Given u = 28 m s⁻¹, v = 0 and s = 98 m. Using v² = u² + 2as, we get a = −4 m s⁻². Time taken, t = (v − u)/a = 7 s.
A motorbike moving with initial velocity 28 m s⁻¹ and constant acceleration stops after travelling 98 m. Find the acceleration of the motorbike and the time taken to come to a stop.
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For the motorbike, initial velocity u = 28 m s⁻¹, final velocity v = 0 and distance s = 98 m. Using v² = u² + 2as:
0 = 28² + 2(a)(98), giving a = −4 m s⁻².
Using v = u + at: 0 = 28 − 4t, so t = 7 s. Thus, acceleration is −4 m s⁻² and stopping time is 7 s.
For more NCERT Solutions of Class 9 Science Exploration Chapter 4 Describing Motion Around Us Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/science/exploration-chapter-4/