NCERT Solutions for Class 9th Science Chapter 4 End-of-chapter questions — Revise, Reflect, Refine

Book page 68–70 Updated on2026-09-08

Q1.

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?

Answer

Break the trip into its four legs. Each leg is 250 m long.

Leg	Journey	Distance
1	home → shop	250 m
2	shop → home (for the bag)	250 m
3	home → shop (again)	250 m
4	shop → home (with provisions)	250 m

total distance travelled = 250 m + 250 m + 250 m + 250 m
= 4 × 250 m = 1000 m = 1 km

For the displacement, take home as the origin and the direction of the shop as positive:

displacement = final position − initial position
= 0 m − 0 m = 0 m

Why it happens: he ends the whole outing standing exactly where he started, so the net change in position is nil — no matter how many times he walked the road. The distance, in contrast, counts every metre of walking, and the forgotten bag cost him an extra 500 m of it.

Q2.

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: (i) the total vertical distance travelled, and (ii) their displacement from the starting point.

Answer

Height of one floor = 3 m. Measure heights above the ground floor, taking upward as positive.

ground floor → fourth floor: rise of 4 floors = 4 × 3 m = 12 m (upward)
fourth floor → second floor: fall of 2 floors = 2 × 3 m = 6 m (downward)

(i) Total vertical distance travelled

= 12 m + 6 m = 18 m

(ii) Displacement from the starting point

height of second floor above ground floor = 2 × 3 m = 6 m
displacement = 6 m − 0 m = 6 m, vertically upward

Why it happens: the downward 6 m is real climbing — the legs feel it — so it adds to the distance. But it cancels part of the upward climb when we ask only "how much higher is the student now than at the start?" Hence 12 + 6 = 18 m for distance, but 12 − 6 = 6 m for displacement.

Tip: the numbers are floor gaps, not floor numbers. From the ground floor to the fourth floor there are 4 gaps, not 5.

Q3.

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?

Answer

Yes, it certainly can be accelerating — if she is changing direction, for example going round a bend or a roundabout.

Why it happens: the speedometer reports only the magnitude of the velocity. Velocity, however, is magnitude and direction, and acceleration is the rate of change of velocity. So a change in direction alone is enough to give a non-zero acceleration even while the needle sits perfectly still.

acceleration ≠ 0 if magnitude of velocity changes, or direction of velocity changes, or both

This is exactly the case of uniform circular motion in Section 4.4: constant speed, continuously turning velocity, and therefore accelerated motion.

The only way her scooter can have zero acceleration is if it moves along a straight road with that constant speedometer reading.

Check it yourself: you feel this acceleration as the sideways push while turning a corner, even though the engine note and the speedometer do not change at all.

Q4.

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.

Answer

Given: u = 0 m s⁻¹ (starts from rest), v = 24 m s⁻¹, t = 6 s

Average acceleration — use Eq. (4.3b):

a = (v − u) / t
= (24 m s⁻¹ − 0 m s⁻¹) / 6 s
= 24 m s⁻¹ / 6 s = 4 m s⁻², in the direction of motion

Distance travelled — use Eq. (4.4b):

s = ut + ½at²
= (0 m s⁻¹ × 6 s) + (½ × 4 m s⁻² × (6 s)²)
= 0 + ½ × 4 × 36 m
= 72 m

Check it yourself: for constant acceleration the average velocity is simply (u + v)/2 = (0 + 24)/2 = 12 m s⁻¹, so s = 12 m s⁻¹ × 6 s = 72 m ✓. Equally, from Eq. (4.4c): v² = u² + 2as → 576 = 0 + 8s → s = 72 m ✓. Three routes, one answer.

Why it happens: the car is speeding up all through, so it covers less ground than a car doing a steady 24 m s⁻¹ (which would manage 144 m) but more than one crawling near zero. Exactly halfway, in fact — because with constant acceleration the velocity rises linearly, so the average velocity is the plain mean of the start and end values.

Q5.

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.

Answer

Given: u = 28 m s⁻¹, v = 0 m s⁻¹ (it stops), s = 98 m

Acceleration — Eq. (4.4c) links u, v, a and s without needing the time:

v² = u² + 2as
(0 m s⁻¹)² = (28 m s⁻¹)² + 2 × a × 98 m
0 = 784 m² s⁻² + 196 m × a
a = − 784 m² s⁻² / 196 m
a = − 4 m s⁻²

The minus sign says the acceleration is directed opposite to the velocity — the motorbike is slowing down.

