NCERT Solutions Class 9 Science Chapter 4 — Describing Motion Around Us

NCERT Class 9 Science — Describing Motion Around Us, Chapter 4, Revise, Reflect, Refine. All 16 questions solved.

Also here. Solutions to the in-text questions for this chapter and the activities for this chapter are on their own pages.

These are the end-of-chapter questions of Chapter 4 in Exploration, the new NCERT science textbook for Class 9 (2026). Questions 1, 2 and 16 separate distance from displacement. Questions 3 and 11 ask what acceleration and rest really mean. Questions 4, 5, 8, 9, 10, 14 and 15 are calculations with the kinematic equations, and Questions 6, 7 and 12 to 15 read or draw position–time and velocity–time graphs. Everything rests on three sets of ideas from the chapter:

• Speed and velocity — average speed $=\dfrac{\text{total distance travelled}}{\text{time interval}}$; average velocity $v_{av} = \dfrac{s}{t}$, where $s$ is the displacement, the change in position, with its direction.
• Acceleration and the kinematic equations — average acceleration $a = \dfrac{v-u}{t}$; for constant acceleration, $v = u + at$, $s = ut + \tfrac{1}{2}at^2$ and $v^2 = u^2 + 2as$.
• Reading graphs — the slope of a position–time graph gives the velocity; the slope of a velocity–time graph gives the acceleration, and the area between a velocity–time graph and the time axis gives the displacement.

Throughout, the direction of motion is taken as positive, so a minus sign on an acceleration means it acts opposite to the velocity, as in the chapter’s Example 4.3. Speeds in $\text{km h}^{-1}$ are changed to $\text{m s}^{-1}$ before they go into an equation with seconds, using $1\ \text{km h}^{-1} = \dfrac{1000\ \text{m}}{3600\ \text{s}}$.

Key insight. Keep two pairs of quantities apart. Distance and speed describe the whole path, every metre of it, whatever the direction; displacement and velocity describe only the change in position from the start to the end, with its direction. The father who walks to the shop and back twice (Question 1), the student who goes up and then partly down (Question 2), and the tip of a clock hand that goes round one and a half times (Question 16) all travel much further than their displacement. On a velocity–time graph the two pairs meet: the area under the graph is the displacement, and when the object never turns back it is also the distance (Questions 12 to 15).

Question 1

My father went to a shop from home which is located at a distance of $250\ \text{m}$ on a straight road. On reaching there, he discovered that he forgot to carry a cloth bag. He came home to take it, went to the shop again, bought provisions and came back home. How much was the total distance travelled by him? What was his displacement from home?

Solution. Distance counts the full length of the path, whichever way he was walking. He made four trips along the same $250\ \text{m}$ of road: home to shop, shop to home, home to shop again, and shop to home again.

$$\text{distance} = 250 + 250 + 250 + 250 = 4\times250 = 1000\ \text{m}$$

Displacement is only the change in position between the start and the end. Take home as the reference point. He started at home and finished at home, so his position did not change at all overall.

$$\text{displacement} = \text{final position} – \text{initial position} = 0\ \text{m} – 0\ \text{m} = 0\ \text{m}$$

Question 2

A student runs from the ground floor to the fourth floor of a school building to collect a book and then comes down to their classroom on the second floor. If the height of each floor is $3\ \text{m}$, find: (i) the total vertical distance travelled, and (ii) their displacement from the starting point.

Solution. Take the ground floor as the reference point and upward as positive. The fourth floor is four floor-heights above the ground floor, so it is $4\times3 = 12\ \text{m}$ up. The second floor is two floor-heights up, $2\times3 = 6\ \text{m}$.

Question 2 (i)

The total vertical distance travelled.

