NCERT Solutions Class 11 Physics Chapter 2 — Motion in a Straight Line

NCERT Class 11 Physics — Motion in a Straight Line, Chapter 2 Exercises. All 18 questions solved.

The exercises at the end of Chapter 2 fall into three groups. Questions 2.1, 2.7, 2.9 to 2.11 and 2.14 are about the ideas themselves — when a body can be treated as a point, the difference between displacement and path length, between velocity and speed, and what “relative velocity” means. Questions 2.5 and 2.6 use the equations of uniformly accelerated motion. Everything else, 2.2 to 2.4, 2.8, 2.12, 2.13 and 2.15 to 2.18, is about reading or drawing graphs of motion. Three sets of results run through the whole exercise:

• Slopes and areas of graphs — the slope of an $x$–$t$ graph is the velocity, $v = \dfrac{dx}{dt}$; the slope of a $v$–$t$ graph is the acceleration, $a = \dfrac{dv}{dt}$; the area under a $v$–$t$ graph is the displacement.
• Uniform acceleration — $v = v_0 + at$, $\ x = v_0t + \tfrac12at^2$, $\ v^2 = v_0^2 + 2ax$.
• Averages — average velocity $=$ displacement $\div$ time interval, average speed $=$ total path length $\div$ time interval; and the velocity of $B$ relative to $A$ is $v_{BA} = v_B – v_A$.

Key insight. A graph of motion is not a picture of the path. An $x$–$t$ graph plots a single number, the position along the line, against time, and the only thing its shape tells you directly is the velocity, through its slope. Read slopes rather than shapes and the graph questions (2.2, 2.16, 2.17, 2.18) settle themselves — and 2.12 and 2.13 are exactly the cases where mistaking the graph for the path leads you wrong.

Question 2.1

In which of the following examples of motion, can the body be considered approximately a point object: (a) a railway carriage moving without jerks between two stations. (b) a monkey sitting on top of a man cycling smoothly on a circular track. (c) a spinning cricket ball that turns sharply on hitting the ground. (d) a tumbling beaker that has slipped off the edge of a table.

Solution. A body can be treated as a point when the distance it moves is much larger than its own size, and all its parts move together, so that one point describes the whole motion.

Question 2.1 (a)

Solution. A carriage some $20\ \text{m}$ long travels kilometres between stations, and moving without jerks, every part of it moves in the same way.

Question 2.1 (b)

Solution. The track is far larger than the cyclist and the monkey, and the monkey is carried along with the cycle, so its position on the track is all that matters.

Question 2.1 (c)

Solution. The interesting part of this motion is the spin and the sharp turn it produces off the ground. The turn happens over a distance comparable to the ball’s size, and it is the rotation of the ball, which a point cannot have, that causes it.

Question 2.1 (d)

Solution. The beaker falls through a height of about a metre, not much more than its own size, and it tumbles, so different parts of it move quite differently.

Question 2.2

The position-time ($x$-$t$) graphs for two children $A$ and $B$ returning from their school $O$ to their homes $P$ and $Q$ respectively are shown in Fig. 2.9. Choose the correct entries in the brackets below; (a) ($A$/$B$) lives closer to the school than ($B$/$A$) (b) ($A$/$B$) starts from the school earlier than ($B$/$A$) (c) ($A$/$B$) walks faster than ($B$/$A$) (d) $A$ and $B$ reach home at the (same/different) time (e) ($A$/$B$) overtakes ($B$/$A$) on the road (once/twice).

Solution. On an $x$–$t$ graph the height of a line is how far the child is from the school, the point where it leaves the time axis is when the child sets out, the slope is the walking speed, and the end of the line is the arrival home. Each part is answered by reading one of these.

Question 2.2 (a)

Solution. $P$ is marked lower on the $x$-axis than $Q$, so $A$’s home is nearer the school.

Question 2.2 (b)

Solution. $A$’s line leaves the time axis at $t = 0$; $B$’s leaves it later.

Question 2.2 (c)

Solution. $B$’s line is steeper, and the slope of an $x$–$t$ graph is the speed.

