Describing Motion Around Us
Class 9 Science chapter notes on motion, distance, displacement, speed, velocity, acceleration, graphs, equations of motion and circular motion — explained for CBSE school exams and Foundation/Olympiad preparation.
Chapter Overview
Motion is one of the most important basic ideas of Physics. It helps us describe how objects change their position with time.
What Students Will Learn
- Meaning of rest and motion
- Uniform and non-uniform motion
- Scalar and vector quantities
- Distance and displacement
- Speed, velocity and acceleration
- Distance-time, displacement-time and velocity-time graphs
- Equations of motion for constant acceleration
- Uniform circular motion and accelerated motion
Why This Chapter Is Important
This chapter builds the foundation for Force, Laws of Motion, Gravitation, Work, Energy and Power. In higher Physics, almost every topic uses the language of motion.
CBSE frequently asks definitions, graph interpretation, formula-based numericals, derivations of equations of motion and conceptual questions on distance versus displacement.
Table of Contents
Use these links to revise topic-wise.
Introduction to Motion
Motion means change in position of an object with time, with respect to a reference point.
An object is said to be in motion if its position changes with time with respect to a chosen reference point or observer.
Reference Point
A reference point is a fixed point used to describe whether an object is moving or at rest. Motion is always described with respect to a reference point.
Example 1: Passenger in a Bus
A passenger sitting inside a moving bus is at rest with respect to another passenger sitting beside him, but he is in motion with respect to a person standing on the road.
Example 2: Earth
We feel at rest on Earth, but Earth is moving around the Sun. This shows that rest and motion are relative terms.
Rest and motion are not absolute. They depend on the observer or reference point.
In advanced Physics, the reference point is part of a reference frame. A full reference frame includes origin, coordinate axes and a clock.
Uniform and Non-Uniform Motion
Motion can be classified according to how equal or unequal the distances covered are in equal time intervals.
Uniform Motion
An object is said to be in uniform motion if it covers equal distances in equal intervals of time, however small those time intervals may be.
Example: A car moving at a constant speed of 40 km/h on a straight road.
Non-Uniform Motion
An object is said to be in non-uniform motion if it covers unequal distances in equal intervals of time.
Example: A bus moving in city traffic because it stops, starts, slows down and speeds up repeatedly.
| Basis | Uniform Motion | Non-Uniform Motion |
|---|---|---|
| Distance covered | Equal distances in equal time intervals | Unequal distances in equal time intervals |
| Speed | Constant | Changing |
| Graph | Straight line distance-time graph | Curved or irregular distance-time graph |
| Example | Train moving steadily on a straight track | Car moving through traffic |
For uniform motion, write “equal distances in equal intervals of time.” Do not write only “same speed,” because the CBSE definition is based on distance and time intervals.
Scalar and Vector Quantities
Physical quantities are classified depending on whether direction is required or not.
A scalar quantity has magnitude only. It does not need direction.
Examples: Distance, speed, time, mass, temperature, volume.
A vector quantity has both magnitude and direction.
Examples: Displacement, velocity, acceleration, force, momentum.
| Scalar | Vector |
|---|---|
| Magnitude only | Magnitude and direction |
| Direction is not mentioned | Direction must be mentioned |
| Added by ordinary arithmetic | Added using vector rules |
| Example: 20 m | Example: 20 m east |
Students often write velocity and speed as the same thing. Speed is scalar, but velocity is vector because it includes direction.
Distance and Displacement
Distance and displacement both describe how far an object has moved, but they are not the same.
Distance is the total length of the actual path travelled by an object.
Nature: Scalar quantity
SI Unit: metre, m
Displacement is the shortest straight-line distance between the initial and final positions of an object, along with direction.
Nature: Vector quantity
SI Unit: metre, m
| Point of Difference | Distance | Displacement |
|---|---|---|
| Meaning | Total path length travelled | Shortest straight-line change in position |
| Quantity type | Scalar | Vector |
| Direction | No direction required | Direction required |
| Can be zero? | Only if object has not moved | Can be zero even after motion |
| Can be negative? | No | Yes, if direction is opposite to chosen positive direction |
| Relation | Always greater than or equal to displacement magnitude | Magnitude is always less than or equal to distance |
If an object returns to its starting point, its initial and final positions are the same. Therefore, displacement becomes zero, even though distance travelled is not zero.
Example: Round Trip
A student walks 100 m from home to a shop and then returns 100 m back home.
- Total distance = 100 m + 100 m = 200 m
- Displacement = 0 m because final position = initial position
Numerical 1: Straight Line Displacement
A boy walks 30 m east and then 20 m west. Find distance and displacement.
Distance = 30 + 20 = 50 m
Displacement = 30 m east − 20 m west = 10 m east
Numerical 2: Perpendicular Displacement
A girl walks 3 m east and then 4 m north. Find distance and displacement.