Solution. Going up from the ground floor to the fourth floor covers $12\ \text{m}$. Coming down from the fourth floor ($12\ \text{m}$) to the second floor ($6\ \text{m}$) covers another $12 – 6 = 6\ \text{m}$. Distance adds up both parts, since both were actually travelled:

$$\text{distance} = 12\ \text{m} + 6\ \text{m} = 18\ \text{m}$$

Question 2 (ii)

Their displacement from the starting point.

Solution. Displacement ignores the trip up and down in between and compares only where the student started and where they finished. They started at the ground floor, position $0\ \text{m}$, and finished on the second floor, position $+6\ \text{m}$:

$$\text{displacement} = 6\ \text{m} – 0\ \text{m} = +6\ \text{m}$$

The plus sign means the displacement is upward.

Question 3

A girl is riding her scooter and finds that its speedometer reading is constant. Is it possible for her scooter to be accelerating and if so, how?

Solution. A speedometer shows only the speed, that is, the magnitude of the velocity. Velocity has a direction as well. Acceleration is the rate of change of velocity, and the velocity changes if its magnitude changes, or its direction changes, or both. So a constant speedometer reading only tells us that the magnitude is not changing.

If she rides along a straight road with a steady reading, neither the magnitude nor the direction of her velocity changes, and she is not accelerating. But if she rides round a bend, or round a circular roundabout, at a steady reading, the direction of her velocity keeps changing. Her velocity is then changing, so she is accelerating even though her speed is not. This is exactly the case of uniform circular motion in the chapter, which is accelerated motion at constant speed.

Question 4

A car starts from rest and its velocity reaches $24\ \text{m s}^{-1}$ in $6\ \text{s}$. Find the average acceleration and the distance travelled in these $6\ \text{s}$.

Solution. “Starts from rest” tells us the initial velocity: $u = 0\ \text{m s}^{-1}$. The final velocity is $v = 24\ \text{m s}^{-1}$ and the time is $t = 6\ \text{s}$. Average acceleration is the change in velocity divided by the time taken:

$$a = \frac{v-u}{t} = \frac{24\ \text{m s}^{-1} – 0\ \text{m s}^{-1}}{6\ \text{s}} = 4\ \text{m s}^{-2}$$

It is positive, so it acts in the direction of the velocity, as it must when the car is speeding up. For the distance, use $s = ut + \tfrac{1}{2}at^2$, treating the acceleration as constant:

$$s = 0\times6 + \frac{1}{2}\times4\times6^2 = \frac{1}{2}\times4\times36 = 72\ \text{m}$$

The car moves in one direction only, so this displacement is also the distance travelled.

Question 5

A motorbike moving with initial velocity $28\ \text{m s}^{-1}$ and constant acceleration stops after travelling $98\ \text{m}$. Find the acceleration of the motorbike and the time taken to come to a stop.

Solution. We know $u = 28\ \text{m s}^{-1}$, the final velocity $v = 0$ (it stops) and $s = 98\ \text{m}$; the time is not given. The equation that links $u$, $v$, $a$ and $s$ without $t$ is $v^2 = u^2 + 2as$:

$$0^2 = 28^2 + 2\times a\times98$$

$$0 = 784 + 196\,a \qquad\Rightarrow\qquad a = -\frac{784}{196} = -4\ \text{m s}^{-2}$$

The minus sign shows that the acceleration acts opposite to the velocity, which is why the motorbike slows down. Now $v = u + at$ gives the time:

$$0 = 28 + (-4)\times t \qquad\Rightarrow\qquad t = \frac{28}{4} = 7\ \text{s}$$

Question 6

Fig. 4.27 shows a position-time graph of two objects A and B that are moving along the parallel tracks in the same direction. Do objects A and B ever have equal velocity? Justify your answer.

Solution. On a position–time graph the velocity is given by the slope, how steeply the line rises: the steeper the line, the faster the position changes. So the question is whether the two lines are ever equally steep.