Time taken — now use Eq. (4.4a):

v = u + at
0 m s⁻¹ = 28 m s⁻¹ + (− 4 m s⁻²) × t
4 t = 28 s
t = 7 s

Check it yourself: s = ut + ½at² = 28 × 7 + ½ × (− 4) × 49 = 196 − 98 = 98 m ✓

Why it happens: choosing Eq. (4.4c) first was the whole trick. The question gives u, v and s but not t, and (4.4c) is the one kinematic equation in which t does not appear. Picking the equation that avoids the unknown you do not want saves a round of substitution every time.

Q6.

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.

Answer

No. Their velocities are never equal.

Both graphs are straight, so each object has a constant velocity — but the slopes differ, so the velocities differ at every instant.

Justification, in two steps:

Both graphs are straight lines. A straight position–time graph means the slope never changes, so each object moves with its own constant velocity throughout.
The line for A is steeper than the line for B at every point. Since the slope of a position–time graph is the velocity, v_A > v_B — and since both are constant, that inequality holds for the whole journey.

v_A = constant, v_B = constant, and slope of A > slope of B
⇒ v_A > v_B at every instant ⇒ they are never equal

Why it happens: the graphs do cross, at about t = 5 s, and that is the trap. Crossing means the two objects have the same position at that instant — A catches up with B — not the same velocity. Equal velocity would need equal slopes, which would make the two lines parallel, and parallel lines never meet. Here the lines meet, which is precisely the proof that they are not parallel.

Tip: B starts ahead of A (its line begins higher on the position axis), but A is faster and overtakes it at t = 5 s. After that A stays ahead and the gap keeps growing.

Q7.

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 s time interval is equal since they have the same initial and final positions. (ii) The average speeds of both over the 10 s time interval are equal since both cover equal distance in equal time. (iii) The average speed of A over the 10 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 s time interval is greater than that of B since B's speed is lower than A's in some segments.

Answer

The correct options are (i) and (ii).

A and B start together and finish together. A keeps a steady velocity; B starts slower and finishes faster — but the two end points are shared.

Reading the graph: both objects begin at the same position at t = 0 s and arrive at the same position at t = 10 s. Both move forward the whole time — neither curve ever falls — so neither turns back.

(i) Correct. Same initial and final positions means the same displacement, over the same 10 s.

average velocity = displacement / time interval
displacement of A = displacement of B ⇒ average velocities are equal

(ii) Correct. Because neither object reverses, the distance travelled equals the magnitude of the displacement for each of them — and those are equal.

distance travelled = |displacement| (no turning back)
⇒ average speed of A = average speed of B

(iii) Wrong — A does not cover a shorter distance; both cover the same distance in the 10 s.

(iv) Wrong — although B is slower than A in the early part of the interval, it is correspondingly faster in the later part, and the two effects cancel over the full 10 s.

Why it happens: average speed and average velocity care only about the two end points and the total time — nothing in between. B's journey is dramatically different from A's moment by moment (B is accelerating, A is not), but the averages cannot see that. To distinguish them you would have to compare their velocities at particular instants, not their averages.

Q8.

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.

Answer

First convert both speeds to SI units, because the time is in seconds.

u = 54 km h⁻¹ = 54 × 1000 m / 3600 s = 15 m s⁻¹
v = 36 km h⁻¹ = 36 × 1000 m / 3600 s = 10 m s⁻¹
t = 36 s

With constant acceleration the average velocity is the plain mean of u and v, so:

s = ½ (u + v) t
= ½ × (15 m s⁻¹ + 10 m s⁻¹) × 36 s
= ½ × 25 m s⁻¹ × 36 s
= 12.5 m s⁻¹ × 36 s = 450 m

Check it yourself, the long way:
a = (v − u)/t = (10 − 15) m s⁻¹ / 36 s = − 0.139 m s⁻²
s = ut + ½at² = 15 × 36 + ½ × (− 0.139) × 1296 = 540 − 90 = 450 m ✓

Why it happens: the driver is decelerating, so the acceleration is negative — its direction is opposite to the motion — but the truck is still going forward, so the distance is positive throughout. Notice that he covers 450 m, most of half a kilometre, just while shedding 18 km h⁻¹. Speed changes on a highway are not instant, which is why the speed-limit sign has to be placed well before the stretch it applies to.

