Motion is the change in position of an object with respect to a reference point over time. Important Questions Class 9 Science Chapter 4 cover Describing Motion Around Us, including distance, displacement, speed, velocity, acceleration, graphs, equations of motion, and uniform circular motion.
A moving train, a runner on a circular track, a braking car, and a falling stone all show motion in different ways.
Class 9 Science Chapter 4 helps students describe motion using exact physics terms. The chapter becomes easier when students separate scalar and vector quantities, read motion graphs correctly, and choose the right equation of motion for each numerical.
Key Takeaways
| Detail |
Information |
| Chapter |
Class 9 Science Chapter 4 |
| Topic |
Describing Motion Around Us |
| Syllabus |
CBSE 2026 |
| Question Types |
VSA, Short Answer, Long Answer, Numericals, Graph-Based, Assertion-Reason, Case Study |
| Key Topics |
Distance, Displacement, Speed, Velocity, Acceleration, Motion Graphs, Equations of Motion |
| High-Value Areas |
Numericals, Velocity-Time Graphs, Equations of Motion, Uniform Circular Motion |
| Best Revision Method |
Learn formulas, practise graphs, then solve mixed numericals |
Class 9 Science Chapter List
Class 9 Science Chapter 4 Describing Motion Around Us: Topics Covered
Class 9 science chapter 4 important questions with answers mostly test concept clarity, graph reading, formula selection, and numerical accuracy.
Revise these topics before starting the question bank.
- Motion and rest
- Position and reference point
- Linear motion and uniform motion
- Distance and displacement
- Average speed and average velocity
- Acceleration and retardation
- Distance-time graphs
- Velocity-time graphs
- Area under velocity-time graph
- Equations of motion
- Uniform circular motion
- Road safety and braking distance
Important Questions Class 9 Science Chapter 4 with Answers
Important Questions Class 9 Science Chapter 4 should be practised in stages. Start with definitions, then short answers, numericals, graph questions, assertion-reason, and case studies.
These class 9 science chapter 4 question answer sets help students revise both theory and numerical patterns.
Very Short Answer Important Questions Class 9 Science Chapter 4
Very short answer questions test definitions, units, and basic distinctions.
Write exact terms and formulas where needed.
Class 9 Science Chapter 4 Important Questions with Answers
Q1. What is motion?
Ans. Motion is the change in position of an object with respect to a fixed reference point over time.
Q2. What is the difference between distance and displacement?
Ans. Distance is the total path length covered by an object.
Displacement is the shortest straight-line distance between the initial and final positions.
Distance is scalar. Displacement is vector.
Q3. Define uniform motion.
Ans. An object is in uniform motion when it covers equal distances in equal time intervals.
Its speed remains constant.
Q4. What is acceleration?
Ans. Acceleration is the rate of change of velocity with time.
a = (v - u) / t
The SI unit of acceleration is m/s².
Q5. What is retardation?
Ans. Retardation is negative acceleration.
It occurs when the velocity of an object decreases with time.
Q6. State the SI unit of speed.
Ans. The SI unit of speed is metre per second, written as m/s.
Q7. What is uniform circular motion?
Ans. Uniform circular motion is motion along a circular path at constant speed.
Velocity changes continuously because direction changes at every point.
Q8. Can displacement be greater than distance?
Ans. No, displacement can never be greater than distance.
Displacement can be equal to distance only when motion is in one straight direction.
Short Answer Questions Class 9 Science Chapter 4
Short answer questions test concept clarity.
Use a formula, example, or comparison table wherever useful.
Speed and Velocity Questions Class 9
Q1. Distinguish between speed and velocity.
Ans.
| Feature |
Speed |
Velocity |
| Meaning |
Distance travelled per unit time |
Displacement per unit time |
| Type |
Scalar |
Vector |
| Formula |
Distance / Time |
Displacement / Time |
| Direction |
Not required |
Required |
| Value |
Always non-negative |
Can be positive, negative, or zero |
Q2. An object travels 4 km east and then 3 km north. Find distance and displacement.
