CBSE · Class 9 · Science

Chapter 4: Describing Motion Around Us

Position, distance and displacement, speed and velocity, acceleration, graphs of motion, the equations of motion and uniform circular motion.

Notes

Open each topic to read it.

Position and motion

Reference point: a fixed point used to describe position.
Position: the distance and direction of an object from the reference point at an instant.
An object is in motion if its position changes with time, and at rest if it does not.
Positions to the right of the origin are taken as positive, and to the left as negative (for motion in a straight line).

Distance and displacement

Distance travelled: the total length of the path. It is a scalar and can never be negative.
Displacement: the net change in position, with direction. It is a vector: the straight-line distance from the first position to the last, with direction from the first position towards the second.
Distance and displacement are equal in magnitude only when the object moves in one direction without turning back.
Both are measured in metres (SI unit).

Speed and velocity

Average speed = total distance travelled / time interval. Scalar.
Average velocity = displacement / time interval. Vector, with direction given by the sign (+ or −) on a straight line.
Instantaneous velocity: velocity at a particular instant. The speedometer reading gives the magnitude of the velocity at an instant.
Unit: metre per second (m s⁻¹). Conversion: 1 km h⁻¹ = 5/18 m s⁻¹, so 36 km h⁻¹ = 10 m s⁻¹.

Acceleration

Average acceleration = change in velocity / time interval = (v − u) / t. Vector. Unit: m s⁻².
Acceleration is positive when it acts in the direction of the velocity (speeding up) and negative when it acts opposite to it (slowing down, also called retardation).
An object can have zero acceleration while moving, for example a bus at constant velocity on a straight highway.
Free-fall acceleration due to gravity is g = 9.8 m s⁻², directed downwards.

Position-time graphs

A straight line with a slope shows constant velocity. The steeper the slope, the greater the velocity.
A horizontal line means the object is at rest.
A curved line shows changing velocity (accelerated motion).
Slope = change in position / change in time = velocity.

Velocity-time graphs

A horizontal line means the velocity is constant, so acceleration is zero.
A straight line with a slope shows constant acceleration. Slope = acceleration.
Area under the velocity-time graph = displacement.
Above the time axis, velocity is positive; below it, velocity is negative.

Equations of motion (constant acceleration)

First equation: v = u + at
Second equation: s = ut + ½ at²
Third equation: v² = u² + 2as
Also: s = ½ (u + v) t and s = vt − ½ at² (derived in the chapter's Journey Beyond).
These are valid only when acceleration is constant, in a straight line. Signs of u, v, a and s give direction.

Uniform circular motion

An object moving in a circle at constant speed is in uniform circular motion.
Speed is constant, but the direction of velocity changes continuously, so the velocity changes and the object is accelerated.
In one revolution of radius R and period T: distance = 2πR; displacement = 0; average speed = 2πR / T; average velocity = 0.

Key terms at a glance

Scalar: has magnitude only. Vector: has magnitude and direction. Retardation: negative acceleration. Uniform motion: constant velocity in a straight line.

Numericals

Worked solutions for the examples in the chapter and selected exercises.

EXAMPLE 4.1 · TWO POSTMEN

Distance between them = 210 yojanas. Speeds = 9 and 5 yojanas per day.

Combined speed = 9 + 5 = 14 yojanas/day. Time = 210 / 14

Answer: 15 days. In 15 days, the first covers 135 yojanas and the second 75 yojanas.

EXAMPLE 4.2 · SARANG IN THE POOL

Pool length = 25 m. Swims to the other end and back (50 m) in 50 s.

Average speed = total distance / time = 50 m / 50 s = 1 m s⁻¹

Average velocity = displacement / time = 0 m / 50 s = 0 m s⁻¹

Answer: speed 1 m s⁻¹, velocity 0 m s⁻¹.

EXAMPLE 4.3 · BUS ON HIGHWAY

(i) Accelerator pressed: u = 36 km h⁻¹ = 10 m s⁻¹, v = 54 km h⁻¹ = 15 m s⁻¹, t = 10 s.

a = (v − u) / t = (15 − 10) / 10 = 0.5 m s⁻² (in the direction of velocity)

(ii) Brakes applied: u = 15 m s⁻¹, v = 0, t = 5 s.

a = (0 − 15) / 5 = −3 m s⁻² (opposite to velocity)

EXAMPLE 4.4 · FREELY FALLING OBJECT

Velocities at 0, 1, 2, 3 and 4 s: 0, 9.8, 19.6, 29.4, 39.2 m s⁻¹.

a = (9.8 − 0) / 1 = (19.6 − 9.8) / 1 = … = 9.8 m s⁻²

Answer: the average acceleration is 9.8 m s⁻² in every interval, directed downwards. This is g.

