Motion and Equations of Motion Solved Examples (Class 9 Physics)
Motion equations (v = u + at, s = ut + 0.5at², v² = u² + 2as) are foundational for solving kinematics problems. These numericals build fluency with constan
TL;DR: Motion equations (v = u + at, s = ut + 0.5at², v² = u² + 2as) are foundational for solving kinematics problems. These numericals build fluency with co…
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Motion equations (v = u + at, s = ut + 0.5at², v² = u² + 2as) are foundational for solving kinematics problems. These numericals build fluency with constan
Motion and Equations of Motion — Solved Numerical Examples (Step by Step)
Example 1: A car accelerates from rest with constant acceleration 2 m/s². How long does it take to reach 20 m/s?
Solution: Initial velocity u = 0, Final velocity v = 20 m/s, Acceleration a = 2 m/s². Using v = u + at: 20 = 0 + 2 × t. Therefore t = 20 / 2 = 10 s.
Example 2: A ball is thrown upward with initial velocity 30 m/s. How high does it go? (g = 10 m/s²)
Solution: Initial velocity u = 30 m/s, Final velocity v = 0 (at highest point), Acceleration a = -10 m/s². Using v² = u² + 2as: 0 = 900 + 2 × (-10) × s. Therefore 20s = 900, so s = 45 m.
Example 3: A cyclist moving at 10 m/s decelerates at 2 m/s². How far does he travel before stopping?
Solution: Initial velocity u = 10 m/s, Final velocity v = 0, Acceleration a = -2 m/s². Using v² = u² + 2as: 0 = 100 + 2 × (-2) × s. Therefore 4s = 100, so s = 25 m.
Example 4: A train starts from rest and accelerates uniformly at 1.5 m/s². What distance does it cover in 8 seconds?
Solution: Initial velocity u = 0, Acceleration a = 1.5 m/s², Time t = 8 s. Using s = ut + 0.5at²: s = 0 + 0.5 × 1.5 × 8² = 0.75 × 64 = 48 m.
Example 5: A stone is dropped from a height. How long does it take to fall 80 m? (g = 10 m/s²)
Solution: Initial velocity u = 0, Distance s = 80 m, Acceleration a = 10 m/s². Using s = ut + 0.5at²: 80 = 0 + 0.5 × 10 × t². Therefore 5t² = 80, so t² = 16, and t = 4 s.
Example 6: A car traveling at 15 m/s accelerates uniformly to 25 m/s in 5 seconds. Find the acceleration and distance covered.
Solution: u = 15 m/s, v = 25 m/s, t = 5 s. Using v = u + at: 25 = 15 + a × 5, so a = 2 m/s². Using s = ut + 0.5at²: s = 15 × 5 + 0.5 × 2 × 25 = 75 + 25 = 100 m.
Tips
- Choose the correct equation based on which variable is unknown.
- Remember: v² = u² + 2as is most useful when time is unknown.
- Downward acceleration is positive; upward motion against gravity has negative acceleration.
Frequently Asked Questions
When should I use which equation of motion?
Use v = u + at when time is involved. Use s = ut + 0.5at² when you need distance and time. Use v² = u² + 2as when time is not given.
Why does a ball thrown upward come back down?
Gravity always acts downward, causing constant downward acceleration. Even when moving upward, the ball is decelerated by gravity until velocity becomes zero, then gravity accelerates it downward.
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