Distance = 3 + 4 = 7 m
Displacement = √(3² + 4²) = √25 = 5 m
Direction is north-east.
Speed and Velocity
Speed tells how fast an object moves. Velocity tells how fast and in which direction an object changes its position.
Speed is the distance travelled by an object per unit time.
Velocity is the displacement of an object per unit time.
| Basis | Speed | Velocity |
|---|---|---|
| Definition | Distance travelled per unit time | Displacement per unit time |
| Quantity type | Scalar | Vector |
| Direction | No direction | Direction required |
| Can be negative? | No | Yes |
| Zero after round trip? | Average speed is not zero | Average velocity can be zero |
Average Speed and Average Velocity
Numerical 3: Speed
A car travels 150 km in 3 hours. Find its average speed.
Average speed = 150 / 3 = 50 km/h
Numerical 4: Average Speed and Velocity in Round Trip
A cyclist travels 2 km east and returns 2 km west in 20 minutes. Find average speed and average velocity.
Total distance = 4 km
Total displacement = 0 km
Total time = 20 min = 1/3 h
Average speed = 4 ÷ 1/3 = 12 km/h
Average velocity = 0 ÷ 1/3 = 0 km/h
Whenever the question says “total path covered,” use distance and speed. Whenever it says “change in position,” use displacement and velocity.
Motion Graphs
Graphs convert motion into a visual form. They help us calculate speed, velocity, acceleration and displacement.
Export these graph diagrams from Canva in PNG format and replace or place them above these HTML graph blocks for a more premium visual look.
Straight inclined line shows uniform speed. Slope gives speed.
Horizontal line means distance from reference point is constant.
Changing slope means speed is changing.
Slope of displacement-time graph gives velocity.
Straight inclined line shows constant acceleration.
Area under velocity-time graph gives displacement.
Important Graph Rules
| Graph | Slope Gives | Area Gives |
|---|---|---|
| Distance-time graph | Speed | Not used at Class 9 level |
| Displacement-time graph | Velocity | Not used at Class 9 level |
| Velocity-time graph | Acceleration | Displacement |
Graph Numerical
A displacement-time graph changes from 0 m to 40 m in 8 s in a straight line. Find velocity.
Velocity = slope = change in displacement / time
v = 40 / 8 = 5 m/s
Acceleration
Acceleration describes how quickly velocity changes with time.
Acceleration is the rate of change of velocity with time.
Here, u = initial velocity, v = final velocity, t = time and a = acceleration.
Positive Acceleration
When velocity increases with time, acceleration is positive.
Example: A car speeding up from rest.
Negative Acceleration / Retardation
When velocity decreases with time, acceleration is negative. It is also called retardation or deceleration.
Example: A car slowing down before a red light.
Numerical 5: Acceleration
A car increases its velocity from 10 m/s to 30 m/s in 5 s. Find acceleration.
a = (v − u) / t
a = (30 − 10) / 5 = 20 / 5 = 4 m/s²
Numerical 6: Retardation
A bike slows down from 20 m/s to 5 m/s in 3 s. Find acceleration.
a = (5 − 20) / 3 = −15 / 3 = −5 m/s²
The negative sign shows retardation.
Do not confuse acceleration with speed. A fast-moving object can have zero acceleration if its velocity is constant.
Kinematic Equations for Motion in a Straight Line
These equations are used only when acceleration is constant and motion is along a straight line.
CBSE often asks derivation of the three equations of motion using graph method and numerical problems based on them.
Symbols Used
| Symbol | Meaning | SI Unit |
|---|---|---|
| u | Initial velocity | m/s |
| v | Final velocity | m/s |
| a | Acceleration | m/s² |
| t | Time | s |
| s | Displacement | m |
Derivation 1: v = u + at
Acceleration is rate of change of velocity.
Multiplying both sides by t:
Rearranging:
Derivation 2: s = ut + ½at²
Average velocity for uniformly accelerated motion:
Displacement = average velocity × time
Using v = u + at:
Derivation 3: v² = u² + 2as
From first equation:
Using displacement formula:
Substitute t:
Numerical 7: First Equation
A car starts from rest and accelerates at 2 m/s² for 10 s. Find final velocity.
u = 0, a = 2 m/s², t = 10 s
v = u + at = 0 + 2 × 10 = 20 m/s
Numerical 8: Second Equation
A body starts from rest and moves with acceleration 3 m/s² for 4 s. Find displacement.
u = 0, a = 3 m/s², t = 4 s
s = ut + ½at²
s = 0 + ½ × 3 × 4² = 1.5 × 16 = 24 m
Numerical 9: Third Equation
A car moving at 10 m/s accelerates at 2 m/s² over a distance of 75 m. Find final velocity.
u = 10 m/s, a = 2 m/s², s = 75 m
v² = u² + 2as
v² = 10² + 2 × 2 × 75 = 100 + 300 = 400
v = 20 m/s
Numerical 10: Retardation Using Third Equation
A vehicle moving at 25 m/s stops after covering 125 m. Find retardation.
u = 25 m/s, v = 0, s = 125 m
v² = u² + 2as
0 = 625 + 2 × a × 125
250a = −625
a = −2.5 m/s²
Retardation = 2.5 m/s²
Motion in a Plane
Motion in a plane means motion in two dimensions, usually along x-axis and y-axis.