Question 2.2 (d)

Solution. Each line ends where the child reaches home: $A$’s at the level of $P$, $B$’s at the level of $Q$. The dashed lines in the figure drop from those two end points to the time axis. $B$’s line reaches $Q$ at $t_B$, well before $A$’s line reaches $P$ at $t_A$ — measured on the figure, $t_B$ is about three-quarters of $t_A$. So they arrive at different times, and $B$, though he set out later, gets home first.

Note on the book’s answer. The answer key prints “same”. That would be the reading if the two lines ended at the same instant, but in Fig. 2.9 as printed they do not: $B$’s line stops at the level of $Q$ while $A$’s is still rising towards $P$. The figure is the data the question is built on, so the answer is “different”.

Question 2.2 (e)

Solution. Where two $x$–$t$ lines cross, the two children are at the same place at the same time. The lines cross once. Before the crossing $A$’s line is higher, so $A$ is ahead; after it $B$’s line is higher. So $B$ catches up with $A$ and passes him, once.

Question 2.3

A woman starts from her home at $9.00$ am, walks with a speed of $5\ \text{km h}^{-1}$ on a straight road up to her office $2.5\ \text{km}$ away, stays at the office up to $5.00$ pm, and returns home by an auto with a speed of $25\ \text{km h}^{-1}$. Choose suitable scales and plot the $x$-$t$ graph of her motion.

Solution. First work out when each stage begins and ends. Walking $2.5\ \text{km}$ at $5\ \text{km h}^{-1}$ takes $\dfrac{2.5}{5} = 0.5\ \text{h}$, so she reaches the office at $9.30$ am. She stays at $x = 2.5\ \text{km}$ until $5.00$ pm. The auto covers $2.5\ \text{km}$ at $25\ \text{km h}^{-1}$ in $\dfrac{2.5}{25} = 0.1\ \text{h} = 6\ \text{min}$, so she is home at $5.06$ pm.

The graph therefore has three straight pieces: a line rising from $0$ to $2.5\ \text{km}$ between $9.00$ and $9.30$ am, a horizontal line at $2.5\ \text{km}$ until $5.00$ pm, while she is at rest, and a steep line falling back to $0$ by $5.06$ pm. The return line has a negative slope, since she is moving back towards home, and it is five times as steep as the outward one, because the auto is five times as fast. The long wait in the office would squeeze both journeys into slivers on a single scale, so the time axis is broken between $10$ am and $4.30$ pm; the scale is the same, $1\ \text{h}$ to $120$ units, on both sides of the break.

Question 2.4

A drunkard walking in a narrow lane takes $5$ steps forward and $3$ steps backward, followed again by $5$ steps forward and $3$ steps backward, and so on. Each step is $1\ \text{m}$ long and requires $1\ \text{s}$. Plot the $x$-$t$ graph of his motion. Determine graphically and otherwise how long the drunkard takes to fall in a pit $13\ \text{m}$ away from the start.

Solution. Each step moves him $1\ \text{m}$ in $1\ \text{s}$, so the graph is made of straight pieces of slope $\pm1\ \text{m s}^{-1}$: up for $5\ \text{s}$, down for $3\ \text{s}$, and repeat. One such cycle takes $8\ \text{s}$ and leaves him $5 – 3 = 2\ \text{m}$ further on.

Graphically. The graph rises in a zigzag whose peaks are at $5, 7, 9, 11, 13\ \text{m}$, reached at $t = 5, 13, 21, 29, 37\ \text{s}$. It first touches the line $x = 13\ \text{m}$ at $t = 37\ \text{s}$.

Otherwise. It is tempting to say that $13\ \text{m}$ at $2\ \text{m}$ per $8\ \text{s}$ takes $52\ \text{s}$, but he does not need to finish a cycle — he falls in as soon as he first reaches $13\ \text{m}$, which happens during a forward run. After $4$ complete cycles, $32\ \text{s}$, he is at $4\times2 = 8\ \text{m}$. His next $5$ forward steps take him from $8\ \text{m}$ to $13\ \text{m}$, arriving at $32 + 5 = 37\ \text{s}$. No earlier peak reaches $13\ \text{m}$: the one before, at $29\ \text{s}$, is only $11\ \text{m}$.

