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Class 9 Science Chapter 4 Important Questions and Answers PDF
Important Questions and Answers for Class 9 Science Exploration Chapter 4 – Describing Motion Around Us | 2026–27
ShareStudents can use these Class 9 Chapter 4 Science important questions to practise motion-related concepts through numerical problems, graphs and real-life situations. The questions cover reference points, acceleration, braking distance, equations of motion and circular motion.
Multiple Choice Questions (MCQs) on Motion, Velocity and Acceleration
Q1. A student walks 40 m east and then 40 m west, returning to the starting point. What are the total distance and displacement?
(a) Distance = 0 m, displacement = 80 m
(b) Distance = 80 m, displacement = 0 m
(c) Distance = 40 m, displacement = 40 m
(d) Distance = 80 m, displacement = 80 m
Answer: (b) Distance = 80 m, displacement = 0 m.
Explanation: The total distance is 40 + 40 = 80 m. Since the student returns to the starting point, the displacement is zero.
Q2. Which physical quantity describes both the rate of change of position and its direction?
(a) Distance
(b) Speed
(c) Velocity
(d) Time
Answer: (c) Velocity.
Explanation: Velocity includes both magnitude and direction. Speed tells us how fast an object moves without specifying its direction.
Q3. What does the slope of a straight-line position–time graph represent?
(a) Acceleration
(b) Velocity
(c) Mass
(d) Distance only
Answer: (b) Velocity.
Explanation: The slope of a position–time graph gives the change in position divided by the corresponding time interval.
Q4. A car moves along a straight road at constant velocity. What is its acceleration?
(a) Positive
(b) Negative
(c) Zero
(d) Increasing continuously
Answer: (c) Zero.
Explanation: Acceleration is the rate of change of velocity. When velocity remains constant, acceleration is zero.
Q5. Which statement is correct about uniform circular motion?
(a) Both speed and velocity remain constant.
(b) Speed remains constant, but velocity changes.
(c) Acceleration is always zero.
(d) The object moves in a straight line.
Answer: (b) Speed remains constant, but velocity changes.
Explanation: The direction of motion changes continuously along a circular path. Since velocity depends on direction, it changes even when speed remains constant.
Very Short Answer Questions on Reference Points and Motion
Q6. What is a reference point? Why is it important for describing motion?
Answer: A reference point is a fixed point used to describe an object's position. For example, a person standing 20 m east of a school gate can describe their position using the gate as a reference point. Motion is identified by checking whether the object's position changes with time relative to that point.
Q7. What is the difference between an instant of time and a time interval?
Answer: An instant of time refers to a particular moment, such as 9:00 a.m. A time interval is the duration between two instants, such as the five minutes between 9:00 a.m. and 9:05 a.m.
Q8. State the SI units of displacement, velocity and acceleration.
Answer:
Q9. Can average velocity be zero even when average speed is not zero? Explain with an example.
Answer: Yes. A swimmer who travels from one end of a pool to the other and returns to the starting point covers a certain distance. However, the total displacement is zero, so the average velocity is zero while the average speed remains positive.
Q10. What does a horizontal line on a position–time graph indicate?
Answer: A horizontal line indicates that the object's position does not change with time. Therefore, the object is at rest relative to the chosen reference point, and its velocity is zero during that interval.
Short Answer and Numerical Questions on Distance, Velocity and Motion Graphs
Q11. Two postmen start walking towards each other from towns 180 yojanas apart. One walks 7 yojanas per day, while the other walks 5 yojanas per day. After how many days will they meet?
Solution:
Distance between the towns = 180 yojanas
Combined distance covered per day = 7 + 5 = 12 yojanas
Time taken = Total distance ÷ Combined daily distance
= 180 ÷ 12
= 15 days
Answer: The two postmen will meet after 15 days.
Explanation: Since both postmen move towards each other, their combined distance covered each day reduces the gap between them. This question is based on the speed concept discussed through an ancient Indian mathematical example in the chapter.
