Where Mindful Learning Begins
Where Mindful Learning Begins
Class 9 Science NCERT solutions for Chapter 4, Describing Motion Around Us, with question answers for every Think It Over, Activity, Pause & Ponder, Worked Example, Revise Reflect Refine and Journey Beyond question (CBSE 2026-27), with full step-by-step working for every numerical. Read your Class 9 Science chapter 4 notes first, then practise here.
This is where Class 9 physics really begins, and these Class 9 Science Chapter 4 Describing Motion Around Us NCERT solutions show every step of the numericals: distance and displacement, average speed and velocity, uniform and non-uniform acceleration, position-time and velocity-time graph questions, and the three equations of motion with solved examples. Use them to practise equations-of-motion numericals, braking-distance problems and graph-reading questions for Class 9 tests.
Last updated 20 October 2026 · Reviewed by Suchita Arora, 13+ years teaching CBSE Maths & Science
Describing Motion Around Us is the first full physics chapter of Class 9 Science, and it is built almost entirely around numericals and graphs. It covers distance vs displacement, average speed vs average velocity, average acceleration, how to read and plot position-time and velocity-time graphs, the three equations of motion for constant acceleration, and uniform circular motion. Every question — from the Think It Over prompts to the hardest Revise Reflect Refine numerical — is solved here with full step-by-step working, as complete Class 9 Science Chapter 4 NCERT solutions with answers.
Why displacement can be smaller than distance travelled, and how average speed differs from average velocity.
Reading position-time and velocity-time graphs — what the slope means, and what the area under the graph gives you.
v = u + at, s = ut + ½at², and v² = u² + 2as — used throughout for braking distance, free fall, and multi-phase journeys.
When the truck ahead applies its brakes, it does not stop at once; it moves some distance before it comes to rest.
Our vehicle also does not stop at once. The driver first takes some reaction time before pressing the brake, and the vehicle moves during this time.
After the brake is pressed, the vehicle moves a further distance (braking distance) before coming to rest.
So the gap must be at least the total distance our vehicle needs to stop: the reaction distance plus the braking distance.
Yes, the distance depends on the speed.
Using \(v^2 = u^2 + 2as\) with \(v = 0\), the braking distance is \(s = \dfrac{u^2}{-2a}\), so it depends on the square of the initial velocity \(u\).
The distance travelled during the reaction time is \(u \times t\), so it also increases with the speed.
For example, in Example 4.8 the braking distance is 28.1 m at 54 km h⁻¹ and 112.5 m at 108 km h⁻¹. Doubling the speed makes the braking distance 4 times.
So the faster we move, the larger the distance we must keep.
Step 1: Yes. The ball goes up along a vertical straight line and falls back along the same straight line, so it is a motion in a straight line.
Step 2: The positions in Fig. 4.5 are O = 0 cm, A = 40 cm, C = 80 cm and B = 140 cm. The ball goes up from O to B and then falls back through C to O.
| S. No. | Position | Total distance travelled by the ball from O till that position | Displacement of the ball from O till that position |
|---|---|---|---|
| 1 | O | 0 cm | 0 cm |
| 2 | A | 40 cm | 40 cm in upward direction |
| 3 | B | 140 cm | 140 cm in upward direction |
| 4 | C | 140 + 60 = 200 cm | 80 cm in upward direction |
| 5 | O | 140 + 140 = 280 cm | 0 cm |
Step 3:
(i) is not true, because the displacement is 0 cm when the ball is back at O.
(ii) is not true, because the magnitude of displacement is never greater than the distance travelled (140 cm and 140 cm at B; 80 cm and 200 cm at C).
(iii) is true, because the magnitude of displacement is equal to the distance when the ball does not turn back (at A and B) and is less when it turns back (at C and O).
(iv) is not true, because at A and B the magnitude of displacement is equal to the total distance travelled.
The speed changes from 0 to 100 km h⁻¹.
Converting to SI units, the final speed is:
Magnitude of average acceleration = \(\dfrac{27.8 - 0}{t}\), where \(t\) is the time taken.