The position axis has no numbers, so measure in grid squares. A starts at the origin and rises $4$ squares in the first $5\ \text{s}$, a slope of $4/5 = 0.8$ squares per second. B starts $1.6$ squares up and rises to the same point, $4$ squares, in the same $5\ \text{s}$, a rise of only $2.4$ squares, a slope of $2.4/5 = 0.48$ squares per second. After $5\ \text{s}$ B’s line is flatter still. A’s line is steeper than B’s at every instant, so A is always faster and their velocities are never equal.

The two lines do cross at $t = 5\ \text{s}$. A crossing means both objects are at the same position at that instant — A, which started behind, draws level with B and then moves ahead. It says nothing about their velocities being equal.

Question 7

A graph in Fig. 4.28 shows the change in position with time for two objects A and B moving in a straight line from 0 to 10 seconds. Choose the correct option(s).

(i) The average velocity of both over the $10\ \text{s}$ time interval is equal since they have the same initial and final positions.

(ii) The average speeds of both over the $10\ \text{s}$ time interval are equal since both cover equal distance in equal time.

(iii) The average speed of A over the $10\ \text{s}$ time interval is lower than that of B since it covers a shorter distance than B in 10 seconds.

(iv) The average speed of A over the $10\ \text{s}$ time interval is greater than that of B since B’s speed is lower than A’s in some segments.

Solution. Read the two things the graph gives us: where each object starts and ends, and whether either ever turns back. Both graphs start at the origin at $0\ \text{s}$ and end at the same top point at $10\ \text{s}$, so A and B have the same initial and the same final position. Both graphs also rise all the way — neither ever slopes downwards — so each object keeps moving in one direction and never comes back.

Statement (i). Average velocity is displacement divided by the time interval. Equal starting and finishing positions give equal displacements, and the time is $10\ \text{s}$ for both, so the average velocities are equal. Statement (i) is correct.

Statement (ii). For an object moving in a straight line in one direction only, the distance travelled equals the magnitude of the displacement. Both objects move in one direction and have equal displacements, so they cover equal distances in the same $10\ \text{s}$, and their average speeds are equal. Statement (ii) is correct.

Statement (iii). A does not cover a shorter distance; the distances are equal. Statement (iii) is wrong.

Statement (iv). B is indeed slower than A for the first part of the interval: in grid squares, B covers $1$ square in the first $4\ \text{s}$ and $2.2$ more by $7.6\ \text{s}$, while A covers $0.6$ squares every second. But in the last $2.4\ \text{s}$ B’s line is the steepest on the graph: B covers $2.8$ squares and catches up exactly. Being slower in some segments does not lower B’s average, because it is faster in others. Statement (iv) is wrong.

This question has more than one correct option: (i) and (ii).

Question 8

A truck driver driving at the speed of $54\ \text{km h}^{-1}$ notices a road sign with a speed limit of $40\ \text{km h}^{-1}$ (Fig. 4.29) for trucks. He slows down to $36\ \text{km h}^{-1}$ in $36\ \text{s}$. What was the distance travelled by him during this time? Assume the acceleration to be constant while slowing down.

Solution. The time is in seconds, so first change both speeds to $\text{m s}^{-1}$:

$$u = 54\ \text{km h}^{-1} = 54\times\frac{1000\ \text{m}}{3600\ \text{s}} = 15\ \text{m s}^{-1}, \qquad v = 36\ \text{km h}^{-1} = 36\times\frac{1000\ \text{m}}{3600\ \text{s}} = 10\ \text{m s}^{-1}$$

Next find the acceleration, since the distance equation needs it:

$$a = \frac{v-u}{t} = \frac{10 – 15}{36} = -\frac{5}{36}\ \text{m s}^{-2}$$

The minus sign shows that it acts against the motion, slowing the truck. Now $s = ut + \tfrac{1}{2}at^2$ with $t = 36\ \text{s}$:

$$s = 15\times36 + \frac{1}{2}\times\left(-\frac{5}{36}\right)\times36^2 = 540 – \frac{1}{2}\times5\times36 = 540 – 90 = 450\ \text{m}$$

The truck never reverses, so this is the distance travelled. Its new speed, $36\ \text{km h}^{-1}$, is below the $40\ \text{km h}^{-1}$ limit.