EXAMPLE 4.8 · BRAKING DISTANCE

a = −4 m s⁻², v = 0. Using v² = u² + 2as: 0 = u² + 2(−4)s, so s = u² / 8.

(i) u = 54 km h⁻¹ = 15 m s⁻¹: s = 225 / 8 = 28.1 m.

(ii) u = 108 km h⁻¹ = 30 m s⁻¹: s = 900 / 8 = 112.5 m.

Doubling the speed makes the stopping distance four times as large.

PAUSE AND PONDER 4 · ROAD TRIP

Drive 200 km north in 3 hours, then 200 km south in 2 hours.

Total distance = 400 km; total time = 5 h

Average speed = 400 / 5 = 80 km h⁻¹

Displacement = 200 km north − 200 km south = 0, so average velocity = 0

REVISE, REFLECT, REFINE 1 · SHOP TRIP

Home to shop 250 m, shop to home 250 m, home to shop 250 m, shop to home 250 m.

Total distance = 4 × 250 m = 1000 m

Displacement = 0 (he ends at home)

REVISE, REFLECT, REFINE 2 · STAIRS

Floor height = 3 m. Ground floor = 0 m; second floor = 6 m; fourth floor = 12 m.

Vertical distance = ground to fourth floor (12 m) + fourth to second floor (12 − 6 = 6 m) = 18 m

Displacement = final height − initial height = 6 m − 0 = 6 m upward

REVISE, REFLECT, REFINE 4 · SCOOTER / CAR FROM REST

u = 0, v = 24 m s⁻¹, t = 6 s.

a = (24 − 0) / 6 = 4 m s⁻²

s = ½ (u + v) t = ½ (0 + 24) × 6 = 72 m

REVISE, REFLECT, REFINE 5 · MOTORBIKE

u = 28 m s⁻¹, v = 0, s = 98 m.

v² = u² + 2as: 0 = 784 + 196a, so a = −4 m s⁻²

t = (v − u) / a = (0 − 28) / (−4) = 7 s

REVISE, REFLECT, REFINE 8 · TRUCK SLOWING DOWN

u = 54 km h⁻¹ = 15 m s⁻¹, v = 36 km h⁻¹ = 10 m s⁻¹, t = 36 s.

s = ½ (u + v) t = ½ (15 + 10) × 36 = 450 m

a = (10 − 15) / 36 ≈ −0.14 m s⁻²

REVISE, REFLECT, REFINE 9 · THREE STAGES

Stage 1: 0 to 20 m s⁻¹ in 5 s; s = ½ (0 + 20) × 5 = 50 m

Stage 2: constant 20 m s⁻¹ for 10 s; s = 20 × 10 = 200 m

Stage 3: 20 to 0 m s⁻¹ in 6 s; s = ½ (20 + 0) × 6 = 60 m

Answer: total distance = 50 + 200 + 60 = 310 m.

REVISE, REFLECT, REFINE 10 · BUS AND OBSTACLE

u = 36 km h⁻¹ = 10 m s⁻¹. Reaction time 0.5 s, braking a = −2.5 m s⁻², obstacle 30 m ahead.

Distance during reaction = 10 × 0.5 = 5 m

Braking distance: 0 = 10² + 2(−2.5)s, so s = 100 / 5 = 20 m

Answer: total = 5 + 20 = 25 m, which is less than 30 m. The bus stops 5 m before the obstacle.

REVISE, REFLECT, REFINE 14 · CAR ON STATE HIGHWAY

Phase 1: constant 6 m s⁻¹ for 2 min (120 s): s = 6 × 120 = 720 m.

Phase 2: u = 6 m s⁻¹, a = 1 m s⁻², t = 6 s. Final velocity = 12 m s⁻¹.

s = ut + ½at² = 6 × 6 + ½ × 1 × 36 = 36 + 18 = 54 m

Answer: total displacement = 720 + 54 = 774 m.

