Kinematics Solved Examples (Class 11 Physics)
Master kinematic equations for describing motion with constant acceleration. These problems cover displacement, velocity, acceleration, and time relationsh
TL;DR: Master kinematic equations for describing motion with constant acceleration. These problems cover displacement, velocity, acceleration, and time relat…
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Master kinematic equations for describing motion with constant acceleration. These problems cover displacement, velocity, acceleration, and time relationsh
Kinematics — Solved Numerical Examples (Step by Step)
Example 1: A car accelerates from rest with an acceleration of 4 m/s². What will be its velocity after 5 seconds?
Solution: Initial velocity u = 0
Acceleration a = 4 m/s²
Time t = 5 s
Using v = u + at
v = 0 + 4 × 5 = 20 m/s
Example 2: A body starts from rest and moves with constant acceleration. If it covers 100 m in 5 seconds, find the acceleration.
Solution: Initial velocity u = 0
Displacement s = 100 m
Time t = 5 s
Using s = ut + (1/2)at²
100 = 0 × 5 + (1/2) × a × 5²
100 = (1/2) × a × 25
100 = 12.5a
a = 8 m/s²
Example 3: A cyclist is moving with velocity 15 m/s. He applies brakes and decelerates at 2 m/s². How long will he take to stop?
Solution: Initial velocity u = 15 m/s
Final velocity v = 0 (stopped)
Acceleration a = -2 m/s² (deceleration)
Using v = u + at
0 = 15 + (-2) × t
2t = 15
t = 7.5 s
Example 4: A ball is thrown upward with initial velocity 30 m/s. What is the maximum height reached? (g = 10 m/s²)
Solution: Initial velocity u = 30 m/s
At maximum height, final velocity v = 0
Acceleration a = -g = -10 m/s²
Using v² = u² + 2as
0² = 30² + 2 × (-10) × s
0 = 900 - 20s
20s = 900
s = 45 m
Example 5: An object moving with initial velocity 20 m/s undergoes constant acceleration for 4 seconds and covers 120 m. Find the acceleration.
Solution: Initial velocity u = 20 m/s
Time t = 4 s
Displacement s = 120 m
Using s = ut + (1/2)at²
120 = 20 × 4 + (1/2) × a × 4²
120 = 80 + (1/2) × a × 16
120 = 80 + 8a
40 = 8a
a = 5 m/s²
Example 6: A stone is dropped from a cliff. What is its velocity after falling 20 m? (g = 10 m/s²)
Solution: Initial velocity u = 0 (dropped)
Displacement s = 20 m
Acceleration a = g = 10 m/s²
Using v² = u² + 2as
v² = 0 + 2 × 10 × 20
v² = 400
v = 20 m/s
Tips
- The four kinematic equations are: v = u + at; s = ut + (1/2)at²; v² = u² + 2as; s = (u + v)t/2.
- Choose the equation that contains the three known quantities and the one unknown you need to find.
- Always define positive direction clearly and use sign convention consistently (upward usually positive, downward negative for vertical motion).
Frequently Asked Questions
Why is there a difference between displacement and distance?
Distance is the total path length traveled (always positive), while displacement is the straight-line change in position (can be positive, negative, or zero).
What does negative acceleration mean?
Negative acceleration (or deceleration) means acceleration is opposite to the velocity direction, causing the object to slow down.
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