Ans. Distance = 4 + 3 = 7 km
Displacement = √(4² + 3²)
= √(16 + 9)
= √25
= 5 km
Q3. What does the slope of a distance-time graph represent?
Ans. The slope of a distance-time graph represents speed.
A steeper slope means greater speed. A horizontal line means the object is at rest.
Q4. What does the area under a velocity-time graph represent?
Ans. The area under a velocity-time graph represents displacement.
For uniform motion, the area is a rectangle. For uniformly accelerated motion, it may be a triangle or trapezium.
Q5. Why is uniform circular motion called accelerated motion?
Ans. In uniform circular motion, speed remains constant but direction changes continuously.
Since velocity depends on direction, velocity changes. Change in velocity means acceleration.
Q6. A car starts from rest and reaches 20 m/s in 5 seconds. Find acceleration.
Ans.
a = (v - u) / t
u = 0 m/s
v = 20 m/s
t = 5 s
a = (20 - 0) / 5
a = 4 m/s²
Long Answer Important Questions Class 9 Science Chapter 4
Long answers should include definitions, formulas, explanation, and conclusion.
For derivations, write symbols clearly.
Equations of Motion Class 9 Science Questions
Q1. Derive the three equations of motion for uniform acceleration.
Ans. Consider an object with initial velocity u, final velocity v, acceleration a, time t, and displacement s.
First equation:
a = (v - u) / t
Rearranging,
v = u + at
Second equation:
Average velocity = (u + v) / 2
Displacement = Average velocity × Time
s = [(u + v) / 2] × t
Using v = u + at,
s = [(u + u + at) / 2] × t
s = [(2u + at) / 2] × t
s = ut + ½at²
Third equation:
From v = u + at,
t = (v - u) / a
Using s = [(u + v) / 2] × t,
s = [(u + v) / 2] × [(v - u) / a]
2as = (v + u)(v - u)
2as = v² - u²
v² = u² + 2as
These equations apply only to uniformly accelerated straight-line motion.
Q2. Describe how velocity-time graphs look in four cases of motion.
Ans.
| Case |
Velocity-Time Graph |
Meaning |
| Uniform motion |
Horizontal straight line |
Velocity is constant |
| Uniform acceleration from rest |
Rising straight line from origin |
Velocity increases uniformly |
| Uniform retardation |
Falling straight line |
Velocity decreases uniformly |
| Non-uniform acceleration |
Curved line |
Acceleration changes with time |
The slope of a velocity-time graph gives acceleration. The area under the graph gives displacement.
Numericals Class 9 Science Chapter 4 Describing Motion Around Us
Numericals in this chapter need correct formula selection.
Write given values, convert units, substitute carefully, and include the final unit.
Average Speed Numericals Class 9 Science Chapter 4
Average speed is calculated using total distance and total time.
Use SI units when the answer is required in m/s.
Q1. A car covers 150 km in 3 hours. Calculate average speed.
Ans.
Average speed = Total distance / Total time
= 150 / 3
= 50 km/h
Q2. A person walks 2 km in 30 minutes. Find speed in m/s.
Ans.
Distance = 2 km = 2000 m
Time = 30 minutes = 1800 s
Speed = Distance / Time
= 2000 / 1800
= 1.11 m/s
Q3. An object travels first 30 m in 5 s and next 30 m in 3 s. Find average speed.
Ans.
Total distance = 30 + 30 = 60 m
Total time = 5 + 3 = 8 s
Average speed = 60 / 8
= 7.5 m/s
Average Velocity Numericals Class 9 Science Chapter 4
Average velocity depends on displacement, not total distance.
Always check direction.
Average Velocity Questions
Q1. A ball moves 40 m east and then 30 m west in 10 seconds. Find average speed and average velocity.
Ans.
Total distance = 40 + 30 = 70 m
Average speed = 70 / 10
= 7 m/s
Displacement = 40 m east - 30 m west
= 10 m east
Average velocity = Displacement / Time
= 10 / 10
= 1 m/s east
Q2. An athlete runs one full lap of a circular track of radius 70 m in 44 seconds. Find average speed and average velocity.
Ans.
