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Chapter 4: Describing Motion Around Us

Grade 9 Science | Chapter 4

Describing Motion Around Us

How do we describe a moving object precisely? This chapter builds the language of motion, from distance and velocity to acceleration, motion graphs and the equations of motion.

Core Concepts

Key Principles

Worked Examples

Practice Sets

Contents

Introduction: Describing Motion

Distance and Displacement

Speed and Velocity

Acceleration

Graphs of Motion

The Equations of Motion

Key Reasoning (Principles)

Worked Examples (10)

Practice Sets A to D

10.

Summary and Exam Quick-Check

1. Introduction: Describing Motion

An object is in motion when its position changes with time compared with a fixed point. To describe motion precisely we need more than the everyday word fast; we need measured quantities such as how far an object travels, in which direction, how quickly, and whether its speed is changing. This chapter sets out those quantities and the graphs and equations that link them.

Some quantities, called scalars, have size only, such as distance and speed. Others, called vectors, have both size and direction, such as displacement, velocity and acceleration. Keeping the two apart is the first step to describing motion correctly.

Core idea

Speed tells you how fast an object moves; velocity tells you how fast and in which direction; acceleration tells you how quickly the velocity is changing.

2. Distance and Displacement

Distance is the total length of the path travelled, a scalar with size only. Displacement is the straight-line distance from start to finish in a stated direction, a vector. If you walk 3 m east and then 3 m back west, the distance travelled is 6 m but the displacement is zero, because you end where you began.

Quantity

Type

Meaning

Distance

Scalar

Total path length, size only

Displacement

Vector

Straight-line gap from start to end, with direction

3. Speed and Velocity

Speed is the distance travelled in unit time, speed = distance ÷ time, measured in metres per second. Velocity is the displacement in unit time, so it carries a direction. Average speed over a journey is the total distance divided by the total time, which can differ from the speed at any single instant.

4. Acceleration

Acceleration is the rate at which velocity changes, acceleration = change in velocity ÷ time taken, measured in metres per second squared. If velocity increases, acceleration is positive; if it decreases, the acceleration is negative and is often called deceleration or retardation.

Watch out

An object can be accelerating even at constant speed, if its direction is changing, because velocity includes direction. A car turning a corner at steady speed is still accelerating.

5. Graphs of Motion

Graphs make motion easy to read. On a distance to time graph, a straight sloping line shows uniform speed and a curve shows changing speed; the steeper the line, the faster the object. On a velocity to time graph, a sloping straight line shows uniform acceleration, and the area under the line gives the displacement.

Diagram 1 – Distance to Time Graphs

Distance to time graph with a straight line for uniform speed and a curve for acceleration

Fig 1. A straight line means uniform speed; an upward curve means the object is speeding up. The slope of the line is the speed.

Diagram 2 – Velocity to Time Graph

Velocity to time graph, straight line from u to v, shaded area under the line equal to displacement

Fig 2. For uniform acceleration the line is straight; the shaded area under it equals the displacement, and the slope equals the acceleration.

6. The Equations of Motion

For motion with uniform acceleration, three equations link the initial velocity u, final velocity v, acceleration a, time t and displacement s. They are v = u + at, s = ut + ½ a t², and v² = u² + 2 a s. These let us find any one quantity when the others are known, and they apply only while the acceleration stays constant.

Equation

Finds

Use When

v = u + at

final velocity

time and acceleration known

s = ut + ½ a t²

displacement

time and acceleration known

v² = u² + 2 a s

velocity or displacement

time not known

7. Key Reasoning (Principles)

Principle 1: Slope of a distance to time graph is speed

Speed is distance divided by time, which is exactly the slope of the line on a distance to time graph. A steeper line therefore means a greater speed.

Principle 2: Area under a velocity to time graph is displacement

On a velocity to time graph, the area between the line and the time axis adds up velocity multiplied by time, which gives the displacement covered.

Principle 3: The equations need uniform acceleration

The three equations of motion were derived assuming acceleration stays constant, so they may not be used when the acceleration changes during the motion.

8. Worked Examples

Example 1

Q: A car travels 240 m in 30 s. Find its average speed.

▶ Show Solution

Average speed = distance ÷ time.

= 240 ÷ 30 = 8 m/s.

Answer: 8 m/s.

Example 2

Q: Convert 36 km/h into metres per second.

▶ Show Solution

36 × 1000 ÷ 3600.

= 10 m/s.

Answer: 10 m/s.

Example 3

Q: A body starts from rest and reaches 20 m/s in 5 s. Find its acceleration.

▶ Show Solution

Acceleration = change in velocity ÷ time.

= (20 minus 0) ÷ 5 = 4 m/s².

Answer: 4 m/s².

Example 4

Q: Use v = u + at with u = 5 m/s, a = 2 m/s², t = 3 s.

▶ Show Solution

v = 5 + 2 × 3.

= 5 + 6 = 11 m/s.

Answer: 11 m/s.

Example 5

Q: A body starts from rest with a = 2 m/s². Find s after 4 s.

▶ Show Solution

s = ut + ½ a t² = 0 + ½ × 2 × 16.

= 16 m.

Answer: 16 m.

Example 6

Q: Using v² = u² + 2as with u = 0, a = 2 m/s², s = 25 m, find v.

▶ Show Solution

v² = 0 + 2 × 2 × 25 = 100.

v = √100 = 10 m/s.

Answer: 10 m/s.

Example 7

Q: An object moves at a steady 15 m/s for 8 s. Find the displacement.

▶ Show Solution

Displacement = velocity × time = 15 × 8.

= 120 m.

Answer: 120 m.

Example 8

Q: A car slows from 20 m/s to rest in 4 s. Find its acceleration.