Q12. A bus increases its velocity from 36 km/h to 72 km/h in 10 seconds. Calculate its average acceleration.
Solution:
Initial velocity, u = 36 km/h = 10 m/s
Final velocity, v = 72 km/h = 20 m/s
Time, t = 10 s
Average acceleration = (v − u)/t
= (20 − 10)/10
= 1 m/s²
Final Answer: The average acceleration is 1 m/s² in the direction of motion.
Q13. A car moving at 20 m/s comes to rest in 5 seconds after applying the brakes. Calculate its average acceleration.
Solution:
Initial velocity, u = 20 m/s
Final velocity, v = 0 m/s
Time, t = 5 s
Acceleration = (v − u)/t
= (0 − 20)/5
= −4 m/s²
Final Answer: The acceleration is −4 m/s².
The negative sign shows that acceleration acts opposite to the car's direction of motion.
Q14. A vehicle has the following position–time data. Calculate its velocity and determine whether it is moving with uniform or non-uniform motion.
Solution:
The vehicle covers 12 m of displacement every 2 seconds.
Velocity = Change in position/Time interval
= 12/2
= 6 m/s
Since the position changes by equal amounts in equal time intervals, the vehicle moves with constant velocity.
Final Answer: Velocity = 6 m/s; the motion is uniform straight-line motion.
Q15. A stone is dropped from rest. Calculate its velocity after 3 seconds. Take the acceleration due to gravity as 9.8 m/s².
Solution:
Initial velocity, u = 0 m/s
Acceleration, a = 9.8 m/s² downward
Time, t = 3 s
Using the first equation of motion:
v = u + at
v = 0 + 9.8 × 3
v = 29.4 m/s
Final Answer: The velocity after 3 seconds is 29.4 m/s downward.
Q16. Two cars A and B have straight-line position–time graphs. The graph of car B is steeper than that of car A. Which car has a greater velocity magnitude if both slopes are positive?
Answer: Car B has a greater velocity magnitude because its position–time graph has a steeper slope.
The slope represents the change in position per unit time. A steeper positive slope indicates that the object covers more displacement in the same time interval.
Long Answer and Application-Based Questions on Equations of Motion and Circular Motion
Q17. Explain the differences between distance, displacement, average speed, average velocity and average acceleration.
Answer:
Distance and speed are scalar quantities because they have magnitude only. Displacement, velocity and acceleration are vector quantities because they have both magnitude and direction.
For example, when a runner completes one lap of a circular track, the distance travelled is the circumference of the track, but the displacement is zero.
Q18. Derive the first and second equations of motion for straight-line motion with constant acceleration.
Answer:
Let:
u = Initial velocity
v = Final velocity
a = Constant acceleration
t = Time interval
s = Displacement
First equation of motion:
Acceleration = Change in velocity ÷ Time
a = (v − u)/t
Multiplying both sides by t:
at = v − u
Therefore:
v = u + at
Second equation of motion:
For constant acceleration, the velocity–time graph is a straight line.
The displacement is equal to the signed area between the graph and the time axis.
For a graph with positive velocities, the area consists of a rectangle and a triangle.
Area of rectangle = ut
Area of triangle = ½ × t × (v − u)
Therefore:
s = ut + ½t(v − u)
Using v − u = at:
s = ut + ½t(at)
Hence:
s = ut + ½at²
These equations are used to calculate velocity and displacement when acceleration is constant.
Q19. A car starts from rest, reaches 20 m/s in 5 seconds, continues at this velocity for 10 seconds and then stops uniformly in 5 seconds. Find the total distance travelled.
Solution:
The motion has three stages.