Distance = Circumference
= 2πr
= 2 × 22/7 × 70
= 440 m
Average speed = 440 / 44
= 10 m/s
Displacement after one full lap = 0
Average velocity = 0 / 44
= 0 m/s
Acceleration Numericals Class 9
Acceleration numericals class 9 usually use a = (v - u) / t.
Positive acceleration means speeding up. Negative acceleration means slowing down.
Acceleration Questions with Answers
Q1. A train starts from rest and reaches 72 km/h in 10 seconds. Find acceleration.
Ans.
u = 0
v = 72 km/h = 20 m/s
t = 10 s
a = (v - u) / t
= (20 - 0) / 10
= 2 m/s²
Q2. A car moving at 30 m/s stops in 5 seconds. Find retardation.
Ans.
u = 30 m/s
v = 0
t = 5 s
a = (v - u) / t
= (0 - 30) / 5
= -6 m/s²
Retardation = 6 m/s²
Q3. A cyclist increases speed from 5 m/s to 15 m/s in 4 seconds. Find acceleration.
Ans.
u = 5 m/s
v = 15 m/s
t = 4 s
a = (v - u) / t
= (15 - 5) / 4
= 2.5 m/s²
Equations of Motion Numericals Class 9 Science Chapter 4
Equations of motion numericals class 9 carry high marks.
Identify the known quantities among u, v, a, s, and t before choosing the formula.
Equations of Motion Numericals Class 9
Q1. A ball is thrown upward at 20 m/s. Find velocity after 2 seconds. Take g = 10 m/s².
Ans.
u = 20 m/s
a = -10 m/s²
t = 2 s
v = u + at
v = 20 + (-10)(2)
v = 0 m/s
The ball is momentarily at rest after 2 seconds.
Q2. A car starts from rest and accelerates at 3 m/s² for 5 seconds. Find distance covered.
Ans.
u = 0
a = 3 m/s²
t = 5 s
s = ut + ½at²
s = 0 + ½ × 3 × 5²
s = 1.5 × 25
s = 37.5 m
Q3. A vehicle moving at 10 m/s accelerates at 2 m/s². Find velocity after covering 100 m.
Ans.
u = 10 m/s
a = 2 m/s²
s = 100 m
v² = u² + 2as
v² = 10² + 2 × 2 × 100
v² = 100 + 400
v² = 500
v = √500
v = 22.36 m/s
Important Questions on Motion in a Straight Line Class 9
Motion in a straight line questions class 9 test reference point, rest, uniform motion, and non-uniform motion.
Do not ignore these concepts because they often appear in assertion-reason questions.
Position and Reference Point Questions Class 9 Science
A reference point is needed to describe whether an object is moving or at rest.
The same object can be at rest for one observer and in motion for another.
Q1. What is a reference point? Why is it needed to describe motion?
Ans. A reference point is a fixed location from which the position of an object is measured.
Motion has no meaning without a reference point. An object may appear at rest relative to one observer and in motion relative to another.
Q2. Give one example where an object is in motion relative to one observer but at rest relative to another.
Ans. A passenger sitting in a moving train is at rest relative to other passengers.
The same passenger is in motion relative to a person standing on the platform.
Rest and Motion Questions Class 9 Science
Rest and motion are relative terms.
They depend on the reference point chosen by the observer.
Rest and Motion Questions
Q1. Can an object be at rest and in motion at the same time?
Ans. Yes, an object can be at rest and in motion at the same time with respect to different reference points.
A tree is at rest for a person standing near it. The same tree appears to move backward for a person sitting in a moving car.
Linear Motion Questions Class 9 Science
Linear motion means motion along a straight line.
It can be uniform or non-uniform.
Uniform and Non-Uniform Motion Questions
Q1. What is the difference between uniform and non-uniform motion?
Ans.
| Feature |
Uniform Motion |
Non-Uniform Motion |
| Speed |
Constant |
Changes |
| Distance in equal time |
Equal distances |
Unequal distances |
| Distance-time graph |
Straight line |
Curved line or changing slope |
| Acceleration |
Zero |
Non-zero |
| Example |
Train moving steadily on a straight track |
Car moving in traffic |
Distance and Displacement Questions Class 9 Science
Distance and displacement questions class 9 test scalar and vector understanding.