REVISE, REFLECT, REFINE 15 · CARS A AND B

Car A: 5 m s⁻¹ in 5 s, so a = 1 m s⁻²; s = ½ × 1 × 25 = 12.5 m.

Car B: 3 m s⁻¹ in 10 s, so a = 0.3 m s⁻²; s = ½ × 0.3 × 100 = 15 m.

Velocity-time graph: A is a straight line from (0, 0) to (5 s, 5 m s⁻¹); B is a straight line from (0, 0) to (10 s, 3 m s⁻¹). Each displacement is the area under its line.

REVISE, REFLECT, REFINE 16 · MINUTE HAND

Minute hand = 7 cm. From 6 PM to 7:30 PM = 90 min = 1.5 revolutions.

Distance = 1.5 × 2π × 7 cm ≈ 66 cm

Displacement: the tip goes from 12 to 6, which is across the dial = diameter = 14 cm

Average speed = 66 cm / 90 min ≈ 0.73 cm min⁻¹ = 66 / 5400 = 11/900 cm s⁻¹

Average velocity = 14 cm / 90 min ≈ 0.16 cm min⁻¹ = 14 / 5400 = 7/2700 cm s⁻¹, from 12 towards 6

NCERT Q&A

Answers to the Pause and Ponder questions and the Revise, Reflect, Refine exercises. Numerical working is in the Numericals tab.

Pause and Ponder

PAUSE AND PONDER 1

In the athlete running back and forth on a straight track (Fig. 4.4), when will the displacement be zero? What is the total distance in that case?

Answer: Displacement is zero when the athlete returns to the starting point O. The total distance is then twice the distance she ran out, for example 2 × 100 m = 200 m if she ran 100 m out and came back.

PAUSE AND PONDER 2

Fuel used by a vehicle depends on (i) total distance travelled or (ii) displacement? Justify.

Answer: (i) total distance travelled. The engine works over the whole path, including turns and back-tracking. Displacement ignores the path taken, so a vehicle that goes out and back uses fuel even though its displacement is zero.

PAUSE AND PONDER 3

A ball rolls down the inclined track in Fig. 4.6 from O to D. Is it a straight-line motion? Are the values of total distance and displacement at A, B, C and D equal or different?

Answer: Yes, it is straight-line motion along the track. The ball moves in one direction and does not turn back, so distance and displacement are equal at every point. From O, the values are: A = 40 cm, B = 50 cm, C = 70 cm, D = 100 cm, for both distance and displacement.

PAUSE AND PONDER 4

A family drives 200 km north in three hours, then 200 km south in two hours. Find the average speed and average velocity for the whole trip.

Answer: Average speed = 400 km / 5 h = 80 km h⁻¹. Average velocity = 0, because the displacement is zero.

PAUSE AND PONDER 5

(i) Under what condition is the magnitude of average velocity equal to average speed? (ii) When is the magnitude of average velocity zero while average speed is not zero?

Answer: (i) When the object moves in a straight line in one direction without turning back, so that distance equals displacement. (ii) When the object returns to its starting point (a closed path), so the displacement is zero but the distance is not.

Revise, Reflect, Refine

QUESTION 1

A man walks 250 m to a shop, returns home, goes back to the shop and returns home. Find the total distance and his displacement.

Answer: Total distance = 1000 m. Displacement = 0, since he ends where he started.

QUESTION 2

A student goes from the ground floor to the fourth floor, then down to the second floor, with each floor 3 m high. Find the total vertical distance and displacement.

Answer: (i) Total vertical distance = 12 m + 6 m = 18 m. (ii) Displacement = 6 m upward from the starting point.

QUESTION 3

A girl's scooter speedometer reads a constant value. Can the scooter be accelerating? If so, how?

Answer: Yes. If the scooter moves along a curved path at constant speed, its velocity changes because the direction changes. This is acceleration, even though the speedometer reading stays constant.

QUESTION 4

A car starts from rest and reaches 24 m s⁻¹ in 6 s. Find the average acceleration and the distance travelled.

Answer: a = 4 m s⁻²; distance = 72 m.

QUESTION 5

A motorbike with initial velocity 28 m s⁻¹ stops after 98 m with constant acceleration. Find the acceleration and the time taken.

Answer: a = −4 m s⁻²; time = 7 s.