Question 2.5

A car moving along a straight highway with speed of $126\ \text{km h}^{-1}$ is brought to a stop within a distance of $200\ \text{m}$. What is the retardation of the car (assumed uniform), and how long does it take for the car to stop?

Solution. Convert the speed first: $v_0 = 126\times\dfrac{1000}{3600} = 35\ \text{m s}^{-1}$. The final speed is $0$ and the distance $200\ \text{m}$, with no time given, so the equation to use is the one without $t$, $v^2 = v_0^2 + 2ax$:

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

The minus sign says the acceleration is opposite to the motion — a retardation of $3.06\ \text{m s}^{-2}$. The time then follows from $v = v_0 + at$:

$$t = \frac{v – v_0}{a} = \frac{0 – 35}{-3.0625} = 11.4\ \text{s}$$

Question 2.6

A player throws a ball upwards with an initial speed of $29.4\ \text{m s}^{-1}$. (a) What is the direction of acceleration during the upward motion of the ball? (b) What are the velocity and acceleration of the ball at the highest point of its motion? (c) Choose the $x = 0\ \text{m}$ and $t = 0\ \text{s}$ to be the location and time of the ball at its highest point, vertically downward direction to be the positive direction of $x$-axis, and give the signs of position, velocity and acceleration of the ball during its upward, and downward motion. (d) To what height does the ball rise and after how long does the ball return to the player’s hands? (Take $g = 9.8\ \text{m s}^{-2}$ and neglect air resistance).

Question 2.6 (a)

Solution. Once the ball leaves the hand, the only force on it (air resistance neglected) is its weight, which acts downwards. So the acceleration is $g$, downwards, all the time — while it rises as well as while it falls. Rising, the ball is slowing down, which is exactly what a downward acceleration does to an upward velocity.

Question 2.6 (b)

Solution. At the highest point the ball stops rising and has not yet begun to fall, so its velocity is momentarily zero. Its acceleration is not zero: gravity still acts. If the acceleration were zero there, the velocity would stay zero and the ball would hang in the air.

Question 2.6 (c)

Solution. With $x = 0$ at the highest point and downwards positive, the ball is always at or below the top, so $x > 0$ throughout, going up and coming down. On the way up it moves upwards, against the positive direction, so $v < 0$; on the way down $v > 0$. The acceleration is $g$ downwards, along the positive direction, so $a > 0$ throughout. On the way up, $v$ and $a$ have opposite signs, and the ball slows down.

Question 2.6 (d)

Solution. Take upwards as positive for this part, so $v_0 = 29.4\ \text{m s}^{-1}$ and $a = -9.8\ \text{m s}^{-2}$. At the top $v = 0$, so $v^2 = v_0^2 + 2ax$ gives the height:

$$h = \frac{v_0^2}{2g} = \frac{(29.4)^2}{2\times9.8} = 44.1\ \text{m}$$

The time to reach the top is $t = \dfrac{v_0}{g} = \dfrac{29.4}{9.8} = 3\ \text{s}$. Coming down, the ball starts from rest and falls the same $44.1\ \text{m}$ with the same acceleration, so the fall also takes $3\ \text{s}$. It is back in the player’s hands after $6\ \text{s}$.

Question 2.7

Read each statement below carefully and state with reasons and examples, if it is true or false; A particle in one-dimensional motion (a) with zero speed at an instant may have non-zero acceleration at that instant (b) with zero speed may have non-zero velocity, (c) with constant speed must have zero acceleration, (d) with positive value of acceleration must be speeding up.

Question 2.7 (a)

Solution. Acceleration is the rate at which velocity is changing, not the velocity itself. A ball thrown straight up has zero speed at the top, as in Question 2.6(b), yet its acceleration there is $9.8\ \text{m s}^{-2}$ — which is why it starts to fall.

Question 2.7 (b)

Solution. Speed is the magnitude of the velocity. If the magnitude is zero, the velocity is zero; a vector of zero length has no direction left to point in.

Question 2.7 (c)

Solution. In one dimension, velocity is either $+v$ or $-v$, where $v$ is the speed. With the speed constant, the velocity could change only by jumping from $+v$ to $-v$, reversing in no time at all. That would take an infinite acceleration, which no real body has. So the velocity itself stays constant, and the acceleration is zero. (A ball bouncing off a wall is not a counter-example: during the brief contact its speed is not constant.)