Did you know? He ends at 36 km h⁻¹, which is below the 40 km h⁻¹ limit — so he is within the law by the end of these 36 s.

Q9.

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.

Answer

Three phases, three areas under one velocity–time graph.

The total distance is the whole area under the graph: a triangle, a rectangle and another triangle.

Phase 1 — speeding up (0 s to 5 s): u = 0, v = 20 m s⁻¹, t = 5 s

s₁ = ½ (u + v) t = ½ × (0 + 20) m s⁻¹ × 5 s = 50 m

Phase 2 — constant velocity (5 s to 15 s): v = 20 m s⁻¹, t = 10 s

s₂ = v × t = 20 m s⁻¹ × 10 s = 200 m

Phase 3 — braking (15 s to 21 s): u = 20 m s⁻¹, v = 0, t = 6 s

s₃ = ½ (u + v) t = ½ × (20 + 0) m s⁻¹ × 6 s = 60 m

total distance = s₁ + s₂ + s₃
= 50 m + 200 m + 60 m = 310 m

Why it happens: the kinematic equations apply only where the acceleration is constant, and here it takes three different values (+4 m s⁻², 0, and −10/3 m s⁻²). So the journey has to be cut at the two points where the acceleration changes, treated separately, and the results added. The graph makes this visible: three simple shapes, one after the other.

Check it yourself: braking takes 6 s from 20 m s⁻¹, so a₃ = (0 − 20)/6 = − 3.33 m s⁻². Then s₃ = 20 × 6 + ½ × (− 3.33) × 36 = 120 − 60 = 60 m ✓

Q10.

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?

Answer

Yes — the bus stops 5 m short of the obstacle.

u = 36 km h⁻¹ = 36 × 1000 m / 3600 s = 10 m s⁻¹

Step 1 — the reaction distance. For 0.5 s the driver has not yet touched the brake, so the bus continues at full speed:

s₁ = u × t_r = 10 m s⁻¹ × 0.5 s = 5 m

Step 2 — the braking distance. Now u = 10 m s⁻¹, v = 0, a = − 2.5 m s⁻². Use Eq. (4.4c):

v² = u² + 2as
(0 m s⁻¹)² = (10 m s⁻¹)² + 2 × (− 2.5 m s⁻²) × s₂
0 = 100 m² s⁻² − 5 m s⁻² × s₂
s₂ = 100 / 5 m = 20 m

Step 3 — total stopping distance.

s = s₁ + s₂ = 5 m + 20 m = 25 m
25 m < 30 m ⇒ the bus stops with 5 m to spare

Why it happens: the reaction distance is easy to forget, and it is a fifth of the whole stopping distance here. During those 0.5 s the brakes do nothing at all — the bus is simply covering ground at 10 m s⁻¹. Had the driver been distracted for 1.5 s instead of 0.5 s, the reaction distance alone would be 15 m and the total 35 m: the bus would have hit the obstacle, with the brakes working exactly as well as before.

Try This: repeat the calculation for a bus at 54 km h⁻¹ (15 m s⁻¹) with the same reaction time and braking. Reaction distance = 7.5 m, braking distance = 225/5 = 45 m, total 52.5 m — nearly double the obstacle distance. The bus would not stop in time.

Q11.

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.

Answer

Both statements can be true at the same time, because rest and motion are always stated with respect to a chosen reference point.

Reference point chosen	Does the object's position change with time?	Verdict
A table, a wall, the ground	No	The object is at rest
The Sun	Yes — it is carried around the Sun with the Earth	The object is in motion
The Earth's axis	Yes — it is carried round once a day by rotation	The object is in motion

Why it happens: the chapter's own definition says an object is in motion if its position with respect to the reference point changes with time. Change the reference point and you can change the answer, without the object doing anything different. So "at rest" is never an absolute property of an object — it is a statement about the object and the reference point together.

So the honest answer is: an object kept on the Earth is at rest with respect to the Earth, and in motion with respect to the Sun. In everyday life we silently take the Earth as our reference point, which is why we call a book on a table "at rest" without feeling we are saying anything incomplete.