Stage 1: Acceleration
Initial velocity = 0 m/s
Final velocity = 20 m/s
Time = 5 s
Distance = ½ × (0 + 20) × 5
= 50 m
Stage 2: Constant velocity
Distance = Velocity × Time
= 20 × 10
= 200 m
Stage 3: Braking
Initial velocity = 20 m/s
Final velocity = 0 m/s
Time = 5 s
Distance = ½ × (20 + 0) × 5
= 50 m
Total distance = 50 + 200 + 50
Final Answer: 300 m.
Q20. A bus travels at 36 km/h when the driver notices an obstacle 30 m ahead. The reaction time is 0.5 seconds, and the braking acceleration is −2.5 m/s². Will the bus stop before reaching the obstacle?
Solution:
Initial speed = 36 km/h = 10 m/s.
Step 1: Calculate the reaction distance.
Reaction distance = Speed × Reaction time
= 10 × 0.5
= 5 m
Step 2: Calculate the braking distance.
Given:
u = 10 m/s
v = 0 m/s
a = −2.5 m/s²
Using:
v² = u² + 2as
0 = 100 + 2(−2.5)s
0 = 100 − 5s
s = 20 m
Step 3: Calculate the total stopping distance.
Total stopping distance = Reaction distance + Braking distance
= 5 + 20
= 25 m
Distance available = 30 m.
Remaining distance = 30 − 25 = 5 m.
Final Answer: Yes. Under the given conditions, the bus stops 5 m before the obstacle.
Q21. Why is uniform circular motion considered accelerated motion even when speed remains constant?
Answer:
Uniform circular motion occurs when an object travels along a circular path at constant speed.
However, velocity depends on both speed and direction. In circular motion, the direction of velocity changes continuously.
For example, a child riding a merry-go-round moves along a circular path. Even if the speed remains constant, the direction of motion changes at every point.
Therefore, the velocity changes, and the object has acceleration.
Q22. A child rides a merry-go-round of radius 3.5 m. If one complete revolution takes 11 seconds, calculate the average speed and average velocity. Use π = 22/7.
Solution:
Given:
Radius, R = 3.5 m
Time, T = 11 s
Step 1: Calculate the distance travelled.
Distance = 2πR
= 2 × 22/7 × 3.5
= 22 m
Step 2: Calculate average speed.
Average speed = Distance ÷ Time
= 22 ÷ 11
= 2 m/s
Step 3: Calculate average velocity.
After one complete revolution, the child returns to the starting point.
Displacement = 0 m
Average velocity = 0/11 = 0 m/s
Final Answer:
Average speed = 2 m/s
Average velocity = 0 m/s
Important Concepts, Formulas and Preparation Tips for Class 9 Science Chapter 4
The Science Class 9 Chapter 4 important questions cover how position changes with time, how velocity and acceleration are measured, and how graphs help describe motion. Revise the following concepts, formulas and preparation tips to understand the chapter and solve numerical problems accurately.
Important Concepts in Describing Motion Around Us
Important Formulas for Class 9 Science Chapter 4
Symbols: u = initial velocity, v = final velocity, a = acceleration, t = time interval, s = displacement, R = radius and T = time for one revolution.
Note: The kinematic equations apply to straight-line motion with constant acceleration.
How to Prepare Class 9 Science Chapter 4 Important Questions
Understand the differences between distance, displacement, speed, velocity and acceleration with real-life examples.
Practise numerical problems using the equations of motion, including unit conversions from km/h to m/s.
Learn to draw and interpret position–time and velocity–time graphs, including their slopes and areas.
Solve application-based questions involving braking distance, falling objects and uniform circular motion.
Revise signed positions, reference points and the direction of velocity using labelled diagrams.
Additional Study Materials for Class 9 Science Chapter 4 – Describing Motion Around Us
Students can use the following study materials alongside the Class 9 Science Chapter 4 important questions and answers to revise motion concepts, practise calculations and improve their understanding of graphs. These resources support chapter-wise revision and help students prepare for school examinations.