Question 2.7 (d)

Solution. Whether a body speeds up or slows down depends on whether its velocity and acceleration have the same sign, not on the sign of the acceleration alone. In Question 2.6(c), with downwards positive, the rising ball has $a > 0$ and $v < 0$, and it is slowing down.

Question 2.8

A ball is dropped from a height of $90\ \text{m}$ on a floor. At each collision with the floor, the ball loses one tenth of its speed. Plot the speed-time graph of its motion between $t = 0$ to $12\ \text{s}$.

Solution. Take $g = 9.8\ \text{m s}^{-2}$ and follow the ball stage by stage, since the graph is a straight line between collisions.

First fall. From rest through $90\ \text{m}$, $x = \tfrac12gt^2$ gives $t_1 = \sqrt{\dfrac{2\times90}{9.8}} = 4.29\ \text{s}$, and the ball hits the floor at $v_1 = gt_1 = 42\ \text{m s}^{-1}$. The speed rises uniformly from $0$ to $42\ \text{m s}^{-1}$.

First bounce. It loses one tenth of its speed, so it leaves the floor at $0.9\times42 = 37.8\ \text{m s}^{-1}$. On the graph the speed drops suddenly from $42$ to $37.8$.

Rise and fall. Going up, the speed falls uniformly to zero in $\dfrac{37.8}{9.8} = 3.86\ \text{s}$, reaching zero at $t = 8.14\ \text{s}$. Coming down, it takes the same $3.86\ \text{s}$ to regain $37.8\ \text{m s}^{-1}$, and hits the floor again at $t = 4.29 + 2\times3.86 = 12.0\ \text{s}$. The next bounce, at $12\ \text{s}$, would cut the speed to $34.0\ \text{m s}^{-1}$, which is where the requested graph ends.

Every sloping piece has slope $\pm9.8\ \text{m s}^{-2}$, the magnitude of $g$. The graph shows speed, which cannot be negative, so the part where the ball rises slopes down towards zero, not below the axis.

Question 2.9

Explain clearly, with examples, the distinction between: (a) magnitude of displacement (sometimes called distance) over an interval of time, and the total length of path covered by a particle over the same interval; (b) magnitude of average velocity over an interval of time, and the average speed over the same interval. [Average speed of a particle over an interval of time is defined as the total path length divided by the time interval]. Show in both (a) and (b) that the second quantity is either greater than or equal to the first. When is the equality sign true? [For simplicity, consider one-dimensional motion only].

Question 2.9 (a)

Solution. The magnitude of displacement compares only the starting and finishing positions, $|x_2 – x_1|$; the path length adds up every stretch actually travelled, forwards and backwards. A man who walks $2.5\ \text{km}$ to a market and back (Question 2.10) has covered a path of $5\ \text{km}$ but has zero displacement. In one dimension, going from $x_1$ to $x_2$ means covering at least the stretch between them, and any backtracking only adds more, so the path length is always at least $|x_2 – x_1|$. The two are equal when the particle moves in one direction only, never turning back.

Question 2.9 (b)

Solution. Both averages divide by the same time interval, so

$$\text{average speed} = \frac{\text{path length}}{\Delta t} \;\ge\; \frac{|\text{displacement}|}{\Delta t} = |\text{average velocity}|$$

follows directly from (a). For the man going to the market and back, the average velocity over the round trip is zero while the average speed is not. Equality again holds only when there is no reversal of direction.

Question 2.10

A man walks on a straight road from his home to a market $2.5\ \text{km}$ away with a speed of $5\ \text{km h}^{-1}$. Finding the market closed, he instantly turns and walks back home with a speed of $7.5\ \text{km h}^{-1}$. What is the (a) magnitude of average velocity, and (b) average speed of the man over the interval of time (i) $0$ to $30$ min, (ii) $0$ to $50$ min, (iii) $0$ to $40$ min? [Note: You will appreciate from this exercise why it is better to define average speed as total path length divided by time, and not as magnitude of average velocity. You would not like to tell the tired man on his return home that his average speed was zero!]