QUESTION 6

Fig. 4.27 shows position-time graphs of two objects moving on parallel tracks in the same direction. Do A and B ever have equal velocity?

Answer: No. On a position-time graph, velocity is the slope. The two straight lines have different slopes, so their velocities are never equal. They cross at a point, which means they are at the same position at that instant, but their velocities differ.

QUESTION 7

Fig. 4.28 shows the position-time graphs of A and B from 0 to 10 s. Choose the correct option(s) about average velocity and average speed.

Answer: (i) and (ii). Both start at the origin and end at the same position, so their displacements are equal and their average velocities are equal. Both move only forward, so distance equals displacement for each, and their average speeds are equal too. Options (iii) and (iv) are wrong.

QUESTION 8

A truck slows from 54 km h⁻¹ to 36 km h⁻¹ in 36 s. Find the distance travelled, assuming constant acceleration.

Answer: 450 m (about 0.14 m s⁻² of deceleration).

QUESTION 9

A car accelerates uniformly from rest to 20 m s⁻¹ in 5 s, travels at 20 m s⁻¹ for 10 s, then stops in 6 s. Find the total distance.

Answer: 50 m + 200 m + 60 m = 310 m.

QUESTION 10

A bus at 36 km h⁻¹ sees an obstacle 30 m ahead. The driver reacts in 0.5 s, then brakes at 2.5 m s⁻². Will the bus stop before the obstacle?

Answer: Yes. The total stopping distance is 5 m + 20 m = 25 m, which is less than 30 m.

QUESTION 11

A student says, "The Earth moves around the Sun." Can an object kept on the Earth be considered at rest?

Answer: Yes and no. It is at rest with respect to the Earth, which is its reference point. It is in motion with respect to the Sun. Rest and motion depend on the chosen reference point.

QUESTION 12

Using the velocity-time graph of a cyclist (Fig. 4.30), find the displacement when the velocity is constant, when it decreases, and in the 120 s interval.

Answer: From the graph: 0–20 s, velocity rises from 0 to 3 m s⁻¹ (triangle area 30 m). 20–100 s, constant 3 m s⁻¹, displacement = 3 × 80 = 240 m. 100–120 s, velocity falls from 3 to 2 m s⁻¹ (area 50 m). Total displacement in 120 s = 320 m. Average acceleration = (2 − 0) / 120 ≈ 0.017 m s⁻² (from 0 to 120 s, the velocity changes from 0 to 2 m s⁻¹). For the decreasing part, shade 100–120 s (50 m).

QUESTION 13

Fig. 4.31 shows a runner's velocity in km h⁻¹ against time over about 6.5 h. Estimate the distance run.

Answer: The velocity stays between about 6.5 and 7.5 km h⁻¹. Distance = area under the graph, roughly the average velocity × time ≈ 7 km h⁻¹ × 6.5 h ≈ 45–50 km. Answers near this value are acceptable, since they are read from the graph.

QUESTION 14

A car moves at 6 m s⁻¹ for 2 minutes, then accelerates at 1 m s⁻² for 6 s. Find the displacement in the 2 min 6 s interval.

Answer: 720 m + 54 m = 774 m. The velocity-time graph is a horizontal line from 0 to 120 s at 6 m s⁻¹, then a straight line rising to 12 m s⁻¹ at 126 s. The area gives the same total.

QUESTION 15

Cars A and B start from rest with constant acceleration. A reaches 5 m s⁻¹ in 5 s; B reaches 3 m s⁻¹ in 10 s. Plot their velocity-time graphs and find each displacement.

Answer: Car A: a = 1 m s⁻², displacement 12.5 m. Car B: a = 0.3 m s⁻², displacement 15 m. Both graphs are straight lines from the origin: A reaches 5 m s⁻¹ at 5 s and B reaches 3 m s⁻¹ at 10 s.

QUESTION 16

Rohan studies from 6 PM to 7:30 PM. The minute hand is 7 cm long. Find the (i) distance, (ii) displacement, (iii) speed and (iv) velocity of the tip of the minute hand.

Answer: (i) Distance ≈ 66 cm. (ii) Displacement = 14 cm (from 12 to 6). (iii) Average speed ≈ 0.73 cm min⁻¹ = 11/900 cm s⁻¹. (iv) Average velocity ≈ 0.16 cm min⁻¹ = 7/2700 cm s⁻¹, directed from 12 towards 6.