CBSE Class 9 Science Chapter-wise Important Questions
Practising important questions from different chapters helps students revise the concepts covered in Class 9 Science Exploration. The chapter-wise questions include conceptual explanations, numerical problems, diagrams and application-based exercises.
Other Important Related Links for CBSE Class 9 Science
Students can explore additional Class 9 Science study materials, including revision notes, textbook solutions, sample papers and practice questions. These resources can be used alongside chapter-wise important questions for regular revision.
FAQs on Important Questions and Answers for Class 9 Science Exploration Chapter 4 – Describing Motion Around Us | 2026–27
1. What are the most important topics in Class 9 Science Chapter 4?
The important topics include distance, displacement, speed, velocity, acceleration, motion graphs, kinematic equations and uniform circular motion. Students should understand these concepts and practise numerical questions based on them.
2. What types of numerical questions are included in Chapter 4 Science Class 9 important questions?
Students should practise numerical problems involving average speed, average velocity, acceleration, final velocity, displacement, braking distance and circular motion. Graph-based calculations are also important.
3. What is the difference between distance and displacement in Class 9 Science?
Distance is the total length of the path travelled by an object, while displacement is the net change in position with direction. Distance cannot be negative, but displacement can be positive, negative or zero depending on the chosen coordinate direction.
4. Why can average velocity be zero even when average speed is not zero?
Average velocity depends on displacement, whereas average speed depends on distance travelled. If an object returns to its starting point, its displacement becomes zero even though it has travelled a certain distance.
5. Which graphs are important for Class 9 Science Chapter 4 examinations?
Position–time and velocity–time graphs are important. Students should understand how the slope of a position–time graph gives velocity and how the slope of a velocity–time graph gives acceleration.
6. What are the three equations of motion in Class 9 Science Chapter 4?
The equations are v = u + at, s = ut + ½at² and v² = u² + 2as. These equations apply to straight-line motion with constant acceleration.
7. Why is acceleration negative when a vehicle slows down?
When a vehicle moves in the positive direction and slows down, its acceleration acts in the opposite direction. Therefore, its acceleration is negative. The sign depends on the chosen coordinate direction.
8. What does the area under a velocity–time graph represent?
The signed area between a velocity–time graph and the time axis represents displacement. For motion with positive velocity throughout the interval, this area also equals the distance travelled.
9. Why is uniform circular motion an accelerated motion?
In uniform circular motion, speed remains constant, but the direction of velocity changes continuously. Since acceleration occurs when velocity changes, the object is accelerating.
10. What is the displacement after one complete revolution around a circular track?
The displacement is zero because the object returns to its starting point. However, the distance travelled equals the circumference of the circular track, which is 2πR.
11. Why can two objects have the same position but different velocities?
Two objects can reach the same location at the same time while moving at different velocities. On a position–time graph, their lines may intersect, but their slopes can be different. This means they have the same position at that instant but different velocities.
12. Are graph-based questions included in the important questions for Class 9 Science Chapter 4?
Yes. Graph-based questions cover uniform and changing velocity, acceleration, slopes and displacement calculations. Students should practise plotting graphs from data tables and interpreting their shapes.
13. Why does the distance required to stop a vehicle increase when its speed increases?
A faster vehicle usually needs more distance to stop because it travels farther during the driver's reaction time and requires a longer braking distance. Under constant braking deceleration, doubling the initial speed makes the braking distance four times as large.
14. What happens to a marble moving in a circular path when the ring guiding its motion is removed?
The marble moves approximately along a straight line tangent to the circular path at the point of release. This happens because the ring no longer provides the force needed to keep changing the marble's direction. The activity helps explain velocity direction during circular motion.
15. Where can students find Science Chapter 4 Class 9 important questions with answers PDF?
Students can practise solved MCQs, numericals, motion graphs and application-based questions on this Vedantu page. The questions cover distance, displacement, velocity, acceleration, equations of motion and uniform circular motion. A downloadable PDF can support offline revision once it is available.