Question 9

A car starts from rest and accelerates uniformly to $20\ \text{m s}^{-1}$ in $5$ seconds. It then travels at $20\ \text{m s}^{-1}$ for $10$ seconds and finally applies the brake (with uniform acceleration) to stop in $6$ seconds. Find the total distance travelled.

Solution. The motion has three stages with different accelerations, so find the distance in each stage separately and add them.

Stage 1, speeding up. $u = 0$, $v = 20\ \text{m s}^{-1}$, $t = 5\ \text{s}$, so $a = \dfrac{20 – 0}{5} = 4\ \text{m s}^{-2}$ and

$$s_1 = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2}\times4\times5^2 = 50\ \text{m}$$

Stage 2, steady speed. There is no acceleration, so distance $=$ speed $\times$ time:

$$s_2 = 20\times10 = 200\ \text{m}$$

Stage 3, braking. $u = 20\ \text{m s}^{-1}$, $v = 0$, $t = 6\ \text{s}$, so $a = \dfrac{0 – 20}{6} = -\dfrac{10}{3}\ \text{m s}^{-2}$ and

$$s_3 = 20\times6 + \frac{1}{2}\times\left(-\frac{10}{3}\right)\times6^2 = 120 – 60 = 60\ \text{m}$$

$$\text{total distance} = 50 + 200 + 60 = 310\ \text{m}$$

The same answer comes from the area under the velocity–time graph: a triangle $\tfrac{1}{2}\times5\times20 = 50\ \text{m}$, a rectangle $10\times20 = 200\ \text{m}$ and a triangle $\tfrac{1}{2}\times6\times20 = 60\ \text{m}$.

Question 10

A bus is travelling at $36\ \text{km h}^{-1}$ when the driver sees an obstacle $30\ \text{m}$ ahead. The driver takes $0.5$ seconds to react before pressing the brake. Once the brake is applied, the velocity of the bus reduces with constant acceleration of $2.5\ \text{m s}^{-2}$. Will the bus be able to stop before reaching the obstacle?

Solution. First change the speed to $\text{m s}^{-1}$: $36\ \text{km h}^{-1} = 36\times\dfrac{1000}{3600} = 10\ \text{m s}^{-1}$. The bus covers ground in two parts.

While the driver reacts. For $0.5\ \text{s}$ nothing has been pressed yet, so the bus keeps its full speed:

$$\text{reaction distance} = 10\ \text{m s}^{-1}\times0.5\ \text{s} = 5\ \text{m}$$

While braking. Now $u = 10\ \text{m s}^{-1}$, $v = 0$, and since the velocity reduces, the acceleration is opposite to the motion: $a = -2.5\ \text{m s}^{-2}$. The time is not needed, so use $v^2 = u^2 + 2as$:

$$0 = 10^2 + 2\times(-2.5)\times s \qquad\Rightarrow\qquad 0 = 100 – 5s \qquad\Rightarrow\qquad s = 20\ \text{m}$$

$$\text{stopping distance} = 5\ \text{m} + 20\ \text{m} = 25\ \text{m}$$

$25\ \text{m}$ is less than $30\ \text{m}$, so the bus stops $5\ \text{m}$ short of the obstacle.

Question 11

A student said, “The Earth moves around the Sun”. In this context, discuss whether an object kept on the Earth can be considered to be at rest.

Solution. The chapter defines rest and motion only with respect to a reference point: an object is at rest if its position with respect to the reference point does not change with time, and in motion if it does. So “Is it at rest?” has no answer until we say what the reference point is.