Sample values (the times differ from car to car; replace them with the times you find on the internet for your cars):
| Car type | Time interval during which the speed goes from 0 to 100 km h⁻¹ | Magnitude of average acceleration (m s⁻²) |
|---|---|---|
| Small hatchback | 13 s | 27.8 ÷ 13 ≈ 2.1 |
| Family sedan | 10 s | 27.8 ÷ 10 ≈ 2.8 |
| SUV | 9 s | 27.8 ÷ 9 ≈ 3.1 |
| Sports car | 5 s | 27.8 ÷ 5 ≈ 5.6 |
Table 4.3 gives the positions of the vehicle:
| Time (s) | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|---|
| Position (m) | 0 | 20 | 40 | 60 | 80 | 100 | 120 |
Time is shown along the x-axis and position along the y-axis.
Scale: x-axis 5 divisions = 1 s; y-axis 5 divisions = 20 m.
The points to be plotted are (0 s, 0 m), (1 s, 20 m), (2 s, 40 m), (3 s, 60 m), (4 s, 80 m), (5 s, 100 m) and (6 s, 120 m).
The first point is the origin itself, because the position is 0 m at 0 s.
On joining the points, we get a straight line.
Step 2: BC represents the change in position, \(s_2 - s_1\). CA represents the change in time, \(t_2 - t_1\).
Step 4: From the graph, A is at \(t_1 = 2\text{ s}\), \(s_1 = 40\text{ m}\) and B is at \(t_2 = 4\text{ s}\), \(s_2 = 80\text{ m}\).
Prediction (step 3): After the ring is lifted, the marble will not continue in the circle. It will move in a straight line.
Observation (step 4): When the ring is lifted, the marble moves in a straight line. It does not continue in a circular motion.
The marble moves in the direction in which it was moving at the instant the ring was removed, i.e., along the tangent to the circle at that point.
Step 5: The same result is obtained every time the activity is repeated.
The displacement is the net change in position from the starting point to the stopping point.
The displacement of the athlete will be zero when she is back at the starting point O.
In Fig. 4.4, she runs from O to A (100 m) and then back from A to O (100 m).
Total distance travelled = OA + AO = 100 m + 100 m = 200 m.
The engine uses fuel for every metre that the vehicle moves along its path.
The fuel used is more if the vehicle travels a longer path.
The displacement only gives the net change in position and does not tell the path travelled.
If a vehicle goes to a place and comes back to the starting point, the displacement is zero, but the fuel used is not zero.
The track is inclined but it is straight.
The ball moves along the straight track in one direction and does not change its direction.
Take O as the origin.
The position of the ball is the distance along the track from O, so it can be marked on a straight line with O as the origin, as in Fig. 4.3.
From Fig. 4.6: OA = 40 cm, AB = 10 cm, BC = 20 cm and CD = 30 cm.
The ball moves in one direction only, so the total distance travelled is equal to the magnitude of displacement at each position.
| Position | Total distance travelled from O | Magnitude of displacement from O |
|---|---|---|
| A | 40 cm | 40 cm |
| B | 40 + 10 = 50 cm | 50 cm |
| C | 50 + 20 = 70 cm | 70 cm |
| D | 70 + 30 = 100 cm | 100 cm |
Given: Leg 1: 200 km north in 3 h; leg 2: 200 km south in 2 h
To find: average speed and average velocity for the entire trip
Total distance travelled = 200 km + 200 km = 400 km
Total time = 3 h + 2 h = 5 h
Displacement = 200 km north – 200 km south = 0 km (the trip ends at the starting point).
The average speed is the total distance travelled divided by the time interval.
The magnitude of average velocity is the magnitude of displacement divided by the time interval.
They are equal when the total distance travelled is equal to the magnitude of displacement.
This happens when the object moves in a straight line in one direction, without turning back.
The average velocity is zero when the displacement is zero.
The average speed is not zero when the total distance travelled is not zero.
This happens when the object returns to its starting point, for example, Sarang swimming one length of the pool and back (Example 4.2).
Distance covered by the two postmen together in one day = 9 yojanas + 5 yojanas = 14 yojanas.
To meet each other, they must cover 210 yojanas together.
Check: in 15 days the first postman covers 9 × 15 = 135 yojanas and the second covers 5 × 15 = 75 yojanas.