In Class 9, most numerical problems are based on motion in a straight line. However, understanding motion in a plane helps in displacement problems and future topics such as projectile motion.
One-Dimensional Motion
Motion along a straight line only.
Example: A train moving on a straight track.
Two-Dimensional Motion
Motion in a plane involving two directions.
Example: A football moving across a field.
In two-dimensional motion, displacement can be found using Pythagoras theorem when two perpendicular components are given.
Uniform Circular Motion
Uniform circular motion is a special type of accelerated motion even when speed remains constant.
When an object moves along a circular path with uniform speed, its motion is called uniform circular motion.
Velocity is a vector quantity. Even if speed remains constant, direction changes continuously in circular motion. Since velocity changes due to change in direction, the object is accelerated.
Examples
- Stone tied to a string and whirled in a circle
- Moon revolving around Earth
- Tip of fan blade rotating
- Athlete running on a circular track
Important Point
In uniform circular motion, speed is constant but velocity is changing because direction is changing at every point.
Numerical 11: Circular Motion Speed
A body moves in a circular path of radius 7 m and completes one round in 11 s. Find speed. Use π = 22/7.
v = 2πr / T
v = 2 × 22/7 × 7 / 11
v = 44 / 11 = 4 m/s
Important Terms
Revise these definitions before exams.
Concept Map / Flowchart
A quick visual connection between important ideas.
Important Formula Sheet
Memorise these formulas with units.
NCERT Focus Section
These lines and concepts must be understood clearly for CBSE exams.
- Motion is described by change in position with time.
- Distance is actual path length; displacement is shortest directed distance.
- Uniform motion means equal distances in equal intervals of time.
- Speed is a scalar quantity; velocity is a vector quantity.
- Acceleration is change in velocity per unit time.
- The slope of a distance-time graph gives speed.
- The slope of a velocity-time graph gives acceleration.
- The area under a velocity-time graph gives displacement.
- Uniform circular motion is accelerated because direction of velocity changes continuously.
Why is a body moving with uniform speed in a circular path said to be accelerated?
Answer: Because velocity changes continuously due to continuous change in direction.
Board Exam Focus
Most repeated question patterns from this chapter.
Very Important Concepts
- Distance versus displacement
- Speed versus velocity
- Uniform versus non-uniform motion
- Graph slope interpretation
- Derivation of three equations of motion
- Uniform circular motion as accelerated motion
Common Question Patterns
- Define and differentiate
- Interpret graph
- Calculate speed, velocity or acceleration
- Use equations of motion
- Explain zero displacement after motion
- Explain circular motion conceptually
Foundation / Olympiad Edge
These points help students go beyond basic board-level learning.
An object can have constant speed but changing velocity if its direction changes. Example: uniform circular motion.
Average speed is always based on total distance, but average velocity depends on displacement. Hence, for a round trip, average velocity is zero.
Negative velocity means motion in the opposite direction of the chosen positive direction. It does not mean the object is slowing down.
Negative acceleration may increase speed if the object is moving in the negative direction. Sign depends on direction, not only on slowing down.
Common Mistakes
Avoid these errors in exams.
Quick Revision Mind Map
Revise the entire chapter in a few minutes.
Practice Questions
Board-level, conceptual and Foundation-style questions for complete preparation.
A. MCQs
- Which of the following is a scalar quantity?
A. Velocity B. Displacement C. Speed D. Acceleration - The SI unit of acceleration is:
A. m/s B. m/s² C. km/h D. m²/s - Slope of velocity-time graph gives:
A. Speed B. Distance C. Acceleration D. Displacement - Area under velocity-time graph gives:
A. Displacement B. Acceleration C. Speed D. Time - If a body returns to its starting point, its displacement is:
A. Maximum B. Zero C. Equal to distance D. Negative always - Uniform circular motion is accelerated because:
A. Speed changes B. Direction changes C. Mass changes D. Radius changes - The first equation of motion is:
A. v = u + at B. s = ut + ½at² C. v² = u² + 2as D. s = vt - A horizontal distance-time graph shows:
A. Uniform speed B. Rest C. Acceleration D. Retardation
B. Fill in the Blanks
- Distance is a ______ quantity.
- Velocity is displacement per unit ______.
