These questions appear inside the chapter itself, not at the end. The book page number is shown against each one so you can find it while reading.
In the example of an athlete running back and forth on a straight track (Fig. 4.4), when will the displacement of the athlete be zero? What will be the total distance travelled in that case?
In this example the athlete starts at the origin O, runs 100 m forward to point A, and then turns and runs back along the same track.
Displacement is zero when the athlete comes back to the starting point O. Displacement is the net change in position between the start and the finish. It becomes zero only when the finishing position is the same as the starting position — that is, when the athlete returns exactly to O, the point from which she started.
Total distance travelled in that case. To return to O, the athlete first runs forward from O to A, a distance of 100 m, and then runs back from A to O, another 100 m. So the total distance travelled is
total distance = OA + AO = 100 m + 100 m = 200 m.
Thus, even though the displacement is zero, the athlete has actually run a total distance of 200 m. This shows clearly that distance travelled and displacement are two different quantities: distance depends on the whole path covered, while displacement depends only on the starting and ending positions.
Fuel used up in a vehicle depends on which of the following? Justify your answer.
Fuel used up depends on (i) the total distance travelled.
The fuel a vehicle burns depends on how much the engine has to work, which in turn depends on the actual length of the path covered by the wheels — that is, the total distance travelled. The more road the vehicle covers, the more fuel it uses, no matter in which direction it goes.
Why not displacement? Displacement is only the straight-line gap between the start and the end, and it takes no account of the real path. For example, if a car drives 5 km away from home and then drives 5 km back home, its displacement is zero — but the engine has run for the full 10 km and has clearly burnt fuel for the whole journey. Since the fuel is consumed over the entire 10 km path and not over the zero displacement, fuel consumption must depend on the total distance travelled, not on the displacement.
A ball rolls down an inclined track as shown in Fig. 4.6. Is its motion a straight-line motion? Assuming the starting point of the ball (O) to be the origin, can its motion from O to D be depicted using a horizontal line as shown in Fig. 4.3? Are the values of total distance travelled and magnitude of displacement from O equal at positions A, B, C and D?
The ball rolls down a straight inclined track, passing in order through the points O, A, B, C and D without ever turning back.
Is it straight-line motion? Yes. The track is straight (even though it is tilted), so the ball moves along a single straight line. Motion along a straight line is called straight-line, or linear, motion — the fact that the line is sloping rather than horizontal does not change this.
Can the motion O to D be shown on a horizontal line like Fig. 4.3? Yes. To describe motion in one dimension we only need to show how far the object is from the origin along its line of travel. It does not matter whether the real track is horizontal or inclined — we can still mark the origin O and the successive positions A, B, C and D as points on a single straight (horizontal) line, exactly as in Fig. 4.3, with distances measured along the track.
Are total distance and magnitude of displacement equal at A, B, C and D? Yes, they are equal at every one of these positions. This is because the ball moves in one direction only (from O towards D) and never turns back. When an object moves without reversing direction, the length of the path it has covered (distance) is exactly the same as the straight-line gap from the origin (magnitude of displacement). So at each of A, B, C and D, the total distance travelled from O equals the magnitude of the displacement from O.
During a family road trip, you drive 200 km north in three hours. Afterwards, you drive 200 km south in two hours. Find the average speed and average velocity for your entire trip.
Given: 200 km north in 3 h, then 200 km south in 2 h.
Total distance travelled = 200 km + 200 km = 400 km.
Total time taken = 3 h + 2 h = 5 h.
Average speed = total distance ÷ total time = 400 km ÷ 5 h = 80 km h⁻¹.
Displacement. You first move 200 km north and then 200 km south, ending up exactly where you started. The 200 km north and the 200 km south cancel out, so the net change in position is
displacement = 200 km (north) − 200 km (south) = 0 km.
Average velocity = displacement ÷ total time = 0 km ÷ 5 h = 0 km h⁻¹.
So the trip has an average speed of 80 km h⁻¹ but an average velocity of zero, because the car returns to its starting point.
Under what condition(s) is the
(i) When the magnitude of average velocity equals the average speed. Average speed is (total distance ÷ time) and the magnitude of average velocity is (magnitude of displacement ÷ time). These two are equal only when the total distance travelled is equal to the magnitude of the displacement. That happens when the object moves along a straight line in one direction only, without turning back. In such motion the path length and the straight-line displacement are the same, so the average speed and the magnitude of average velocity come out equal.
(ii) When the magnitude of average velocity is zero but the average speed is not. The magnitude of average velocity is zero when the displacement is zero, which happens when the object returns to its starting point. But if it has actually moved along some path to get back, it has covered a real distance, so the average speed is not zero. For example, an athlete who runs once around a circular track and stops at the start has zero displacement (zero average velocity) but a non-zero average speed, because a real distance was covered.