Take a book lying on a table. With respect to a reference point on the Earth, such as a corner of the room, the book’s distance and direction stay the same as time passes, so it is at rest. With respect to the Sun, the book is carried along with the Earth as the Earth goes round the Sun, so its position keeps changing and it is in motion. Both statements are true at the same time; they simply use different reference points. When we say everyday objects are at rest, we are, without saying so, using a reference point on the Earth.

Question 12

The velocity-time graph from $0\ \text{s}$ to $120\ \text{s}$ for a cyclist is shown in Fig. 4.30. Shade the areas (in different colours) representing the displacement of the cyclist (i) while cyclist is moving with constant velocity, (ii) when the velocity of cyclist is decreasing. Also, calculate the displacement and average acceleration in the $120\ \text{s}$ time interval.

Solution. Read the graph first. The velocity rises steadily from $0$ at $0\ \text{s}$ to $3\ \text{m s}^{-1}$ at $20\ \text{s}$, stays at $3\ \text{m s}^{-1}$ from $20\ \text{s}$ to $100\ \text{s}$, and then falls steadily to $2\ \text{m s}^{-1}$ at $120\ \text{s}$. The displacement in any time interval is the area between the graph and the time axis over that interval.

Question 12 (i)

Displacement while the cyclist moves with constant velocity.

Solution. The velocity is constant where the graph is flat, from $20\ \text{s}$ to $100\ \text{s}$. The area under this part is the rectangle shaded blue in the figure:

$$\text{area} = \text{velocity}\times\text{time interval} = 3\ \text{m s}^{-1}\times(100 – 20)\ \text{s} = 3\times80 = 240\ \text{m}$$

Question 12 (ii)

Displacement while the velocity is decreasing.

Solution. The velocity decreases from $100\ \text{s}$ to $120\ \text{s}$. The area under this part, shaded red, is a trapezium. Split it, as the chapter does, into a rectangle of height $2\ \text{m s}^{-1}$ and a triangle on top of it of height $3 – 2 = 1\ \text{m s}^{-1}$, both $20\ \text{s}$ wide:

$$\text{area} = 2\times20 + \frac{1}{2}\times20\times1 = 40 + 10 = 50\ \text{m}$$

Question 12, displacement and average acceleration

Displacement and average acceleration over the whole $120\ \text{s}$.

Solution. The total displacement is the whole area under the graph: the triangle for the first $20\ \text{s}$ plus the two areas already found.

$$\text{triangle} = \frac{1}{2}\times20\times3 = 30\ \text{m}$$

$$\text{displacement} = 30 + 240 + 50 = 320\ \text{m}$$

Average acceleration uses only the velocities at the start and end of the interval, not what happens in between. At $0\ \text{s}$ the velocity is $u = 0$; at $120\ \text{s}$ it is $v = 2\ \text{m s}^{-1}$.

$$a = \frac{v-u}{t} = \frac{2 – 0}{120} = \frac{1}{60}\ \text{m s}^{-2} \approx 0.017\ \text{m s}^{-2}$$

It is positive, so it is in the direction of motion: overall, over the $120\ \text{s}$, the cyclist has gained speed.

Question 13

A girl is preparing for her first marathon by running on a straight road. She uses a smartwatch to calculate her running speed at different intervals. The graph (Fig. 4.31) depicts her velocity versus time. Estimate the distance she ran based on the graph.

Solution. The area under a velocity–time graph gives the displacement. She runs along a straight road and her velocity is positive throughout, so she never turns back, and the displacement is also the distance she ran. Because the velocity is in $\text{km h}^{-1}$ and the time in hours, the area comes out directly in kilometres.

