📅 Published on: 19.12.2025
Motion is one of the most fundamental topics in Class 9 Physics and forms the base for understanding many advanced concepts in science. The chapter Motion deals with the study of the movement of objects and introduces important physical quantities such as distance, displacement, speed, velocity, acceleration, and time. Numerical problems from this chapter help students apply theoretical concepts to real-life situations.
Motion Class 9 extra questions (numericals) are specially designed to strengthen students’ problem-solving skills and improve their understanding of formulas and units. These questions go beyond textbook exercises and cover a wide variety of situations, including uniform and non-uniform motion, motion along a straight line, circular motion, average speed, average velocity, and graphical interpretation.
Practicing extra numerical questions from the Motion chapter helps students:
- Develop a clear conceptual understanding
- Improve calculation speed and accuracy
- Avoid common mistakes in unit conversion
- Prepare effectively for school exams, unit tests, and board examinations
Regular practice of Class 9 Motion numericals builds confidence and enables students to tackle both basic and higher-order problems with ease.
Connect with Us
Motion class 9 extra questions numerical
CLASS IX PHYSICS
MOTION
NUMERICALS FOR PRACTICE
1. A particle is moving up an inclined plane. Its velocity changes from 25m/s to 10m/s in 5 seconds. What is its acceleration?
Answer
Initial velocity, u = 25 m/s
Final velocity, v = 10 m/s
Time, t = 5 s
Acceleration, a = (v − u) / t
a = (10 − 25) / 5
a = −15 / 5
a = −3 m/s²
Acceleration = −3 m/s²
2. The velocity changes from 35m/s to 60m/s in 3 seconds. What is its acceleration?
Answer
u = 35 m/s
v = 60 m/s
t = 3 s
a = (v − u) / t
a = (60 − 35) / 3
a = 25 / 3
a = 8.33 m/s²
Acceleration = 8.33 m/s²
3. A body covered a distance of 4 metre along a semicircular path. Calculate the magnitude of displacement of the body, and the ratio of distance to displacement?
Answer
Length of semicircle = πr
πr = 4
r = 4 / π
Displacement = Diameter = 2r = 8 / π m ≈ 2.55 m
Ratio of distance to displacement:
4 : (8 / π) = π : 2
Displacement ≈ 2.55 m
Distance : Displacement = π : 2
4. A particle moving with an initial velocity of 8m/s is subjected to a uniform acceleration of 2.5m/s². Find the displacement in the next 4 sec.
Answer
u = 8 m/s
a = 2.5 m/s²
t = 4 s
s = ut + ½at²
s = (8 × 4) + ½ × 2.5 × 16
s = 32 + 20
s = 52 m
Displacement = 52 m
5. A train is travelling at a speed of 40 km/h. Brakes are applied so as to produce a uniform acceleration of −0.5 m/s². Find how far the train will go before it is brought to rest.
Answer
v = 0
a = −0.5 m/s²
v² = u² + 2as
0 = (11.11)² + 2(−0.5)s
0 = 123.45 − s
s = 123.45 m
Stopping distance ≈ 123.5 m
6. A Truck covers 90km at a uniform speed of 30km/hr. what should be its speed for the next 120km if the average speed for the entire journey is 60km/h?
Answer
Average speed = 60 km/h
Total time = 210 / 60 = 3.5 h
Time for first part = 90 / 30 = 3 h
Remaining time = 3.5 − 3 = 0.5 h
Required speed = 120 / 0.5 = 240 km/h
Required speed = 240 km/h
7. A stone is thrown in a vertically upward direction with a velocity of 15 m/s. If the acceleration of the stone during its motion is 8m/s² in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?
Answer
a = −8 m/s²
v = 0
Using v² = u² + 2as
0 = 225 − 16s
s = 14.06 m
Using v = u + at
0 = 15 − 8t
t = 1.875 s
Height attained ≈ 14.06 m
Time taken ≈ 1.88 s
8. A person goes to market, makes purchases and comes back at a constant slower speed. Draw displacement-time and velocity time graphs of the person?