Did you know? The Earth carries you around the Sun at about 30 km s⁻¹, and spins you eastward at roughly 0.46 km s⁻¹ at the equator. You feel none of it, because everything around you — the ground, the air, your chair — moves along with you.

Q12.

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 (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 s time interval.

Answer

Reading Fig. 4.30 point by point: the velocity rises from 0 to 3 m s⁻¹ between 0 s and 20 s, stays at 3 m s⁻¹ from 20 s to 100 s, and falls to 2 m s⁻¹ between 100 s and 120 s.

Blue: the constant-velocity stretch (20 s – 100 s). Rose: the stretch where the velocity is decreasing (100 s – 120 s). Green: the initial speeding-up.

(i) Constant velocity — 20 s to 100 s. Shade the rectangle under the flat part of the graph.

displacement = area of rectangle = v × t
= 3 m s⁻¹ × (100 s − 20 s) = 3 × 80 m = 240 m

(ii) Velocity decreasing — 100 s to 120 s. Shade the trapezium under the falling part.

displacement = ½ (sum of parallel sides) × width
= ½ × (3 m s⁻¹ + 2 m s⁻¹) × 20 s = ½ × 5 × 20 m = 50 m

Total displacement in the 120 s. Add the first stretch too (0 s to 20 s, a triangle):

0 s – 20 s: ½ × 20 s × 3 m s⁻¹ = 30 m
20 s – 100 s: 240 m
100 s – 120 s: 50 m
total displacement = 30 m + 240 m + 50 m = 320 m

Average acceleration over the 120 s. This depends only on the velocities at the two ends:

a_av = (v_final − v_initial) / t
= (2 m s⁻¹ − 0 m s⁻¹) / 120 s
= 0.0167 m s⁻² ≈ 0.017 m s⁻², in the direction of motion

Why it happens: the average acceleration is tiny, and positive, even though the cyclist actually slowed down over the last 20 s. That is because "average" compares only the first and last readings: she began at rest and ended at 2 m s⁻¹, a net gain. The graph shows three quite different accelerations along the way (+0.15 m s⁻², 0, and −0.05 m s⁻²) — the single average hides all of that.

Q13.

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.

Answer

She runs on a straight road in one direction, so the distance she covers is the area under the velocity–time graph. Read the seven marked points off Fig. 4.31 (one small square = 0.2 h across and 0.25 km h⁻¹ up):

Time (h)	0	0.6	1.6	3.0	4.6	5.6	6.6
Velocity (km h⁻¹)	7.0	7.0	7.5	7.5	7.0	6.5	6.5

Area under the runner's velocity–time graph. Because both axes are in km and h, the area comes out directly in kilometres.

Split the area into six strips and use the trapezium rule for each:

0 → 0.6 h: 7.0 × 0.6 = 4.2 km
0.6 → 1.6 h: ½(7.0 + 7.5) × 1.0 = 7.25 km
1.6 → 3.0 h: 7.5 × 1.4 = 10.5 km
3.0 → 4.6 h: ½(7.5 + 7.0) × 1.6 = 11.6 km
4.6 → 5.6 h: ½(7.0 + 6.5) × 1.0 = 6.75 km
5.6 → 6.6 h: 6.5 × 1.0 = 6.5 km
total ≈ 4.2 + 7.25 + 10.5 + 11.6 + 6.75 + 6.5 = 46.8 km ≈ 47 km

Check it yourself, the quick way: her velocity never leaves the band 6.5 – 7.5 km h⁻¹, so it averages about 7 km h⁻¹ over roughly 6.6 h. That gives 7 × 6.6 ≈ 46 km — the same answer to the accuracy a graph can give. Since we are reading points by eye, quoting "about 47 km" is honest; quoting 46.8 km would claim more precision than the graph carries.

Why it happens: the units do the bookkeeping for us. Height in km h⁻¹ times width in h gives km, so no conversion is needed at all. And note how far this is — a full marathon is 42.2 km, so she has comfortably covered the distance in her training run.

Q14.

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.

Answer

Convert the time and find the final velocity first, so the graph can be drawn to scale.

2 min = 120 s, so the whole interval is 120 s + 6 s = 126 s
final velocity after accelerating: v = u + at = 6 m s⁻¹ + (1 m s⁻² × 6 s) = 12 m s⁻¹

The displacement is the whole shaded area: a long rectangle for the steady stretch and a narrow trapezium for the 6 s of acceleration.