Solution. The outward walk takes $\dfrac{2.5}{5} = 0.5\ \text{h} = 30\ \text{min}$, and the walk back $\dfrac{2.5}{7.5} = \dfrac13\ \text{h} = 20\ \text{min}$, so he is home at $50$ min. For each interval, find the displacement and the path length, then divide by the time.

Question 2.10 (i)

Solution. In the first $30$ min he walks straight to the market without turning, so displacement and path length are both $2.5\ \text{km}$, and both averages are $\dfrac{2.5\ \text{km}}{0.5\ \text{h}} = 5\ \text{km h}^{-1}$.

Question 2.10 (ii)

Solution. After $50$ min he is back home: the displacement is zero, so the average velocity is zero. The path length is $2.5 + 2.5 = 5\ \text{km}$ in $\dfrac{50}{60}\ \text{h}$, so the average speed is $5\div\dfrac56 = 6\ \text{km h}^{-1}$ — not the $6.25$ that averaging $5$ and $7.5$ would give, because he spends longer at the slower speed.

Question 2.10 (iii)

Solution. By $40$ min he has walked back for $10$ min at $7.5\ \text{km h}^{-1}$, covering $7.5\times\dfrac{10}{60} = 1.25\ \text{km}$, so he is $2.5 – 1.25 = 1.25\ \text{km}$ from home. The time is $\dfrac{40}{60} = \dfrac23\ \text{h}$.

$$|\bar v| = \frac{1.25}{2/3} = \frac{15}{8} = 1.875\ \text{km h}^{-1}, \qquad \text{average speed} = \frac{2.5 + 1.25}{2/3} = \frac{45}{8} = 5.625\ \text{km h}^{-1}$$

Question 2.11

In Exercises 2.9 and 2.10, we have carefully distinguished between average speed and magnitude of average velocity. No such distinction is necessary when we consider instantaneous speed and magnitude of velocity. The instantaneous speed is always equal to the magnitude of instantaneous velocity. Why?

Solution. Instantaneous quantities are limits over a vanishingly small interval $\Delta t$. In a short enough interval the particle has no time to turn back, so it moves in one direction only — which, by Question 2.9, is exactly the condition for the path length to equal the magnitude of the displacement. Dividing both by $\Delta t$ and letting $\Delta t \to 0$, the two limits are the same: instantaneous speed $= |dx/dt| = |v|$.

Question 2.12

Look at the graphs (a) to (d) (Fig. 2.10) carefully and state, with reasons, which of these cannot possibly represent one-dimensional motion of a particle.

Solution. Each graph breaks a rule that every real motion obeys.

(a) A vertical line drawn at many values of $t$ cuts this curve in more than one place, so it gives the particle two or three different positions at the same instant. The loops also make time run backwards along the curve.

(b) The circle likewise gives two values of velocity — one positive, one negative — at each instant, so the particle would be moving in both directions at once; and it too would need time to run backwards.

(c) Part of the curve lies below the time axis, which would mean a negative speed. Speed is a magnitude and cannot be negative.

(d) The total path length falls from its peak back to zero. Path length only accumulates: it can stay constant (at rest) or increase, but never decrease.

The arrows drawn on graphs (a) and (b) change nothing, since time can only run one way along the $t$-axis whatever the arrows suggest.

Question 2.13

Figure 2.11 shows the $x$-$t$ plot of one-dimensional motion of a particle. Is it correct to say from the graph that the particle moves in a straight line for $t < 0$ and on a parabolic path for $t > 0$? If not, suggest a suitable physical context for this graph.

Solution. No. An $x$–$t$ graph is not the path of the particle. The particle moves along the $x$-axis all the time — the motion is one-dimensional — and the graph only records where on that line it is at each instant. For $t < 0$ the graph lies along the time axis: the particle sits at rest at $x = 0$. For $t > 0$ the graph is a parabola, $x \propto t^2$, which is the signature of motion from rest with uniform acceleration, $x = \tfrac12at^2$.

A suitable context: a stone held at the top of a tower and released at $t = 0$, with $x$ measured downwards from the point of release. Before $t = 0$ it is at rest; afterwards it falls straight down with acceleration $g$.

