Kinematics - Solved Numerical - Physics 9th

Q1: Calculate the acceleration of a bus that speeds up from 20 m/s to 40 m/s in 8 seconds

Given: u = 20 m/s, v = 40 m/s, t = 8 s
Formula: a = (v - u) / t
Solution: a = (40 - 20) / 8 = 20 / 8 = 2.5 m/s²
Answer: Acceleration = 2.5 m/s²

Q2: A bus is moving on a road with 15 m/s and it accelerates at 5 m/s². Find the final velocity after 6 seconds

Given: u = 15 m/s, a = 5 m/s², t = 6 s
Formula: v = u + at
Solution: v = 15 + 5 × 6 = 15 + 30 = 45 m/s
Answer: Final velocity = 45 m/s

Q3: A car starts moving from rest with an acceleration of 5 m/s². Find out the time to travel 50 m distance

Given: u = 0, a = 5 m/s², s = 50 m
Formula: s = ut + ½at² → 50 = 0 + ½×5×t² → 50 = 2.5t²
Solution: t² = 50 / 2.5 = 20 → t = √20 ≈ 4.47 s
Answer: t ≈ 4.47 seconds

Q4: A ball is dropped from a height of 50 m. What will be its velocity before touching the ground?

Given: u = 0, s = 50 m, a = g = 9.8 m/s²
Formula: v² = u² + 2as
Solution: v² = 0 + 2×9.8×50 = 980 → v = √980 ≈ 31.3 m/s
Answer: v ≈ 31.3 m/s

See Also: Kinematics MCQs

Q5: If a body is thrown upward with vertical velocity 50 m/s. Calculate maximum height which body can reach

Given: u = 50 m/s, v = 0 m/s, a = -9.8 m/s²
Formula: v² = u² + 2as
Solution: 0 = 50² + 2×(-9.8)×s → s = 2500 / (2×9.8) = 2500 / 19.6 ≈ 127.6 m
Answer: Maximum height ≈ 127.6 m

Q6: A ball falls down from top of height 70 m. How much time the ball will take to reach the ground?

Given: u = 0, s = 70 m, a = 9.8 m/s²
Formula: s = ut + ½at² → 70 = 0 + 0.5×9.8×t² → 70 = 4.9t²
Solution: t² = 70 / 4.9 ≈ 14.29 → t ≈ √14.29 ≈ 3.78 s
Answer: Time ≈ 3.78 seconds