Kinematics (Motion in a Straight Line) Solved Examples (Class 11 Physics)
Kinematics deals with the description of motion without considering forces. These solved examples cover displacement, velocity, acceleration, and equations
TL;DR: Kinematics deals with the description of motion without considering forces. These solved examples cover displacement, velocity, acceleration, and equa…
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Kinematics deals with the description of motion without considering forces. These solved examples cover displacement, velocity, acceleration, and equations
Kinematics (Motion in a Straight Line) — Solved Numerical Examples (Step by Step)
Example 1: A car travels 120 m in the first 5 seconds with uniform acceleration starting from rest. Calculate (a) acceleration and (b) velocity after 5 seconds.
Solution: Given: Distance s = 120 m, time t = 5 s, initial velocity u = 0 (starts from rest)
(a) Using equation of motion: s = ut + (1/2)at²
120 = 0 × 5 + (1/2) × a × 5²
120 = (1/2) × a × 25
120 = 12.5a
a = 120/12.5 = 9.6 m/s²
(b) Using v = u + at
v = 0 + 9.6 × 5
v = 48 m/s
Example 2: A train accelerates uniformly from 18 km/h to 54 km/h in 10 seconds. Find the acceleration and distance covered.
Solution: Given: Initial velocity u = 18 km/h = 18 × (5/18) = 5 m/s
Final velocity v = 54 km/h = 54 × (5/18) = 15 m/s
Time t = 10 s
Acceleration: v = u + at
15 = 5 + a × 10
10 = 10a
a = 1 m/s²
Distance: s = ut + (1/2)at²
s = 5 × 10 + (1/2) × 1 × 10²
s = 50 + 50 = 100 m
Alternatively: v² = u² + 2as
15² = 5² + 2 × 1 × s
225 = 25 + 2s
s = 100 m
Example 3: A ball is thrown upward with velocity 30 m/s. How long does it take to reach maximum height? (Take g = 10 m/s²)
Solution: Given: Initial velocity u = 30 m/s (upward), acceleration a = -g = -10 m/s² (downward), at maximum height v = 0
Using v = u + at
0 = 30 + (-10) × t
0 = 30 - 10t
10t = 30
t = 3 seconds
Example 4: A car moving at 20 m/s comes to rest in 4 seconds with uniform deceleration. Calculate the deceleration and distance covered.
Solution: Given: Initial velocity u = 20 m/s, final velocity v = 0, time t = 4 s
Deceleration: v = u + at
0 = 20 + a × 4
a = -20/4 = -5 m/s² (negative sign indicates deceleration)
Deceleration = 5 m/s²
Distance: s = ut + (1/2)at²
s = 20 × 4 + (1/2) × (-5) × 4²
s = 80 + (1/2) × (-5) × 16
s = 80 - 40 = 40 m
Example 5: Two cars A and B start from the same point. Car A moves with uniform velocity 10 m/s. Car B starts from rest with uniform acceleration 2 m/s². When will they have the same velocity?
Solution: Car A: velocity = 10 m/s (constant), so vA = 10 m/s at all times
Car B: starts from rest with acceleration, vB = 0 + 2t = 2t
When velocities are equal:
vA = vB
10 = 2t
t = 5 seconds
Verification: At t = 5 s, vB = 2 × 5 = 10 m/s ✓
Example 6: A stone is dropped from a cliff and reaches the ground in 4 seconds. Find the height of the cliff and final velocity of the stone. (g = 10 m/s²)
Solution: Given: Initial velocity u = 0 (dropped), time t = 4 s, g = 10 m/s²
Height: h = ut + (1/2)gt²
h = 0 × 4 + (1/2) × 10 × 4²
h = 0 + 5 × 16
h = 80 m
Final velocity: v = u + gt
v = 0 + 10 × 4
v = 40 m/s
Tips
- Always convert units to SI system (km/h to m/s) before using equations of motion.
- Pay attention to sign convention: upward is positive, downward is negative (or vice versa, but be consistent).
- For uniformly accelerated motion, use the three equations: v = u + at, s = ut + (1/2)at², and v² = u² + 2as.
- In free fall problems, g is always 10 m/s² or 9.8 m/s² and acts downward.
Frequently Asked Questions
What is the difference between velocity and speed?
Speed is a scalar quantity that measures how fast something is moving (only magnitude). Velocity is a vector quantity that includes both magnitude and direction. For example, 50 km/h is speed, but 50 km/h northward is velocity.
Why do we use three equations of motion?
The three equations are derived from definitions of velocity and acceleration. Each equation is useful depending on which variables are known and which need to be found. If time is unknown, use v² = u² + 2as; if final velocity is zero, use v = u + at.
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