Question 2.14

A police van moving on a highway with a speed of $30\ \text{km h}^{-1}$ fires a bullet at a thief’s car speeding away in the same direction with a speed of $192\ \text{km h}^{-1}$. If the muzzle speed of the bullet is $150\ \text{m s}^{-1}$, with what speed does the bullet hit the thief’s car? (Note: Obtain that speed which is relevant for damaging the thief’s car).

Solution. The muzzle speed is the bullet’s speed relative to the gun, and the gun is moving with the van. So relative to the ground the bullet moves at $150\ \text{m s}^{-1}$ plus the van’s speed, $30\ \text{km h}^{-1} = 8.33\ \text{m s}^{-1}$:

$$v_{\text{bullet}} = 150 + 8.33 = 158.33\ \text{m s}^{-1}$$

What damages the car is how fast the bullet approaches it, which is the bullet’s velocity relative to the car. The car moves the same way at $192\ \text{km h}^{-1} = 53.33\ \text{m s}^{-1}$:

$$v_{\text{bullet, car}} = 158.33 – 53.33 = 105\ \text{m s}^{-1}$$

Question 2.15

Suggest a suitable physical situation for each of the following graphs (Fig 2.12):

Question 2.15 (a)

Solution. Read the $x$–$t$ graph piece by piece. At first $x$ is constant at zero: the body is at rest. Then $x$ rises along a straight line, uniform motion in the positive direction, until $A$, where it abruptly reverses. It comes back along a straight line of smaller slope — uniform motion in the negative direction at a lower speed — passes its starting point at $B$, and finally stops at a negative value of $x$ and stays there. A ball fits this: at rest on a smooth floor, kicked towards a wall, rebounding with reduced speed, rolling back past where it started, and stopped by the opposite wall.

Question 2.15 (b)

Solution. Every line has the same negative slope, so between the jumps the acceleration is constant and directed downwards if upwards is positive — free fall under gravity. Along each line the velocity decreases through zero (the top of a flight) and becomes negative (falling). Then it jumps suddenly to a positive value, a little smaller each time: the body has hit the floor and rebounded upwards with less speed than it arrived with.

Question 2.15 (c)

Solution. The acceleration is zero — uniform motion — except for one short, large pulse. That is a large force acting for a very short time, after which the body moves uniformly again with a changed velocity.

Question 2.16

Figure 2.13 gives the $x$-$t$ plot of a particle executing one-dimensional simple harmonic motion. (You will learn about this motion in more detail in Chapter 13). Give the signs of position, velocity and acceleration variables of the particle at $t = 0.3\ \text{s}$, $1.2\ \text{s}$, $-1.2\ \text{s}$.

Solution. The curve crosses the time axis every second, so the period is $2\ \text{s}$, and just after $t = 0$ it dips below the axis. Read each sign off the graph: the sign of $x$ is whether the curve is above or below the axis, and the sign of $v$ is whether it is rising or falling. For the acceleration, simple harmonic motion has $a = -\omega^2x$, so $a$ always has the sign opposite to $x$ — the particle is always pulled back towards $x = 0$.

At $t = 0.3\ \text{s}$ the curve is below the axis and still falling towards its trough at about $0.5\ \text{s}$: $x < 0$, $v < 0$, $a > 0$. At $t = 1.2\ \text{s}$ it has just crossed upwards at $t = 1\ \text{s}$ and is rising towards the crest at $1.5\ \text{s}$: $x > 0$, $v > 0$, $a < 0$. At $t = -1.2\ \text{s}$ it is below the axis, past the trough at $-1.5\ \text{s}$ and rising towards the crossing at $-1\ \text{s}$: $x < 0$, $v > 0$, $a > 0$.

Question 2.17

Figure 2.14 gives the $x$-$t$ plot of a particle in one-dimensional motion. Three different equal intervals of time are shown. In which interval is the average speed greatest, and in which is it the least? Give the sign of average velocity for each interval.

Solution. In an $x$–$t$ graph the average velocity over an interval is the change in $x$ divided by the time, which is the slope of the chord across the interval. Within each shaded interval the particle does not turn back, so the average speed is the magnitude of that slope. Since the three intervals are equal, it is enough to compare how much $x$ changes across each.