135 + 75 = 210 yojanas, which is the total distance.
Given: time interval = 50 s; length of the pool = 25 m (Fig. 4.7)
To find: average speed and average velocity in 50 s
Sarang swims from one end to the other end and back.
Total distance travelled = 25 m + 25 m = 50 m
Displacement = 0 m, because he is back at the starting point.
\(u = 36\text{ km h}^{-1} = 36 \times \dfrac{1000}{3600} = 10\text{ m s}^{-1}\)
\(v = 54\text{ km h}^{-1} = 15\text{ m s}^{-1}\), \(t = 10\text{ s}\)
The speed is increasing, so the acceleration is in the direction of the velocity.
\(u = 54\text{ km h}^{-1} = 15\text{ m s}^{-1}\), \(v = 0\), \(t = 5\text{ s}\)
The minus sign shows that the acceleration is opposite to the direction of velocity (the speed is decreasing).
The velocity at different instants (Fig. 4.10) is 0 m s⁻¹ at 0 s, 9.8 m s⁻¹ at 1 s, 19.6 m s⁻¹ at 2 s, 29.4 m s⁻¹ at 3 s and 39.2 m s⁻¹ at 4 s.
Average acceleration in each interval = change in velocity ÷ time interval.
The average acceleration is the same in every interval, so it is constant.
The velocity increases in the direction of motion, so the acceleration is in the direction of motion (vertically downwards).
This acceleration is called the acceleration due to gravity, \(g\).
Scale: x-axis 5 divisions = 2 s; y-axis 5 divisions = 5 m.
Points to be plotted: (0 s, 0 m), (2 s, 1 m), (4 s, 4 m), (6 s, 9 m), (8 s, 16 m), (10 s, 25 m) and (12 s, 36 m).
The points do not fall on a straight line, so they are joined by a smooth curve.
The distance covered in equal intervals of time keeps increasing, so the velocity of the vehicle is increasing.
In Fig. 4.15, the position of the vehicle is 40 m from the origin at 0 s, 1 s, 2 s and 3 s.
The position does not change with time.
So the vehicle is not moving.
Draw lines parallel to the axes (Fig. 4.16b) and compare the displacements of A and B in the same time interval \(t\).
The displacement of B is more than the displacement of A in the same time interval.
So the slope of the line for B is steeper than the slope of the line for A.
Average velocity = displacement ÷ time interval, so the object with the larger displacement has the larger velocity.
Given: \(a = -4\text{ m s}^{-2}\), \(v = 0\text{ m s}^{-1}\)
Let the initial velocity be \(u\) and the distance travelled be \(s\). Using \(v^2 = u^2 + 2as\):
\(u = 54 \times \dfrac{1000}{3600} = 15\text{ m s}^{-1}\)
\(u = 108 \times \dfrac{1000}{3600} = 30\text{ m s}^{-1}\)
The shop is 250 m from home.
Path of the father: home to shop (250 m), shop to home (250 m), home to shop (250 m), shop to home (250 m).
Total distance travelled = 250 m × 4 = 1000 m.
He starts from home and ends at home, so the displacement = 0 m.
Ground floor to fourth floor = 4 × 3 m = 12 m (upwards).
Fourth floor to second floor = 2 × 3 m = 6 m (downwards).
Total distance = 12 m + 6 m = 18 m.
The starting point is the ground floor and the final position is the second floor.
Displacement = 2 × 3 m = 6 m, directed upwards.
Yes, it is possible.
The speedometer shows only the speed, i.e., the magnitude of the velocity.
Acceleration is non-zero if the velocity changes, and the velocity changes if its magnitude or its direction (or both) changes.
If the girl takes a turn or rides along a curved path at a constant speed, the direction of her velocity changes continuously.
So the scooter is accelerating even though the speedometer reading is constant.
Given: \(u = 0\), \(v = 24\text{ m s}^{-1}\), \(t = 6\text{ s}\)
To find: average acceleration \(a\) and distance \(s\)
Using \(s = ut + \tfrac{1}{2}at^2\):
Given: \(u = 28\text{ m s}^{-1}\), \(v = 0\), \(s = 98\text{ m}\)
To find: acceleration \(a\) and time \(t\)
Using \(v^2 = u^2 + 2as\):
Using \(v = u + at\):
Velocity is given by the slope of a position-time graph.