Read the corners off the grid. Each small square is $0.2\ \text{h}$ wide and $0.25\ \text{km h}^{-1}$ tall, and the corners fall at $(0\ \text{h}, 7.0)$, $(0.6, 7.0)$, $(1.6, 7.5)$, $(3.0, 7.5)$, $(4.6, 7.0)$, $(5.6, 6.5)$ and $(6.6, 6.5)$. Dashed lines through the corners divide the area into six strips. A flat strip is a rectangle; a sloping strip is a rectangle with a triangle on top, as in the chapter.

$$\begin{array}{c|c|l|c} \text{Strip} & \text{Time} & \text{Area} & \text{km} \\ \hline 1 & 0 \text{ to } 0.6\ \text{h} & 7.0\times0.6 & 4.2 \\ 2 & 0.6 \text{ to } 1.6\ \text{h} & 7.0\times1.0 + \tfrac{1}{2}\times1.0\times0.5 & 7.25 \\ 3 & 1.6 \text{ to } 3.0\ \text{h} & 7.5\times1.4 & 10.5 \\ 4 & 3.0 \text{ to } 4.6\ \text{h} & 7.0\times1.6 + \tfrac{1}{2}\times1.6\times0.5 & 11.6 \\ 5 & 4.6 \text{ to } 5.6\ \text{h} & 6.5\times1.0 + \tfrac{1}{2}\times1.0\times0.5 & 6.75 \\ 6 & 5.6 \text{ to } 6.6\ \text{h} & 6.5\times1.0 & 6.5 \end{array}$$

$$\text{distance} = 4.2 + 7.25 + 10.5 + 11.6 + 6.75 + 6.5 = 46.8\ \text{km}$$

Since the corners are read off a graph, this is an estimate: about $47\ \text{km}$ in $6.6$ hours. As a rough check, her velocity stays between $6.5$ and $7.5\ \text{km h}^{-1}$, and $7\ \text{km h}^{-1}\times6.6\ \text{h} = 46.2\ \text{km}$, close to the area found.

Question 14

On entering a state highway, a car continues to move with a constant velocity of $6\ \text{m s}^{-1}$ for $2$ minutes and then accelerates with a constant acceleration $1\ \text{m s}^{-2}$ for $6$ seconds. Find the displacement of the car on the state highway in the $2\ \text{min}\ 6\ \text{s}$ time interval by drawing a velocity-time graph for its motion.

Solution. First put all the times in seconds: $2\ \text{min} = 120\ \text{s}$, so the whole interval runs from $0$ to $126\ \text{s}$. To draw the graph we also need the velocity at the end. During the last $6\ \text{s}$, $u = 6\ \text{m s}^{-1}$ and $a = 1\ \text{m s}^{-2}$, so

$$v = u + at = 6 + 1\times6 = 12\ \text{m s}^{-1}$$

The graph is a horizontal line at $6\ \text{m s}^{-1}$ from $0$ to $120\ \text{s}$, then a straight line rising to $12\ \text{m s}^{-1}$ at $126\ \text{s}$.

The displacement is the area under the graph. For the first $120\ \text{s}$ it is the blue rectangle:

$$6\times120 = 720\ \text{m}$$

For the last $6\ \text{s}$ it is the narrow red trapezium: a rectangle of height $6\ \text{m s}^{-1}$ with a triangle of height $12 – 6 = 6\ \text{m s}^{-1}$ on top.

$$6\times6 + \frac{1}{2}\times6\times6 = 36 + 18 = 54\ \text{m}$$

$$\text{displacement} = 720 + 54 = 774\ \text{m}$$

As a check, $s = ut + \tfrac{1}{2}at^2 = 6\times6 + \tfrac{1}{2}\times1\times6^2 = 54\ \text{m}$ for the last part agrees.

Question 15

Two cars A and B start moving with a constant acceleration from rest, in a straight line. Car A attains a velocity of $5\ \text{m s}^{-1}$ in $5\ \text{s}$. Car B attains a velocity of $3\ \text{m s}^{-1}$ in $10\ \text{s}$. Plot the velocity-time graphs for both the cars in the same graph. Using the graph, calculate the displacement in the two time intervals mentioned. (Hint: Calculate the acceleration in both cases. Then calculate their velocities at five instants of time to plot the graph.)