The most common trap is a round trip, where distance is non-zero but displacement is zero.
Distance vs Displacement Questions Class 9
Distance is the total path length.
Displacement is the shortest distance between the starting point and final point.
Q1. A person walks 3 km north and then 4 km east. Find distance and displacement.
Ans.
Distance = 3 + 4
= 7 km
Displacement = √(3² + 4²)
= √(9 + 16)
= √25
= 5 km
Direction is northeast from the starting point.
Q2. Can displacement be zero while distance is non-zero?
Ans. Yes.
If a person completes one full round of a circular track, the distance covered is the circumference of the track.
The displacement is zero because the starting and ending positions are the same.
Average Speed and Average Velocity Questions Class 9
Average speed uses total distance.
Average velocity uses displacement.
Speed and Velocity Questions Class 9
Q1. Why is average velocity zero for a round trip but average speed is not?
Ans. Average velocity = Displacement / Time.
In a round trip, displacement is zero because the start and end points are the same.
Average speed = Distance / Time.
Distance is non-zero because the object actually travels along a path.
Acceleration Questions Class 9 Science Chapter 4
Acceleration is one of the most numerical-heavy parts of this chapter.
Learn the sign of acceleration carefully.
Uniform Acceleration Questions Class 9
Uniform acceleration means velocity changes equally in equal time intervals.
It appears as a straight line on a velocity-time graph.
Q1. Define uniform acceleration and give one example.
Ans. Uniform acceleration is when velocity changes by equal amounts in equal time intervals.
Example: A freely falling object near Earth’s surface accelerates uniformly at about 9.8 m/s² downward.
Q2. How does uniform acceleration appear on a velocity-time graph?
Ans. Uniform acceleration appears as a straight line on a velocity-time graph.
The slope of the line gives acceleration.
Retardation Questions Class 9 Science
Retardation is negative acceleration.
It means the object slows down with time.
Retardation Questions
Q1. What is retardation? How does it appear on a velocity-time graph?
Ans. Retardation is negative acceleration.
On a velocity-time graph, it appears as a straight line with negative slope.
The line falls towards the time axis.
Acceleration Due to Gravity Questions Class 9
Near Earth’s surface, freely falling bodies accelerate downward due to gravity.
Use g = 9.8 m/s² or 10 m/s² if the question says so.
Acceleration Due to Gravity Questions
Q1. A stone is dropped from rest. Find velocity after 3 seconds. Take g = 10 m/s².
Ans.
u = 0
a = 10 m/s²
t = 3 s
v = u + at
v = 0 + 10 × 3
v = 30 m/s downward
Q2. Why does every object fall with the same acceleration near Earth’s surface?
Ans. Gravitational acceleration near Earth’s surface is almost constant.
If air resistance is ignored, all objects fall with the same acceleration, regardless of mass.
Graph-Based Questions Class 9 Science Chapter 4
Graph questions test reading, interpretation, and calculation.
Velocity time graph questions class 9 are especially important because they combine slope and area.
Distance-Time Graph Questions Class 9
A distance-time graph shows how distance changes with time.
The slope gives speed.
Q1. What does each distance-time graph shape indicate?
Ans.
| Graph Shape |
Meaning |
| Horizontal line |
Object is at rest |
| Straight line with positive slope |
Uniform motion |
| Steeper straight line |
Higher speed |
| Upward curve |
Speed is increasing |
| Downward curve |
Speed is decreasing |
Q2. Two objects A and B are shown on a distance-time graph. A has a steeper slope than B. Which object is faster?
Ans. Object A is faster.
A steeper slope means more distance covered per unit time.
Velocity-Time Graph Questions Class 9
A velocity-time graph shows how velocity changes with time.
Its slope gives acceleration.
Velocity Time Graph Questions Class 9
Q1. Describe the velocity-time graph for an object thrown vertically upward and caught at the same point.
Ans. The graph starts at positive velocity.