Both graphs in Fig. 4.27 are straight lines, so both objects move with constant velocities.
The two lines are not parallel; the line for A is steeper than the line for B, so the slopes are different.
The lines cross at 5 s. This only means that A and B are at the same position at 5 s. It does not mean that their velocities are equal.
In Fig. 4.28, the position of both A and B keeps increasing with time, so neither object turns back.
A and B start from the same position and are at the same position at 10 s.
So their displacements are equal, and the average velocities (displacement ÷ time) are equal.
Both move in one direction only, so the distance travelled is equal to the displacement.
The distances are equal and the time is the same (10 s), so the average speeds are equal.
A does not cover a shorter distance than B; the distances are equal.
The speed of B is lower than that of A in some segments, but it is higher in the last segment.
The total distance and the time are the same, so the average speeds are equal, not different.
Given: \(u = 54\text{ km h}^{-1} = 15\text{ m s}^{-1}\), \(v = 36\text{ km h}^{-1} = 10\text{ m s}^{-1}\), \(t = 36\text{ s}\)
To find: distance travelled \(s\)
The acceleration is constant. Using \(v = u + at\):
Using \(s = ut + \tfrac{1}{2}at^2\):
The journey has three parts: speeding up, constant velocity and braking.
\(u = 0\), \(v = 20\text{ m s}^{-1}\), \(t = 5\text{ s}\)
\(a = \dfrac{20 - 0}{5} = 4\text{ m s}^{-2}\)
\(v = 20\text{ m s}^{-1}\), \(t = 10\text{ s}\)
\(u = 20\text{ m s}^{-1}\), \(v = 0\), \(t = 6\text{ s}\)
\(a = \dfrac{0 - 20}{6} = -\dfrac{10}{3}\text{ m s}^{-2}\)
Given: \(u = 36\text{ km h}^{-1} = 10\text{ m s}^{-1}\); obstacle at 30 m; reaction time = 0.5 s; \(a = -2.5\text{ m s}^{-2}\) after braking
To find: whether the bus stops before the obstacle
During the reaction time of 0.5 s, the bus moves with constant velocity:
After the brake is applied, \(v = 0\). Using \(v^2 = u^2 + 2as\):
Total distance travelled by the bus before it stops = 5 m + 20 m = 25 m.
25 m is less than 30 m.
Rest and motion are described with respect to a reference point.
An object kept on the Earth does not change its position with respect to the Earth, so it is at rest with respect to the Earth.
The Earth moves around the Sun, and the object moves along with the Earth. With respect to the Sun, the position of the object changes with time, so it is in motion.
So the same object is at rest with respect to one reference point and in motion with respect to another.
From Fig. 4.30: the velocity increases from 0 to 3 m s⁻¹ in 0 – 20 s, stays at 3 m s⁻¹ from 20 s to 100 s, and decreases from 3 m s⁻¹ to 2 m s⁻¹ in 100 – 120 s.
The rectangle between the graph and the time axis from 20 s to 100 s is shaded (teal).
The area between the graph and the time axis from 100 s to 120 s is shaded (gold).
Displacement = area under the velocity-time graph:
0 – 20 s (triangle): ½ × 20 s × 3 m s⁻¹ = 30 m
20 – 100 s (rectangle): 3 m s⁻¹ × 80 s = 240 m
100 – 120 s (trapezium): ½ × (3 + 2) m s⁻¹ × 20 s = 50 m
Average acceleration in 120 s: initial velocity = 0, final velocity = 2 m s⁻¹.
Fig. 4.31 shows the velocity of the girl (in km h⁻¹) against time (in h).
The area under a velocity-time graph gives the distance (the girl runs in one direction).
Reading the graph, the velocity is 7.0 km h⁻¹ at 0 h and 0.6 h, 7.5 km h⁻¹ at 1.6 h and 3.0 h, 7.0 km h⁻¹ at 4.6 h, and 6.5 km h⁻¹ at 5.6 h and 6.6 h.