- Negative acceleration is also called ______.
- The slope of displacement-time graph gives ______.
- The speed of circular motion is given by ______.
C. True / False
- Displacement can be zero even if distance is not zero.
- Speed is a vector quantity.
- Uniform circular motion is not accelerated.
- Acceleration can be negative.
- Distance is always less than displacement.
D. Assertion-Reason Questions
Options: A. Both assertion and reason are true and reason explains assertion. B. Both are true but reason does not explain assertion. C. Assertion is true but reason is false. D. Assertion is false but reason is true.
- Assertion: Displacement can be zero even if distance travelled is not zero.
Reason: Displacement depends only on initial and final positions. - Assertion: Speed and velocity always have the same value.
Reason: Speed is scalar and velocity is vector. - Assertion: A body moving with uniform speed in a circle is accelerated.
Reason: Its velocity changes due to continuous change in direction. - Assertion: Slope of velocity-time graph gives acceleration.
Reason: Acceleration is change in velocity per unit time.
E. Very Short Answer Questions
- Define motion.
- What is a reference point?
- Write the SI unit of velocity.
- What is retardation?
- When is displacement equal to distance?
F. Short Answer Questions
- Differentiate between distance and displacement.
- Differentiate between speed and velocity.
- Why is uniform circular motion called accelerated motion?
- What does the slope of a distance-time graph represent?
- Explain uniform and non-uniform motion with examples.
G. Long Answer Questions
- Derive the three equations of motion for uniformly accelerated motion.
- Explain distance-time graph and velocity-time graph with diagrams.
- Explain the concept of acceleration and retardation with examples and numericals.
- Explain scalar and vector quantities. Give examples and compare speed with velocity.
H. Case Study Questions
Case Study 1: School Bus Motion
A school bus starts from rest, accelerates uniformly for some time, moves with constant velocity on a straight road, and then slows down before stopping near the school gate.
- What is the initial velocity of the bus?
- Which part of motion shows positive acceleration?
- Which part shows retardation?
- What will be the slope of velocity-time graph during constant velocity?
Case Study 2: Circular Track
An athlete runs one complete round of a circular track of radius 70 m in 44 s with uniform speed.
- What is the displacement after one complete round?
- What is the distance covered?
- Is the athlete’s velocity constant?
- Why is the motion accelerated?
I. Numericals
- A car travels 100 m in 5 s. Find its speed.
- A body starts from rest and reaches 20 m/s in 4 s. Find acceleration.
- A train moving at 10 m/s accelerates at 1 m/s² for 20 s. Find final velocity.
- A body starts from rest and accelerates at 4 m/s² for 5 s. Find displacement.
- A car moving at 15 m/s stops in 5 s. Find acceleration.
- A cyclist covers 300 m in 60 s. Find average speed.
- A person walks 6 m east and 8 m north. Find distance and displacement.
- A body has initial velocity 5 m/s, acceleration 2 m/s² and displacement 60 m. Find final velocity.
Answer Key
Use this key for quick checking after practice.
A. MCQs
1. C, 2. B, 3. C, 4. A, 5. B, 6. B, 7. A, 8. B
B. Fill in the Blanks
1. scalar, 2. time, 3. retardation/deceleration, 4. velocity, 5. v = 2πr/T
C. True / False
1. True, 2. False, 3. False, 4. True, 5. False
D. Assertion-Reason
1. A, 2. D, 3. A, 4. A
E. Very Short Answers
- Motion is change in position with time.
- A reference point is a fixed point used to describe position or motion.
- m/s
- Retardation is negative acceleration.
- When motion is along a straight line in one direction.
I. Numerical Answers
- 20 m/s
- 5 m/s²
- 30 m/s
- 50 m
- −3 m/s²
- 5 m/s
- Distance = 14 m, Displacement = 10 m
- v² = 5² + 2 × 2 × 60 = 265, so v = √265 ≈ 16.28 m/s
One-Page Final Revision Sheet
Perfect for last-minute revision before class tests and CBSE exams.
Describing Motion Around Us — Final Revision
Core Definitions
- Motion: Change in position with time.
- Distance: Total path length travelled.
- Displacement: Shortest directed distance between initial and final positions.
- Speed: Distance per unit time.
- Velocity: Displacement per unit time.
- Acceleration: Rate of change of velocity.
Must-Know Formulas
Graph Rules
- Slope of distance-time graph = speed.
- Slope of displacement-time graph = velocity.
- Slope of velocity-time graph = acceleration.
- Area under velocity-time graph = displacement.
Most Important Exam Points
- Displacement can be zero even after motion if final position equals initial position.
- Distance is never negative.
- Velocity can be positive, negative or zero.
- Uniform circular motion is accelerated due to continuous change in direction.
- Kinematic equations apply only for constant acceleration.
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