Answer
Straight line upward while going to market,
Horizontal line while shopping,
Straight line downward with smaller slope while returning.
Velocity-time graph shows:
Constant positive velocity,
Zero velocity during shopping,
Constant negative velocity of smaller magnitude during return.
9. Rahul runs for 8 min. at a uniform speed 5 km/h. At what speed should he run for the next 10 min. so that the average speed comes 15km/hr?
Answer
Speed₁ = 5 km/h
Distance₁ = 5 × (8/60) = 2/3 km
Total time = 18/60 = 0.3 h
Total distance = 15 × 0.3 = 4.5 km
Distance₂ = 4.5 − 2/3 = 3.83 km
Time₂ = 10/60 h
Speed₂ = 3.83 ÷ (10/60) = 23 km/h
Required speed ≈ 23 km/h
MOTION – NUMERICALS (Q.10 to Q.15)
10. A particle was at rest from 1 a.m. It moved at a uniform speed 40km/hr from 1.30 a.m. to 2:00 a.m. Find the average speed between
(a) 1.00 a.m. and 2.00 a.m.
Answer (a)
Distance covered = 0
From 1.30 a.m. to 2.00 a.m.
Time = 30 min = 0.5 h
Speed = 40 km/h
Distance covered = 40 × 0.5 = 20 km
Total time = 1 hour
Average speed = Total distance / Total time
= 20 / 1 = 20 km/h
(b) 1.15 a.m. and 2.00 a.m.
Answer (b)
Time = 15 min
Distance = 0
From 1.30 a.m. to 2.00 a.m.
Time = 30 min = 0.5 h
Distance = 40 × 0.5 = 20 km
Total time = 45 min = 0.75 h
Average speed = 20 / 0.75 = 26.67 km/h
(c) 1.30 a.m. and 2.00 a.m.
Answer (c)
Speed = 40 km/h
Average speed = 40 km/h
11. An object moves along a circular path of diameter 16cm with constant speed. If it takes 4 min. to move from a point on the path to the diametrically opposite point. Find
(a) The distance covered by the object
Answer (a)
Radius = 8 cm
Distance covered = Half circumference
= πr = 3.14 × 8
= 25.12 cm
(b) The speed
Answer (b)
Speed = Distance / Time
= 25.12 / 4
= 6.28 cm/min
(c) The displacement
Answer (c)
= 16 cm
(d) average velocity.
Answer (d)
= 16 / 4
= 4 cm/min
12. A particle with a velocity of 5m/s at t=0 moves along a straight line with a constant acceleration of 0.2m/s². Find the displacement of the particle in 15s?
Answer
a = 0.2 m/s²
t = 15 s
s = ut + ½at²
s = (5 × 15) + ½ × 0.2 × 225
s = 75 + 22.5
s = 97.5 m
13. A particle is pushed along a horizontal surface in such a way that it starts with a velocity of 15m/s. Its velocity decreases at a uniform rate of 0.5m/s².
(a) Find the time it will take to come to rest.
Answer (a)
v = 0
a = −0.5 m/s²
v = u + at
0 = 15 − 0.5t
t = 30 s
(b) Find the distance covered by it before coming to rest?
Answer (b)
0 = 225 − s
s = 225 m
14. A train accelerated from 30km/hr to 80km/hr in 5 minutes. How much distance does it cover in this period? Assume that the tracks are straight?
Answer
v = 80 km/h = 22.22 m/s
t = 5 min = 300 s
Distance = (u + v)/2 × t
= (8.33 + 22.22)/2 × 300
= 15.27 × 300
= 4581 m ≈ 4.58 km
15. A cyclist moving on a circular track of radius 100m completes one revolution in 8 minutes. What is his
(a) average speed
Answer (a)
Distance in one revolution = 2πr
= 2 × 3.14 × 100
= 628 m
Time = 8 min = 480 s
Average speed = 628 / 480
= 1.31 m/s
(b) average velocity in one full revolution?
Answer (b)
Average velocity = Displacement / Time
= 0 / 480
= 0 m/s