Area of each part (average velocity × time interval):
| Time interval | Average velocity (km h⁻¹) | Time interval (h) | Distance (km) |
|---|---|---|---|
| 0 – 0.6 h | 7.0 | 0.6 | 4.2 |
| 0.6 – 1.6 h | (7.0 + 7.5) ÷ 2 = 7.25 | 1.0 | 7.25 |
| 1.6 – 3.0 h | 7.5 | 1.4 | 10.5 |
| 3.0 – 4.6 h | (7.5 + 7.0) ÷ 2 = 7.25 | 1.6 | 11.6 |
| 4.6 – 5.6 h | (7.0 + 6.5) ÷ 2 = 6.75 | 1.0 | 6.75 |
| 5.6 – 6.6 h | 6.5 | 1.0 | 6.5 |
Given: constant velocity \(u = 6\text{ m s}^{-1}\) for 2 min = 120 s; then \(a = 1\text{ m s}^{-2}\) for 6 s
To find: displacement in 2 min 6 s
The velocity-time graph is a horizontal line at 6 m s⁻¹ from 0 s to 120 s. Then the velocity increases uniformly for 6 s.
Final velocity after 6 s: \(v = u + at = 6 + 1 \times 6 = 12\text{ m s}^{-1}\).
Displacement = area under the graph.
Area of the rectangle (0 – 120 s) = 6 × 120 = 720 m
Area of the trapezium (120 – 126 s) = ½ × (6 + 12) × 6 = 54 m
Car A: \(u = 0\), \(v = 5\text{ m s}^{-1}\), \(t = 5\text{ s}\)
| t (s) | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|
| Velocity of A (m s⁻¹) | 0 | 1 | 2 | 3 | 4 | 5 |
Car B: \(u = 0\), \(v = 3\text{ m s}^{-1}\), \(t = 10\text{ s}\)
| t (s) | 0 | 2.5 | 5 | 7.5 | 10 |
|---|---|---|---|---|---|
| Velocity of B (m s⁻¹) | 0 | 0.75 | 1.5 | 2.25 | 3 |
Displacement = area under the velocity-time graph.
Car A in 5 s: area of the triangle = ½ × 5 × 5 = 12.5 m
Car B in 10 s: area of the triangle = ½ × 10 × 3 = 15 m
Given: radius \(r = 7\text{ cm}\); time interval = 6 PM to 7:30 PM = 1.5 h = 90 min = 5400 s
The minute’s hand makes one revolution in 60 min.
Number of revolutions in 90 min = 90 ÷ 60 = 1.5
Distance in one revolution = circumference = \(2\pi r\).
After one complete revolution the tip is back at its starting position.
The extra half revolution takes the tip to the point diametrically opposite to the starting point (from the 12 mark to the 6 mark).
No, the speeds of the numbers and the letters are not the same.
The numbers are on a circle of radius 7 cm and the letters are on a circle of radius 4 cm.
The disc makes one revolution in the same time \(T\) for both. Speed = circumference ÷ time, so:
The numbers on the outer circle move faster than the letters on the inner circle.
On spinning the disc fast, the fast-moving numbers blur and fade, but the slower-moving letters can still be seen.
The readings keep changing and are small but not zero.
Our hand is never perfectly still; it makes tiny involuntary movements, which the accelerometer detects.
The readings are almost zero and nearly steady.
The floor does not move, so the phone is not accelerating.
Difference: the readings are larger and keep changing on the palm, and are close to zero on the floor.
(a) To derive \(s = vt - \tfrac{1}{2}at^2\)
From Eq. (4.4a), \(v = u + at\), so \(u = v - at\).
Substitute this in Eq. (4.4b), \(s = ut + \tfrac{1}{2}at^2\):
(b) To derive \(s = \tfrac{1}{2}(u + v)t\) using the area of a trapezium
For uniform acceleration the velocity-time graph is a straight line from \(u\) at 0 to \(v\) at time \(t\) (Fig. 4.19).
The displacement is the area under this graph, which is a trapezium.
The parallel sides of the trapezium are \(u\) and \(v\), and its height (the distance between them along the time axis) is \(t\).