Velocity decreases linearly because gravity acts downward. It becomes zero at the highest point.
Then velocity becomes negative as the object falls back.
The graph is a straight line crossing the time axis.
Displacement from Velocity-Time Graph Questions Class 9
The area under a velocity-time graph gives displacement.
Use rectangle, triangle, or trapezium area depending on the graph.
Graph Numerical Questions
Q1. A velocity-time graph shows a straight line from (0, 0) to (4, 20). Find displacement.
Ans.
Area under graph = Area of triangle
= ½ × base × height
= ½ × 4 × 20
= 40 m
Q2. A velocity-time graph shows a rectangle from t = 0 to t = 5 s at v = 10 m/s. Find displacement.
Ans.
Area under graph = Area of rectangle
= 5 × 10
= 50 m
Equations of Motion Important Questions Class 9
The three equations of motion apply only to uniformly accelerated straight-line motion.
Pick the equation that contains the unknown and the given values.
v = u + at Questions Class 9 Science
Use v = u + at when displacement is not needed.
Q1. A motorbike starts from rest and accelerates at 4 m/s². Find velocity after 6 seconds.
Ans.
u = 0
a = 4 m/s²
t = 6 s
v = u + at
v = 0 + 4 × 6
v = 24 m/s
Q2. A train moving at 90 km/h decelerates at 2.5 m/s². Find time taken to stop.
Ans.
u = 90 km/h = 25 m/s
v = 0
a = -2.5 m/s²
v = u + at
0 = 25 + (-2.5)t
2.5t = 25
t = 10 s
s = ut + ½at² Questions Class 9 Science
Use s = ut + ½at² when displacement, time, and acceleration are involved.
Second Equation of Motion Questions
Q1. A stone is dropped from a building. Find distance fallen in 4 seconds. Take g = 10 m/s².
Ans.
u = 0
a = 10 m/s²
t = 4 s
s = ut + ½at²
s = 0 + ½ × 10 × 4²
s = 5 × 16
s = 80 m
Q2. A car starts from rest and accelerates at 2 m/s² for 8 seconds. Find distance covered.
Ans.
u = 0
a = 2 m/s²
t = 8 s
s = ut + ½at²
s = 0 + ½ × 2 × 8²
s = 64 m
v² = u² + 2as Questions Class 9 Science
Use v² = u² + 2as when time is not given.
Third Equation of Motion Questions
Q1. A ball rolls down a slope with initial velocity 2 m/s and acceleration 3 m/s². Find velocity after 10 m.
Ans.
u = 2 m/s
a = 3 m/s²
s = 10 m
v² = u² + 2as
v² = 2² + 2 × 3 × 10
v² = 4 + 60
v² = 64
v = 8 m/s
Q2. A vehicle moving at 20 m/s decelerates at 4 m/s². Find stopping distance.
Ans.
u = 20 m/s
v = 0
a = -4 m/s²
v² = u² + 2as
0 = 20² + 2(-4)s
0 = 400 - 8s
8s = 400
s = 50 m
Uniform Circular Motion Important Questions Class 9
Uniform circular motion is important because speed stays constant but velocity changes.
The direction of velocity changes at every point.
Velocity Direction in Circular Motion Questions Class 9
In circular motion, velocity acts along the tangent to the path.
This tangent direction changes continuously.
Q1. In which direction does velocity act at any point in uniform circular motion?
Ans. Velocity acts along the tangent to the circular path at that point.
As the object moves around the circle, the tangent direction changes continuously.
Q2. Give two examples of uniform circular motion.
Ans.
- A stone tied to a string and whirled in a circle at constant speed
- The Earth revolving around the Sun at nearly constant orbital speed
Why Uniform Circular Motion is Accelerated Questions Class 9
Acceleration is linked to change in velocity, not only change in speed.
That is why uniform circular motion is accelerated.
Circular Motion Concept Questions
Q1. Why is uniform circular motion accelerated even though speed is constant?
Ans. Acceleration is the rate of change of velocity.
Velocity has magnitude and direction. In uniform circular motion, speed remains constant but direction changes continuously.
Since velocity changes, acceleration exists.