Solution. Both cars start from rest, so $u = 0$. The accelerations are

$$a_A = \frac{5 – 0}{5} = 1\ \text{m s}^{-2}, \qquad a_B = \frac{3 – 0}{10} = 0.3\ \text{m s}^{-2}$$

With $u = 0$, $v = u + at$ becomes $v = at$, which gives the velocity at any instant. Five instants for each car:

$$\begin{array}{l|ccccc} t\ (\text{s}) \text{ for A} & 1 & 2 & 3 & 4 & 5 \\ \hline v_A\ (\text{m s}^{-1}) & 1 & 2 & 3 & 4 & 5 \end{array} \qquad \begin{array}{l|ccccc} t\ (\text{s}) \text{ for B} & 2 & 4 & 6 & 8 & 10 \\ \hline v_B\ (\text{m s}^{-1}) & 0.6 & 1.2 & 1.8 & 2.4 & 3.0 \end{array}$$

Plotting these, together with the origin, gives two straight lines. A’s is steeper, because A has the larger acceleration.

The displacement in each interval is the area of the triangle under that car’s line.

$$s_A = \frac{1}{2}\times5\ \text{s}\times5\ \text{m s}^{-1} = 12.5\ \text{m} \qquad (0 \text{ to } 5\ \text{s})$$

$$s_B = \frac{1}{2}\times10\ \text{s}\times3\ \text{m s}^{-1} = 15\ \text{m} \qquad (0 \text{ to } 10\ \text{s})$$

B covers more ground despite its smaller acceleration because it accelerates for twice as long. The equation $s = \tfrac{1}{2}at^2$ agrees: $\tfrac{1}{2}\times1\times25 = 12.5\ \text{m}$ and $\tfrac{1}{2}\times0.3\times100 = 15\ \text{m}$.

Question 16

Rohan studies science from 6 PM to 7:30 PM at home. Consider the tip of the minute’s hand of the wall clock. During the given time interval, what is its: (i) distance travelled, (ii) displacement, (iii) speed, and (iv) velocity. The length of the minute’s hand is $7\ \text{cm}$ (Fig. 4.32).

Solution. The tip of the minute hand moves on a circle of radius $R = 7\ \text{cm}$, the length of the hand, and goes round once every hour. From 6 PM to 7:30 PM is $1\tfrac{1}{2}$ hours, so the tip makes $1\tfrac{1}{2}$ revolutions. At 6 PM the minute hand points to 12 (its tip is at $P$); at 7:30 PM it points to 6 (its tip is at $Q$). The time interval in seconds is $1.5\times3600 = 5400\ \text{s}$. Take $\pi = \dfrac{22}{7}$.

Question 16 (i)

Distance travelled.

Solution. One revolution covers the circumference, $2\pi R$. One and a half revolutions cover one and a half times that:

$$\text{distance} = 1.5\times2\pi R = 1.5\times2\times\frac{22}{7}\times7 = 1.5\times44 = 66\ \text{cm}$$

Question 16 (ii)

Displacement.

Solution. Displacement is the straight line from where the tip starts, $P$ at the 12, to where it ends, $Q$ at the 6. These two points are at opposite ends of a diameter of the circle, so

$$PQ = 2R = 2\times7 = 14\ \text{cm}$$

in the direction from the 12 towards the 6, straight down the face of the clock. The complete first revolution adds nothing to the displacement, because it brings the tip back to $P$; only the extra half-turn counts.

Question 16 (iii)

Speed.

Solution. Average speed is total distance divided by the time interval:

$$\text{speed} = \frac{66\ \text{cm}}{5400\ \text{s}} = \frac{11}{900}\ \text{cm s}^{-1} \approx 0.012\ \text{cm s}^{-1}$$

The tip moves with uniform circular motion, so its speed is the same at every instant, and this value is also its speed at any moment: $2\pi R$ in one hour gives $44\ \text{cm}$ in $3600\ \text{s}$, which is again $\dfrac{11}{900}\ \text{cm s}^{-1}$.