Area of a trapezium = ½ × (sum of parallel sides) × height.
Table 4.4 gives positions 0, 1, 4, 9, 16, 25 and 36 m at times 0, 2, 4, 6, 8, 10 and 12 s.
Scale 1: x-axis 5 divisions = 2 s, y-axis taking up to 120 m.
Scale 2: x-axis 5 divisions = 2 s, y-axis 5 divisions = 5 m (as in Fig. 4.12).
Comparison: the same data give the same curve, but with Scale 1 the graph occupies only a small part of the sheet and the bending of the curve is hard to see.
With Scale 2 the points are spread over the whole sheet and it is easy to read the values.
A graph plotting app automatically selects a scale that fits the data on the screen.
Water on the road reduces the grip of the tyres, so the retardation is smaller.
Worn tyres grip the road less, so the retardation is smaller.
A heavier vehicle is harder to slow down, so the retardation by the same brakes is smaller.
The driver sees the obstacle late, so the reaction time effectively increases.
The visibility is poor, so the driver notices the obstacle late.
The road becomes slippery and visibility is poor.
The distance travelled in the reaction time = velocity × reaction time.
A longer reaction time means a longer distance before the brake is applied.
Motion can be described precisely with just five linked quantities — displacement, time, initial velocity, final velocity, and acceleration — and the three kinematic equations that connect them let you predict exactly where an object will be and how fast it will be moving, whether it's a braking car, a falling ball, or a child on a merry-go-round.
Here are extra practice questions for Class 9 Chapter 4, Describing Motion Around Us. These are numerical problems on motion in a straight line — apart from the NCERT questions — to make you confident in your understanding of the chapter. Attempt the questions yourself first, and then cross-check your answers below.
Given: \(u = 0\), \(a = 0.1\text{ m s}^{-2}\), \(t = 2\text{ min} = 120\text{ s}\)
To find: (a) speed acquired, (b) distance travelled
Given: \(u = 90\text{ km h}^{-1} = 25\text{ m s}^{-1}\), \(v = 0\), \(a = -0.5\text{ m s}^{-2}\)
To find: distance \(s\) before the train stops
Using \(v^2 = u^2 + 2as\):
Given: \(u = 0\) (starts from rest), \(a = 4\text{ m s}^{-2}\), \(t = 10\text{ s}\)
To find: distance \(s\)
Using \(s = ut + \tfrac{1}{2}at^2\):
Given: \(u = 5\text{ m s}^{-1}\) (upward), \(a = -10\text{ m s}^{-2}\), \(v = 0\) at the highest point
To find: height \(s\) and time \(t\)
Using \(v^2 = u^2 + 2as\):
Using \(v = u + at\):
Given: bus: \(u_1 = 54\text{ km h}^{-1} = 15\text{ m s}^{-1}\); car: \(u_2 = 108\text{ km h}^{-1} = 30\text{ m s}^{-1}\); \(a = -5\text{ m s}^{-2}\); \(v = 0\)
Using \(v^2 = u^2 + 2as\), the stopping distance is \(s = \dfrac{u^2}{2 \times 5}\).
Using \(v = u + at\), the time to stop is \(t = \dfrac{u}{5}\).
Given: \(u = 4\text{ m s}^{-1}\) at 0 s; \(v = 14\text{ m s}^{-1}\) at 10 s
Distance = area under the graph (a trapezium):
Check using \(s = ut + \tfrac{1}{2}at^2\): \(4 \times 10 + \tfrac{1}{2} \times 1 \times 100 = 90\text{ m}\).
Time at which the velocity is 9 m s⁻¹, using \(v = u + at\):
Car A overtakes car B when both have travelled the same distance from the point.
Let the time taken be \(t\).
Car A: \(u = 0\), \(a = 2\text{ m s}^{-2}\)
Car B: constant velocity 10 m s⁻¹
Equating the distances:
Next up is Chapter 5 — Exploring Mixtures and their Separation. Or explore the full chapter list, browse the Class 9 Science hub, or revise with the Class 9 Science Notes, or test yourself with the free Class 9 Science Sample Paper and Marking Scheme.
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