Class 9 Science Chapter 4 Assertion Reason Questions
Assertion-reason questions test whether the reason correctly explains the assertion.
Read both statements separately before choosing the option.
Directions:
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Class 9 Science Chapter 4 Assertion Reason Questions
Q1. Assertion (A): A body can have zero displacement but non-zero distance. Reason (R): Displacement depends on starting and ending positions, not the path taken.
Ans. (a)
Both A and R are true, and R correctly explains A.
A body returning to its starting point has zero displacement, but the path length is non-zero.
Q2. Assertion (A): Uniform circular motion is accelerated motion. Reason (R): In uniform circular motion, the direction of velocity changes continuously even though speed is constant.
Ans. (a)
Both A and R are true, and R correctly explains A.
Change in direction means change in velocity.
Q3. Assertion (A): The slope of a distance-time graph gives the velocity of the object. Reason (R): Slope = change in distance / change in time, which equals speed for linear motion.
Ans. (c)
A is false, but R is true.
The slope of a distance-time graph gives speed, not velocity.
Q4. Assertion (A): A freely falling object has uniform acceleration. Reason (R): Gravitational acceleration near Earth’s surface is approximately constant at 9.8 m/s² downward.
Ans. (a)
Both A and R are true, and R correctly explains A.
Q5. Assertion (A): Average velocity of a round trip is always zero. Reason (R): Displacement is zero for a round trip because start and end points are the same.
Ans. (a)
Both A and R are true, and R correctly explains A.
Average velocity = Displacement / Time.
Case Study Questions Class 9 Science Chapter 4
Case study questions connect motion concepts to real situations.
Road safety, sports, and vehicle motion are common case study contexts.
Case Study 1: Braking Distance and Road Safety
A school safety poster shows that a car at 60 km/h takes 40 m to stop after brakes are applied. The same car at 120 km/h takes 160 m to stop. The relationship between speed and stopping distance is not linear. It follows v² = u² + 2as.
Q1. Why does stopping distance increase much more than speed when a vehicle goes faster?
Ans. Stopping distance depends on the square of speed.
When speed doubles, v² becomes four times. So, stopping distance increases four times, not two times.
Q2. A car at 30 m/s decelerates at 6 m/s² when brakes are applied. Find stopping distance.
Ans.
u = 30 m/s
v = 0
a = -6 m/s²
v² = u² + 2as
0 = 30² + 2(-6)s
0 = 900 - 12s
12s = 900
s = 75 m
Q3. What type of motion does a car undergo during braking?
Ans. A car undergoes uniformly retarded motion during braking if deceleration is constant.
Its velocity decreases at a constant rate.
Case Study 2: Circular Motion in Athletics
A sprinter runs around a circular track of radius 50 m at a constant speed of 8 m/s. The coach notes that even though speed is constant, the athlete’s velocity changes continuously.
Q1. Why does velocity change even though speed stays at 8 m/s?
Ans. Velocity is a vector quantity.
It has magnitude and direction. On a circular track, direction changes at every point.
So, velocity changes even though speed remains constant.
Q2. Calculate time taken for one full lap.
Ans.
Circumference = 2πr
= 2 × 3.14 × 50
= 314 m
Time = Distance / Speed
= 314 / 8
= 39.25 s
Q3. What is displacement after one full lap?
Ans. Displacement is zero.
The sprinter returns to the starting point.
Important Definitions and Formulas Class 9 Science Chapter 4
| Term |
Definition / Formula |
| Distance |
Total path length covered by an object |
| Displacement |
Shortest straight-line distance between start and end point |
| Speed |
Distance / Time |
| Velocity |
Displacement / Time |
| Acceleration |
a = (v - u) / t |
| First equation of motion |
v = u + at |
| Second equation of motion |
s = ut + ½at² |
| Third equation of motion |
v² = u² + 2as |
| Retardation |
Negative acceleration |
| Uniform circular motion |
Motion along a circular path at constant speed |
| Slope of distance-time graph |
Speed |
| Slope of velocity-time graph |
Acceleration |
| Area under velocity-time graph |
Displacement |