Question 16 (iv)

Velocity.

Solution. Over the interval, the average velocity is displacement divided by the time interval:

$$v_{av} = \frac{14\ \text{cm}}{5400\ \text{s}} = \frac{7}{2700}\ \text{cm s}^{-1} \approx 0.0026\ \text{cm s}^{-1}$$

directed like the displacement, from the 12 towards the 6. It is much smaller than the speed because the tip travels $66\ \text{cm}$ of curved path but ends only $14\ \text{cm}$ from where it began. At any single instant the tip’s velocity points along the tangent to the circle, and its direction keeps changing.

Common mistakes

• Question 1: counting only one round trip, or giving the distance as the displacement. He walks the $250\ \text{m}$ road four times, so the distance is $1000\ \text{m}$, not $500\ \text{m}$. And because he ends at home, the displacement is zero however far he walked; students who write $1000\ \text{m}$ for both have treated displacement as another word for distance.
• Question 2: miscounting the floors. The fourth floor is four floor-heights above the ground floor ($12\ \text{m}$), not five; the ground floor itself is floor zero. Students then also give the displacement as $18\ \text{m}$ or as “$6\ \text{m}$ down”; it is $6\ \text{m}$ up, because the second floor is above the starting point.
• Question 5: dropping the minus sign. Writing $a = 4\ \text{m s}^{-2}$ and then using $v = u + at$ gives $0 = 28 + 4t$, a negative time. The motorbike is slowing down, so its acceleration is opposite to its velocity, $a = -4\ \text{m s}^{-2}$, and the time comes out as a sensible $7\ \text{s}$.
• Question 6: reading the crossing point as “equal velocity”. Where two position–time lines cross, the objects are at the same position. Velocity is the slope, and A’s line is steeper everywhere, so the velocities are never equal.
• Question 7: rejecting option (ii) because B’s graph “is longer” or bent. The graph is not the path the object travels; it is position plotted against time. Both objects move in one direction only, so each distance equals the displacement, and the displacements are equal.
• Questions 8 and 10: leaving the speed in $\text{km h}^{-1}$. With the time in seconds, $36\ \text{km h}^{-1}$ must first become $10\ \text{m s}^{-1}$. Mixing units in Question 10 gives a braking distance of $259.2\ \text{m}$, which is nonsense.
• Question 10: forgetting the reaction distance. During the $0.5\ \text{s}$ before the brake is pressed, the bus is still moving at full speed and covers $5\ \text{m}$. Leaving this out gives $20\ \text{m}$, which happens to give the same “yes” here but is the wrong stopping distance.
• Question 12: averaging over only part of the motion. Average acceleration over $120\ \text{s}$ uses only the velocities at $0\ \text{s}$ and $120\ \text{s}$: $(2 – 0)/120$. Students who use the braking part, $(2 – 3)/20$, or the first $20\ \text{s}$, $3/20$, are answering a different question.
• Question 16: thinking the displacement is zero. One full revolution would bring the tip back to where it began, but $1\tfrac{1}{2}$ hours is one and a half revolutions, so the tip ends at the opposite side of the circle, $14\ \text{cm}$ away — a diameter, not the $7\ \text{cm}$ radius.

Practise next

• The in-text Pause and Ponder questions of this chapter, five across two boxes, which turn on the one distinction the end-of-chapter questions assume you have settled: total distance against displacement, and the average speed that follows from each.
• Chapter 6, How Forces Affect Motion — the next physics chapter explains why velocity changes, using the same acceleration and the same kinematic equations met here for braking and speeding up.
• Chapter 7, Work, Energy, and Simple Machines — the speeds calculated in this chapter become kinetic energy, and the stopping distances of Questions 5 and 10 reappear as work done by a braking force.