Part 1 — the rectangle (0 s to 120 s):

s₁ = v × t = 6 m s⁻¹ × 120 s = 720 m

Part 2 — the trapezium (120 s to 126 s):

s₂ = ½ (u + v) t = ½ × (6 m s⁻¹ + 12 m s⁻¹) × 6 s
= ½ × 18 × 6 m = 54 m

total displacement = s₁ + s₂ = 720 m + 54 m = 774 m, in the direction of motion

Check it yourself: for part 2 you can also use s = ut + ½at² = 6 × 6 + ½ × 1 × 36 = 36 + 18 = 54 m ✓

Why it happens: the graph makes the two contributions visually honest. Two minutes of steady driving cover 720 m; six seconds of hard acceleration add only 54 m. It is the width of the area that dominates here, not the height — a useful reminder that a dramatic change in velocity over a short time can still contribute very little displacement.

Q15.

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 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).

Answer

Step 1 — the two accelerations. Both start from rest, so u = 0 for each.

Car A: a_A = (v − u)/t = (5 m s⁻¹ − 0)/5 s = 1 m s⁻²
Car B: a_B = (3 m s⁻¹ − 0)/10 s = 0.3 m s⁻²

Step 2 — velocities at five instants, from v = u + at = at:

Car A: t (s)	1	2	3	4	5
v = 1 × t (m s⁻¹)	1.0	2.0	3.0	4.0	5.0
Car B: t (s)	2	4	6	8	10
v = 0.3 × t (m s⁻¹)	0.6	1.2	1.8	2.4	3.0

Both lines start at the origin because both cars start from rest. A is much steeper (larger acceleration) but its line stops at 5 s.

Step 3 — displacements, as areas of the two triangles.

Car A, in 5 s: s_A = ½ × base × height = ½ × 5 s × 5 m s⁻¹ = 12.5 m
Car B, in 10 s: s_B = ½ × 10 s × 3 m s⁻¹ = 15 m

Check it yourself: s = ut + ½at² gives the same numbers — A: ½ × 1 × 5² = 12.5 m ✓ B: ½ × 0.3 × 10² = 15 m ✓

Why it happens: A accelerates more than three times as hard as B, yet B ends up ahead — because B was given twice as long. This is the point of the question: displacement depends on both acceleration and the time it acts for (s = ½at², so time counts twice over). Comparing accelerations alone would have given the wrong ranking.

Q16.

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 cm (Fig. 4.32).

Answer

The tip moves on a circle of radius R = 7 cm = 0.07 m. From 6:00 PM to 7:30 PM is 90 minutes, and the minute hand takes 60 minutes per revolution.

number of revolutions = 90 min / 60 min = 1.5
time interval t = 90 min = 90 × 60 s = 5400 s

In 90 minutes the tip goes round one and a half times, finishing diametrically opposite where it started.

(i) Distance travelled — 1.5 times the circumference:

distance = 1.5 × 2πR
= 1.5 × 2 × (22/7) × 7 cm
= 1.5 × 44 cm = 66 cm = 0.66 m

(ii) Displacement — after 1.5 revolutions the tip is exactly opposite its starting point, so the straight-line gap is one diameter:

displacement = 2R = 2 × 7 cm = 14 cm = 0.14 m,
directed from the 12 mark towards the 6 mark (vertically downward across the dial)

(iii) Speed — distance ÷ time:

speed = 0.66 m / 5400 s = 1.2 × 10⁻⁴ m s⁻¹

(iv) Velocity — displacement ÷ time:

velocity = 0.14 m / 5400 s = 2.6 × 10⁻⁵ m s⁻¹,
directed from the 12 mark towards the 6 mark

Why it happens: the tip travels nearly five times as far as it is finally displaced, because most of its path curls back on itself. Had Rohan studied for a whole number of hours — say 6 PM to 8 PM — the hand would have come back to the 12 mark and both the displacement and the average velocity would have been exactly zero, while the speed stayed the same.

Tip: convert to metres and seconds before dividing, and the answers come out directly in SI units. Both results are extremely small — the tip of a minute hand creeps along at about a tenth of a millimetre per second.

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