▶ Show Solution

a = (0 minus 20) ÷ 4 = minus 5 m/s².

The minus sign shows it is slowing down.

Answer: minus 5 m/s² (deceleration).

Example 9

Q: A runner goes 50 m east, then 50 m back west. Find the distance and the displacement.

▶ Show Solution

Distance = 50 + 50 = 100 m (total path).

Displacement = 0 m, since start and end are the same point.

Answer: Distance 100 m; displacement 0 m.

Example 10

Q: A car covers 100 m in 20 s, then 60 m in 10 s. Find the average speed for the whole trip.

▶ Show Solution

Total distance = 100 + 60 = 160 m; total time = 20 + 10 = 30 s.

Average speed = 160 ÷ 30 = 5.33 m/s (to 3 figures).

Answer: About 5.33 m/s.

9. Practice Sets A to D

Set A – Multiple Choice (Basic)

1. The SI unit of acceleration is: (a) m/s (b) m/s² (c) m (d) s

2. Which is a vector quantity? (a) distance (b) speed (c) displacement (d) time

3. On a distance to time graph, the slope gives the: (a) acceleration (b) speed (c) time (d) distance

4. A straight, sloping line on a velocity to time graph shows: (a) rest (b) uniform speed (c) uniform acceleration (d) no motion

5. 36 km/h equals: (a) 6 m/s (b) 10 m/s (c) 36 m/s (d) 100 m/s

▶ Reveal Answers

1. (b) m/s².

2. (c) displacement.

3. (b) speed.

4. (c) uniform acceleration.

5. (b) 10 m/s.

Set B – Short Answer (Understanding)

1. State one difference between distance and displacement.

2. Define acceleration and give its SI unit.

3. What does the area under a velocity to time graph represent?

4. When is an object said to have uniform velocity?

5. Why can an object accelerate while moving at constant speed?

▶ Reveal Answers

1. Distance is total path length (scalar); displacement is the straight-line gap from start to end with direction (vector).

2. Acceleration is the rate of change of velocity; its SI unit is metre per second squared (m/s²).

3. It represents the displacement covered by the object.

4. When it covers equal displacements in equal intervals of time in the same direction.

5. Because velocity includes direction, so changing direction changes velocity even if speed stays the same.

Set C – Application and Reasoning

1. A train moves 360 m in 40 s. Find its average speed.

2. A body accelerates from 4 m/s to 16 m/s in 6 s. Find its acceleration.

3. Find the displacement of a body moving at a steady 12 m/s for 9 s.

4. Using v = u + at, find v when u = 0, a = 3 m/s², t = 5 s.

5. A cyclist travels 100 m north then 100 m south. State the distance and displacement.

▶ Reveal Answers

1. 360 ÷ 40 = 9 m/s.

2. (16 minus 4) ÷ 6 = 2 m/s².

3. 12 × 9 = 108 m.

4. v = 0 + 3 × 5 = 15 m/s.

5. Distance 200 m; displacement 0 m.

Set D – Higher Order (Challenge)

1. A car starts from rest with a = 2 m/s². Find its velocity and displacement after 10 s.

2. Using v² = u² + 2as, find the distance a body moving at 10 m/s takes to stop with a = minus 2 m/s².

3. Explain, using a graph, why a steeper distance to time line means a faster object.

4. A bus covers 90 m in 6 s, then 90 m in 9 s. Find the average speed and explain why it is not 15 m/s.

5. Why can the equations of motion not be used when acceleration keeps changing?

▶ Reveal Answers

1. v = u + at = 0 + 2 × 10 = 20 m/s; s = ut + ½ a t² = 0 + ½ × 2 × 100 = 100 m.

2. 0 = 100 + 2 × (minus 2) × s, so 4s = 100, giving s = 25 m.

3. The slope of a distance to time line equals speed, so a steeper line shows more distance per unit time, that is a greater speed.

4. Total distance 180 m in 15 s, so average speed = 12 m/s; it is not 15 m/s because the two parts took different times, so a simple average of speeds is wrong.

5. The equations were derived assuming constant acceleration, so they do not hold once the acceleration varies.

Chapter Summary

Scalars and Vectors

Distance and speed have size only; displacement, velocity and acceleration also have direction.

Speed and Velocity

Speed = distance ÷ time; velocity is speed with a direction.

Acceleration

Rate of change of velocity, in m/s²; negative when slowing down.

Motion Graphs

Slope of a distance to time graph is speed; area under a velocity to time graph is displacement.

Equations of Motion

v = u + at, s = ut + ½ a t², v² = u² + 2as, for uniform acceleration.

Average Speed

Total distance divided by total time, not the average of separate speeds.

Quantity

Unit

Symbol

Speed

metre per second

m/s

Acceleration

metre per second squared

m/s²

Displacement

metre

8-Point Exam Quick-Check

Distance is a scalar; displacement is a vector with direction.

Speed = distance divided by time; velocity is speed with a direction.

Acceleration is the rate of change of velocity, measured in m/s squared.

The slope of a distance to time graph gives the speed.

The area under a velocity to time graph gives the displacement.

The equations of motion are v = u + at, s = ut + half a t squared, v squared = u squared + 2as.

These equations apply only when the acceleration is uniform.

Average speed is total distance over total time, not the average of speeds.

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Class 9 Science Chapter 4: Describing Motion Around Us, Complete Notes and Practice

This revision guide follows the NCERT 2026 to 27 Exploration syllabus and builds the language of motion, covering distance and displacement, speed and velocity, acceleration, distance to time and velocity to time graphs, and the three equations of motion for uniform acceleration, with two graphs, ten worked examples and graded practice. Visit SchoolRevise.com to revise, practise and excel.