My father went to a shop from home which is located at a distance of 250 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?
Tracing the journey. Each trip between home and the shop is 250 m. Father made the following moves:
Total distance travelled = 250 + 250 + 250 + 250 = 4 × 250 m = 1000 m (1 km).
Displacement from home. His journey begins at home and ends at home, so his final position is exactly the same as his starting position. The net change in position is therefore zero.
Displacement = 0 m.
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 m, find:
Setting up the heights. Each floor is 3 m high. Taking the ground floor as the start:
Going up from the ground floor to the fourth floor means climbing 4 floors: 4 × 3 m = 12 m upward.
Then coming down from the fourth floor to the second floor means descending 2 floors: 2 × 3 m = 6 m downward.
(i) Total vertical distance travelled = distance up + distance down = 12 m + 6 m = 18 m.
(ii) Displacement from the starting point. The student starts at the ground floor and finally stops at the second floor. The straight-line gap between these is 2 floors above the ground: 2 × 3 m = 6 m, directed upward.
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?
Yes, the scooter can still be accelerating.
A speedometer shows only the speed of the scooter — that is, the magnitude of its velocity. It says nothing about the direction of motion. Acceleration is the rate of change of velocity, and velocity changes if either its magnitude or its direction changes.
So even if the speedometer reading stays constant, the scooter is accelerating whenever it changes direction — for example, while going round a bend or turning along a curved road. During such a turn the speed is unchanged but the direction of the velocity keeps changing, and this continuous change of direction is itself an acceleration. (If the girl rides in a perfectly straight line at a constant speedometer reading, then the velocity is truly constant and the acceleration is zero.)
A car starts from rest and its velocity reaches 24 m s⁻¹ in 6 s. Find the average acceleration and the distance travelled in these 6 s.
Given: initial velocity u = 0 (starts from rest), final velocity v = 24 m s⁻¹, time t = 6 s.
Average acceleration. Using a = (v − u) ÷ t:
a = (24 − 0) ÷ 6 = 24 ÷ 6 = 4 m s⁻².
Distance travelled. Using s = ut + ½ a t²:
s = (0 × 6) + ½ × 4 × (6)² = 0 + ½ × 4 × 36 = 72 m.
(The same answer follows from s = ½(u + v)t = ½ × (0 + 24) × 6 = 72 m.)
A motorbike moving with initial velocity 28 m s⁻¹ and constant acceleration stops after travelling 98 m. Find the acceleration of the motorbike and the time taken to come to a stop.
Given: initial velocity u = 28 m s⁻¹, final velocity v = 0 (it stops), distance s = 98 m.
Acceleration. Using v² = u² + 2as:
(0)² = (28)² + 2 × a × 98
0 = 784 + 196a ⇒ a = −784 ÷ 196 = −4 m s⁻².
The minus sign shows the acceleration acts opposite to the motion — it is a retardation (the brakes slowing the bike down).
Time taken to stop. Using v = u + at:
0 = 28 + (−4) × t ⇒ 4t = 28 ⇒ t = 7 s.
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.
No, objects A and B never have equal velocity.
On a position-time graph, the velocity of an object is the slope (steepness) of its line. Both A and B are shown as straight lines, so each object moves with its own constant velocity that does not change with time.
The two lines have different slopes — line B is steeper than line A — so B is moving with a larger (constant) velocity than A throughout. Since one slope is always greater than the other, the two velocities are never equal at any instant.
The point where the two lines cross only means that at that instant A and B are at the same position (one overtakes the other). It does not mean they have the same velocity; velocity is given by the slope, and the slopes stay different. Hence A and B never have equal velocity.
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 s to 10 seconds. Choose the correct option(s).
Correct options: (i) and (ii).
From the graph, both A and B start at the same position (the origin) and end at the same position at t = 10 s. So over the 10 s interval both have the same displacement. Since average velocity = displacement ÷ time and the time is the same (10 s) for both, their average velocities are equal — option (i) is correct.
Both objects move in a straight line in one direction without turning back, so the position of each keeps increasing. For such motion the distance covered equals the magnitude of the displacement. As both cover the same net distance in the same 10 s, their average speeds are also equal — option (ii) is correct.
Why (iii) and (iv) are wrong. Both A and B travel the same total distance in the 10 s (they begin and end together), so neither one covers a shorter or longer distance than the other. Their average speeds are equal, not one greater than the other. (What differs is only how they move in between: B moves slowly at first and fast later, while A moves steadily — but the average speed over the whole interval is the same.)