In interval 3 the curve falls steeply, so $x$ changes the most — the average speed is greatest — and it falls, so the average velocity is negative. In interval 2, just before the maximum, the curve is nearly flat and $x$ changes very little — the average speed is least — though $x$ still rises slightly, so the average velocity is positive. In interval 1 the curve rises moderately: positive average velocity, and an average speed between the other two.

Question 2.18

Figure 2.15 gives a speed-time graph of a particle in motion along a constant direction. Three equal intervals of time are shown. In which interval is the average acceleration greatest in magnitude? In which interval is the average speed greatest? Choosing the positive direction as the constant direction of motion, give the signs of $v$ and $a$ in the three intervals. What are the accelerations at the points $A$, $B$, $C$ and $D$?

Solution. The motion is along a constant direction, taken as positive, so the velocity equals the speed, and $v > 0$ in all three intervals. The average acceleration over an interval is the change in speed divided by the time — the slope of the chord across it — and since the intervals are equal, it is enough to compare how much the speed changes across each.

Greatest average acceleration in magnitude. In interval 2 the curve falls steeply from $B$ towards $C$, so the speed changes most there. Interval 1 is on a gentle rise, and interval 3 sits on the peak $D$, where the speed barely changes.

Greatest average speed. Interval 3 includes $D$, the highest point of the graph, so the speed throughout it is the largest.

Signs of $a$. In interval 1 the speed is increasing, so $a > 0$; in interval 2 it is decreasing, so $a < 0$. Interval 3 is centred on $D$: the speed rises in its first half and falls in its second, and measured on the figure it ends very slightly higher than it began, so the average acceleration is small and positive, $a > 0$, though much smaller than in the other two intervals.

At $A$, $B$, $C$ and $D$. The tangent to the curve is horizontal at each of these points — $B$ and $D$ are maxima, $C$ a minimum, and the curve is flat at $A$ — so the rate of change of speed, the acceleration, is zero at each.

Common mistakes

• Question 2.2(d): answering “same” without reading the ends of the lines. In an $x$–$t$ graph a child is home where the line ends. In Fig. 2.9 $B$’s line ends at $Q$ before $A$’s reaches $P$, so they arrive at different times — whatever the key says.
• Question 2.4: dividing $13\ \text{m}$ by $2\ \text{m}$ per cycle. That gives $52\ \text{s}$, but the drunkard falls in the first time he reaches $13\ \text{m}$, partway through a forward run: after four cycles he is at $8\ \text{m}$, and five more steps take him in at $37\ \text{s}$.
• Question 2.5: putting $126$ straight into the equations. The distance is in metres, so the speed must be in metres per second first: $126\ \text{km h}^{-1} = 35\ \text{m s}^{-1}$. Without the conversion the answer is out by a factor of about $13$.
• Question 2.6(b): saying the acceleration is zero at the top. The velocity is zero there, but gravity has not switched off. With zero acceleration the ball would stay at the top for ever.
• Question 2.7(d): reading “positive acceleration” as “speeding up”. The sign of $a$ depends on which direction was chosen as positive. A body speeds up only when $v$ and $a$ have the same sign; the rising ball of Question 2.6(c) has $a > 0$ and is slowing down.
• Question 2.10(ii): averaging the two speeds. $(5 + 7.5)/2 = 6.25\ \text{km h}^{-1}$ is wrong because the man spends $30$ min at the slower speed and only $20$ at the faster. Total path over total time gives $6\ \text{km h}^{-1}$. The opposite slip is to give zero, which is the average velocity.
• Question 2.13: reading the parabola as the path. An $x$–$t$ graph plots position against time; the particle itself moves along one straight line throughout, and the curve only says it is accelerating uniformly from rest.
• Question 2.14: using $150\ \text{m s}^{-1}$ as the bullet’s ground speed. The muzzle speed is relative to the gun, which is moving with the van, so the van’s $8.33\ \text{m s}^{-1}$ is added first; the damage then depends on the speed relative to the car, $105\ \text{m s}^{-1}$.

Practise next

• Chapter 3, Motion in a Plane — the same velocity, acceleration and relative-velocity ideas extended to two dimensions with vectors, including projectiles.
• Chapter 1, Units and Measurement — the unit conversions (km h⁻¹ to m s⁻¹) and significant figures that every numerical answer here relies on.