Motion — Previous Year Questions (Class 9 Science)
Motion is change in position of an object with time. Understand distance, displacement, speed, velocity, acceleration, and equations of motion.
TL;DR: Motion is change in position of an object with time. Understand distance, displacement, speed, velocity, acceleration, and equations of motion.
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Motion is change in position of an object with time. Understand distance, displacement, speed, velocity, acceleration, and equations of motion.
Motion — Previous Year Questions with Solutions
Q (2023, 2 marks): A car travels 200 km west and then 150 km east. Calculate distance and displacement.
Answer: Distance = 200 + 150 = 350 km (total path length)
Displacement = 200 - 150 = 50 km west (straight-line distance from start to end)
Note: Distance is always ≥ |Displacement|
Q (2022, 3 marks): Define acceleration. A car accelerates from rest to 72 km/h in 10 seconds. Find acceleration.
Answer: Acceleration: Rate of change of velocity with time. a = (v - u) / t
Initial velocity u = 0 (from rest)
Final velocity v = 72 km/h = 72 × (1000/3600) = 20 m/s
Time t = 10 s
Acceleration a = (20 - 0) / 10 = 2 m/s²
Q (2023, 2 marks): Using the first equation of motion (v = u + at), find final velocity if u = 5 m/s, a = 2 m/s², t = 6 s.
Answer: Using v = u + at
v = 5 + 2(6)
v = 5 + 12
v = 17 m/s
Q (2021, 3 marks): An object moving at 10 m/s comes to rest in 5 seconds with uniform deceleration. Calculate displacement using s = ut + (1/2)at².
Answer: Initial velocity u = 10 m/s
Final velocity v = 0 (comes to rest)
Time t = 5 s
First find acceleration: v = u + at
0 = 10 + a(5)
a = -2 m/s² (negative, indicating deceleration)
Displacement: s = ut + (1/2)at²
s = 10(5) + (1/2)(-2)(5)²
s = 50 - 25
s = 25 m
Q (2022, 3 marks): Derive the second equation of motion: s = ut + (1/2)at²
Answer: Starting from first equation: v = u + at
Average velocity = (u + v) / 2 = (u + u + at) / 2 = u + (1/2)at
Displacement = Average velocity × Time
s = (u + (1/2)at) × t
s = ut + (1/2)at²
This is the second equation of motion
Q (2023, 5 marks): A ball is thrown upward with initial velocity 20 m/s. Taking g = 10 m/s², find time to reach maximum height and maximum height reached.
Answer: At maximum height, final velocity v = 0
Initial velocity u = 20 m/s
Acceleration a = -g = -10 m/s²
Time to reach maximum height: v = u + at
0 = 20 - 10t
t = 2 s
Maximum height: s = ut + (1/2)at²
s = 20(2) + (1/2)(-10)(2)²
s = 40 - 20
s = 20 m
Alternatively using v² = u² + 2as:
0 = (20)² + 2(-10)s
0 = 400 - 20s
s = 20 m ✓
Frequently Asked Questions
What is the difference between distance and displacement?
Distance is the total length of path traveled (scalar, always positive). Displacement is the straight-line distance from start to end position (vector, can be positive/negative). Example: Circular path of 100 m has distance = 100 m but displacement = 0.
What are the three equations of motion and when are they used?
The three equations of motion (for uniform acceleration): (1) v = u + at (relates velocity, acceleration, time), (2) s = ut + (1/2)at² (relates displacement, time), (3) v² = u² + 2as (relates velocity, acceleration, displacement without time). Use based on which variables you have.
More Class 9 Science PYQs
- Chemical Reactions and Equations
- Light (Reflection and Refraction)
- Electricity
- Acids, Bases and Salts
- Metals and Non-metals
- Life Processes
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