A truck driver driving at the speed of 54 km h⁻¹ notices a road sign with a speed limit of 40 km h⁻¹ (Fig. 4.29) for trucks. He slows down to 36 km h⁻¹ in 36 s. What was the distance travelled by him during this time? Assume the acceleration to be constant while slowing down.
First convert the speeds to m s⁻¹ (multiply km h⁻¹ by 1000÷3600, i.e. by 5÷18):
u = 54 km h⁻¹ = 54 × (5÷18) = 15 m s⁻¹.
v = 36 km h⁻¹ = 36 × (5÷18) = 10 m s⁻¹.
Time, t = 36 s.
Distance travelled. Since the acceleration is constant, use s = ½(u + v)t:
s = ½ × (15 + 10) × 36 = ½ × 25 × 36 = 450 m.
So the truck covers 450 m while it slows down from 54 km h⁻¹ to 36 km h⁻¹.
A car starts from rest and accelerates uniformly to 20 m s⁻¹ in 5 seconds. It then travels at 20 m s⁻¹ for 10 seconds and finally applies the brake (with uniform acceleration) to stop in 6 seconds. Find the total distance travelled.
The journey has three stages; find the distance in each and add them.
Stage 1 — speeding up. u = 0, v = 20 m s⁻¹, t = 5 s.
s₁ = ½(u + v)t = ½ × (0 + 20) × 5 = 50 m.
Stage 2 — constant speed. Speed = 20 m s⁻¹ for 10 s.
s₂ = speed × time = 20 × 10 = 200 m.
Stage 3 — braking to rest. u = 20 m s⁻¹, v = 0, t = 6 s.
s₃ = ½(u + v)t = ½ × (20 + 0) × 6 = 60 m.
Total distance travelled = s₁ + s₂ + s₃ = 50 + 200 + 60 = 310 m.
A bus is travelling at 36 km h⁻¹ when the driver sees an obstacle 30 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 m s⁻². Will the bus be able to stop before reaching the obstacle?
Convert the speed: u = 36 km h⁻¹ = 36 × (5÷18) = 10 m s⁻¹.
The total stopping distance has two parts: the distance during the driver’s reaction time, plus the distance while braking.
1. Reaction distance. During the 0.5 s reaction time the bus keeps moving at 10 m s⁻¹:
s₁ = speed × time = 10 × 0.5 = 5 m.
2. Braking distance. Now u = 10 m s⁻¹, v = 0, retardation a = −2.5 m s⁻². Using v² = u² + 2as:
0 = (10)² + 2 × (−2.5) × s₂ ⇒ 0 = 100 − 5 s₂ ⇒ s₂ = 20 m.
Total stopping distance = s₁ + s₂ = 5 + 20 = 25 m.
Conclusion: The bus needs 25 m to stop, and the obstacle is 30 m away. Since 25 m < 30 m, yes, the bus will stop before reaching the obstacle — with about 5 m to spare.
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.
Whether an object is “at rest” or “in motion” is not absolute — it depends on the reference point (the frame of reference) we choose. An object is at rest if its position does not change with respect to the chosen reference point, and in motion if its position does change.
With respect to the Earth’s surface, an object kept on the ground does not change its position — so, taking the Earth as the reference point, the object is at rest.
With respect to the Sun, the object is carried along with the Earth as the Earth revolves around the Sun. Its position keeps changing relative to the Sun, so, taking the Sun as the reference point, the same object is in motion.
So the object can be described as being at rest and as being in motion at the same time, depending on which reference point we use. This is why motion and rest are always described as being relative — there is no such thing as absolute rest.
The velocity-time graph from 0 s to 120 s for a cyclist is shown in Fig. 4.30. Shade the areas (in different colours) representing the displacement of the cyclist
Also, calculate the displacement and average acceleration in the 120 s time interval.
Shading the areas. On a velocity-time graph the displacement is the area between the line and the time axis. The graph has three parts: the velocity rises from 0 to 3 m s⁻¹ (0–20 s), stays constant at 3 m s⁻¹ (20–100 s), and then falls from 3 to 2 m s⁻¹ (100–120 s). In Fig. 4.30 above, the green rectangle (20–100 s) is the displacement at constant velocity, and the orange region (100–120 s) is the displacement while the velocity is decreasing.
Displacement in the 120 s = total area under the graph, found part by part:
Total displacement = 30 + 240 + 50 = 320 m.
Average acceleration in the 120 s = (final velocity − initial velocity) ÷ total time.
The velocity is 0 at t = 0 and 2 m s⁻¹ at t = 120 s, so
average acceleration = (2 − 0) ÷ 120 = 0.017 m s⁻² (about 1÷60 m s⁻²).
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 running distance based on the graph.
Idea: On a velocity-time graph the distance is the area under the graph. Since the girl runs on a straight road in one direction, this area gives the running distance directly.
