Class 9 · Physics · Lahore

Chapter 2: Kinematics

Kinematics describes motion — rest, speed, velocity and acceleration — without asking what causes it. This chapter's exercise is solved line by line below.

ReadyExercise 2.1

questions: 9

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Question list 1 / 9

Concept 4

Numerical 5

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Concept

1 Concept

Define rest and motion.

1 A body is said to be at rest if it does not change its position with respect to its surroundings.

2 A body is said to be in motion if it changes its position with respect to its surroundings.

3 Both are relative. A passenger sitting in a moving bus is at rest with respect to the bus, but in motion with respect to the road.

2 Concept

Differentiate between scalar and vector quantities.

1 A scalar quantity is described completely by its magnitude together with a proper unit. Mass, time, distance and speed are scalars.

2 A vector quantity needs both a magnitude and a direction to be described completely. Displacement, velocity, force and acceleration are vectors.

3 In writing, a vector is shown in bold type or with an arrow drawn above the symbol.

3 Concept

Differentiate between speed and velocity.

1 Speed is the distance covered in unit time. It is a scalar, so it has magnitude only.

2 Velocity is the displacement covered in unit time. It is a vector, so it carries a direction as well.

3 A car going round a circular track at a steady 40 km h⁻¹ has constant speed but changing velocity, because its direction of motion keeps changing.

4 Concept

What is meant by uniform acceleration?

1 A body has uniform acceleration if its velocity changes by equal amounts in equal intervals of time, however small those intervals are taken.

2 A body falling freely near the surface of the earth is the standard example: its velocity increases by about 10 m s⁻¹ every second.

Numerical

5 Numerical

A train moves with a uniform velocity of 36 km h⁻¹ for 10 s. Find the distance it covers.

v = 36 km h⁻¹t = 10 s

CONVERT v = 36 × 1000 ÷ 3600 = 10 m s⁻¹

FORMULA S = v × t

PUT VALUES S = 10 × 10

Answer Distance = 100 m

6 Numerical

A car starting from rest attains a velocity of 20 m s⁻¹ in 10 s. Find its acceleration and the distance it covers.

vᵢ = 0 v_f = 20 m s⁻¹t = 10 s

FORMULA a = (v_f − vᵢ) ÷ t

PUT VALUES a = (20 − 0) ÷ 10 = 2 m s⁻²

THEN S = vᵢt + ½at² = 0 + ½(2)(10)²

Answer Acceleration and distance = a = 2 m s⁻² , S = 100 m

7 Numerical

A bicycle moving at 10 m s⁻¹ slows down at 0.5 m s⁻². How long does it take to stop, and how far does it travel in that time?

vᵢ = 10 m s⁻¹ v_f = 0 a = −0.5 m s⁻²

FORMULA t = (v_f − vᵢ) ÷ a

PUT VALUES t = (0 − 10) ÷ (−0.5) = 20 s

THEN 2aS = v_f² − vᵢ² → S = (0 − 100) ÷ (2 × −0.5)

Answer Time and distance = t = 20 s , S = 100 m

8 Numerical

A body falls freely from rest. Taking g = 10 m s⁻², find its velocity and the height fallen after 5 s.

vᵢ = 0 g = 10 m s⁻² t = 5 s

FORMULA v_f = vᵢ + g t

PUT VALUES v_f = 0 + 10 × 5 = 50 m s⁻¹

THEN h = vᵢt + ½g t² = 0 + ½(10)(5)²

Answer Velocity and height = v = 50 m s⁻¹ , h = 125 m

9 Numerical

A car covers 100 m in 5 s. Find its average speed.

S = 100 mt = 5 s

FORMULA v = S ÷ t

PUT VALUES v = 100 ÷ 5

Answer Average speed = 20 m s⁻¹

This sheet was checked against the Lahore 2024 marking scheme.