⁻¹ (ii) 0, 6 km h⁻¹ (iii) 15/8, 45/8 km h⁻¹

Instantaneous values are limits over vanishingly small time intervals. In such a tiny interval the particle cannot turn back, so the magnitude of displacement equals the path length. Hence the two ratios, and their limits, are equal.

For an infinitesimal interval, |displacement| = path length.

All four are impossible.

(a) For some instants the curve gives two values of x: a particle cannot be at two places at the same time.

(b) For some instants there are two values of v (one positive, one negative): a particle cannot move in two directions at once.

(c) Speed is never negative.

(d) Total path length can never decrease with time.

None of the four can represent one-dimensional motion.

No. An x–t graph shows position against time; it is not the path of the particle. For t < 0 the particle is at rest at x = 0; for t > 0 it moves along a straight line with $x\propto t^2$, i.e. with uniform acceleration from rest.

Context: a body held at the top of a tower and released (dropped) at t = 0, with x measured downwards.

No; e.g. a stone dropped from a tower at t = 0.

Van: $30\times \frac{5}{18}=8.33{\text{m s}}^{-1}$, so the bullet's speed relative to the ground $=150+8.33=158.33{\text{m s}}^{-1}$.

Car: $192\times \frac{5}{18}=53.33{\text{m s}}^{-1}$.

The damage depends on the speed of the bullet relative to the car: $158.33-53.33=105{\text{m s}}^{-1}$.

105 m s⁻¹

(a) A ball at rest on a smooth floor is kicked; it rebounds from a wall with reduced speed and moves to the opposite wall, which stops it.

(b) A ball thrown up with some velocity, rebounding from the floor with a smaller speed after each hit (the slope of each line is −g).

(c) A cricket ball moving uniformly is hit by a bat for a very short time (a large force for a short time).

(a) A kicked ball rebounding from a wall and stopped by the opposite wall (b) a ball bouncing with reduced speed (c) a ball struck by a bat for a short time

In SHM, $a=-{\omega }^{2}x$, so a always has the sign opposite to x. Velocity is the slope of the x–t graph.

t = 0.3 s: the curve is below the axis and still going down: x < 0, v < 0, a > 0.

t = 1.2 s: above the axis and rising: x > 0, v > 0, a < 0.

t = −1.2 s: below the axis and rising: x < 0, v > 0, a > 0.

0.3 s: x < 0, v < 0, a > 0; 1.2 s: x > 0, v > 0, a < 0; −1.2 s: x < 0, v > 0, a > 0

For equal intervals, the average speed is greatest where the graph is steepest and least where it is flattest.

Interval 3 is the steepest part, interval 2 (near the peak) the flattest. The graph rises in intervals 1 and 2 (v > 0) and falls in interval 3 (v < 0).

Greatest in 3, least in 2; average velocity > 0 in 1 and 2, < 0 in 3.

The magnitude of acceleration is the steepness of the graph: steepest in interval 2. The speed is highest in interval 3.

The motion is always along the positive direction, so v > 0 in all three. The speed rises in 1 and 3 (a > 0) and falls in 2 (a < 0).

At A, B, C and D the tangent is horizontal, so the acceleration is zero.

|a| greatest in 2; speed greatest in 3; v > 0 in 1, 2, 3; a > 0 in 1 and 3, a < 0 in 2; a = 0 at A, B, C, D.