Reading the graph. Her velocity stays close to 7–7.5 km h⁻¹ for the first few hours and then eases down to about 6.5 km h⁻¹, while the run lasts for roughly 6.5 hours. We can estimate the area by taking an average velocity over the whole run.
The velocity mostly lies between 6.5 and 7.5 km h⁻¹, so a fair average velocity is about 6.5–7 km h⁻¹. Taking the running time as about 6.5 h:
estimated distance = average velocity × time ≈ (about 6.5 km h⁻¹) × (about 6.5 h) ≈ about 42 km.
So the area under the graph works out to roughly 40–42 km — which is very close to the full marathon distance (about 42.2 km), a sensible result for a girl training for her first marathon. (Because the value is read off a graph, an estimate in the range of about 40–45 km is acceptable.)
On entering a state highway, a car continues to move with a constant velocity of 6 m s⁻¹ for 2 minutes and then accelerates with a constant acceleration 1 m s⁻² for 6 seconds. Find the displacement of the car on the state highway in the 2 min 6 s time interval by drawing a velocity-time graph for its motion.
The motion has two stages. Find the displacement (area under the velocity-time graph) for each and add them.
Stage 1 — constant velocity. Velocity = 6 m s⁻¹ for 2 minutes = 120 s. On the graph this is a horizontal line at 6 m s⁻¹, and the area beneath it is a rectangle:
s₁ = velocity × time = 6 × 120 = 720 m.
Stage 2 — accelerating. u = 6 m s⁻¹, a = 1 m s⁻², t = 6 s. The final velocity is v = u + at = 6 + 1 × 6 = 12 m s⁻¹. On the graph this is a sloping line rising from 6 to 12 m s⁻¹, and the area beneath it is a trapezium:
s₂ = ½(u + v)t = ½ × (6 + 12) × 6 = ½ × 18 × 6 = 54 m.
Total displacement in the 2 min 6 s (126 s) = s₁ + s₂ = 720 + 54 = 774 m.
Two cars A and B start moving with a constant acceleration from rest in a straight line. Car A attains a velocity of 5 m s⁻¹ in 5 s. Car B attains a velocity of 3 m s⁻¹ in 10 s. Plot the velocity-time graphs for both the cars in the same graph. Using the graph, calculate the displacement mentioned in the two time intervals.
Step 1 — find each car’s acceleration. Both start from rest (u = 0), using a = (v − u) ÷ t:
Car A: aᵤ = (5 − 0) ÷ 5 = 1 m s⁻². Car B: aᵇ = (3 − 0) ÷ 10 = 0.3 m s⁻².
Step 2 — the graphs. For each car the velocity-time graph is a straight line starting from the origin. Car A’s line rises from (0 s, 0) to (5 s, 5 m s⁻¹); Car B’s line rises from (0 s, 0) to (10 s, 3 m s⁻¹). Car A’s line is steeper, showing its larger acceleration.
Step 3 — displacement = area of the triangle under each line.
Car A, in its 5 s: sᵤ = ½ × base × height = ½ × 5 × 5 = 12.5 m.
Car B, in its 10 s: sᵇ = ½ × base × height = ½ × 10 × 3 = 15 m.
So although Car A is faster, Car B covers a slightly greater displacement (15 m) because it travels for a longer time than Car A (12.5 m).
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
The length of the minute’s hand is 7 cm (Fig. 4.32).
How far the hand turns. From 6:00 to 7:30 is 90 minutes. The minute’s hand takes 60 minutes for one full round, so in 90 minutes it makes 90÷60 = 1½ revolutions. The tip moves on a circle of radius r = 7 cm. (Take π = 22÷7.)
(i) Distance travelled = 1½ × circumference = 1.5 × 2πr = 1.5 × 2 × (22÷7) × 7 = 1.5 × 44 = 66 cm.
(ii) Displacement. At 6:00 the minute’s hand points to 12 (top). After 1½ turns, at 7:30 it points to 6 (bottom). The tip has moved from the top of the circle to the bottom, so the straight-line gap is the diameter:
displacement = 2r = 2 × 7 = 14 cm (directed straight down, from the 12 towards the 6).
Time in seconds: 90 minutes = 90 × 60 = 5400 s.
(iii) Speed = distance ÷ time = 66 ÷ 5400 ≈ 0.0122 cm s⁻¹ (that is, about 0.73 cm per minute).
(iv) Velocity = displacement ÷ time = 14 ÷ 5400 ≈ 0.0026 cm s⁻¹, directed downward (from the 12 towards the 6). It is much smaller than the speed because the actual path (66 cm) is far longer than the straight-line displacement (14 cm).
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