Motion Solved Examples (Class 9 Physics)
Motion describes the change in position of an object over time. Displacement, velocity, and acceleration are key concepts. Equations of motion help predict
TL;DR: Motion describes the change in position of an object over time. Displacement, velocity, and acceleration are key concepts. Equations of motion help pr…
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Motion describes the change in position of an object over time. Displacement, velocity, and acceleration are key concepts. Equations of motion help predict
Motion — Solved Numerical Examples (Step by Step)
Example 1: A car travels 100 km in 2 hours. Calculate its average speed.
Solution: Given:
Distance traveled = 100 km
Time taken = 2 hours
Using the formula: Average speed = Total distance / Total time
Average speed = 100 / 2
Average speed = 50 km/h
Example 2: A cyclist starts from rest and accelerates at 2 m/s^2 for 5 seconds. Find the final velocity.
Solution: Given:
Initial velocity u = 0 m/s (starts from rest)
Acceleration a = 2 m/s^2
Time t = 5 seconds
Using the first equation of motion: v = u + at
v = 0 + 2 × 5
v = 10 m/s
Example 3: An object is thrown upward with an initial velocity of 20 m/s. How high will it rise? (g = 10 m/s^2)
Solution: Given:
Initial velocity u = 20 m/s
Final velocity v = 0 m/s (at maximum height)
Acceleration a = -10 m/s^2 (upward motion, gravity acts downward)
Using the third equation of motion: v^2 = u^2 + 2as
(0)^2 = (20)^2 + 2 × (-10) × s
0 = 400 - 20s
20s = 400
s = 20 m
Example 4: A truck starting from rest reaches a velocity of 30 m/s in 10 seconds. Find the distance covered.
Solution: Given:
Initial velocity u = 0 m/s
Final velocity v = 30 m/s
Time t = 10 seconds
First find acceleration using: v = u + at
30 = 0 + a × 10
a = 3 m/s^2
Now find distance using: s = ut + (1/2)at^2
s = 0 × 10 + (1/2) × 3 × (10)^2
s = 0 + 1.5 × 100
s = 150 m
Example 5: A ball is dropped from a height of 45 m. How long will it take to reach the ground? (g = 10 m/s^2)
Solution: Given:
Initial velocity u = 0 m/s (dropped from rest)
Height s = 45 m
Acceleration a = 10 m/s^2
Using the second equation of motion: s = ut + (1/2)at^2
45 = 0 × t + (1/2) × 10 × t^2
45 = 5t^2
t^2 = 9
t = 3 seconds
Example 6: A car moving at 15 m/s is brought to rest in 5 seconds. Find the deceleration.
Solution: Given:
Initial velocity u = 15 m/s
Final velocity v = 0 m/s (brought to rest)
Time t = 5 seconds
Using the first equation of motion: v = u + at
0 = 15 + a × 5
a × 5 = -15
a = -3 m/s^2
The deceleration is 3 m/s^2 (negative acceleration)
Example 7: An object moving with velocity 8 m/s experiences an acceleration of 4 m/s^2 for 3 seconds. Find the distance covered.
Solution: Given:
Initial velocity u = 8 m/s
Acceleration a = 4 m/s^2
Time t = 3 seconds
Using the second equation of motion: s = ut + (1/2)at^2
s = 8 × 3 + (1/2) × 4 × (3)^2
s = 24 + 2 × 9
s = 24 + 18
s = 42 m
Tips
- The three equations of motion are: v = u + at, s = ut + (1/2)at^2, and v^2 = u^2 + 2as. Choose the equation that has the quantities you know and need.
- Remember that upward motion has negative acceleration (gravity acts downward at -g), while downward motion has positive acceleration.
- Displacement and distance are different: displacement is the change in position (vector), while distance is the total path traveled (scalar).
Frequently Asked Questions
What is the difference between speed and velocity?
Speed is a scalar quantity (magnitude only) representing how fast an object is moving. Velocity is a vector quantity (magnitude and direction) representing the rate of change of displacement. Average speed = total distance / time, while average velocity = total displacement / time.
When should I use which equation of motion?
Use v = u + at when you need to find velocity and have time. Use s = ut + (1/2)at^2 when you need to find displacement. Use v^2 = u^2 + 2as when you do not have time but have displacement. Choose based on what is given and what you need.
More Physics Solved Examples
- Electricity
- Light Reflection and Refraction
- Force and Laws of Motion
- Gravitation
- Work and Energy
- Kinematics (Motion in a Straight Line)
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