If the error in the measurement of the momentum of a particle is +100%, what will be the percentage error in the measurement of its kinetic energy? MCQ with Answer and Explanation
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An athlete runs 200 m in 20 seconds, then returns 100 m back in 10 seconds. The average speed for entire motion is
Correct Answer: Option B
Explanation:
Total distance = 200 m (forward) + 100 m (back) = 300 m. Total time = 20 s + 10 s = 30 s. Average speed = total distance / total time = 300/30 = 10 m/s. Average velocity would be displacement (200-100=100 m) / 30 s = 3.33 m/s. But average speed is 10 m/s. Wait, 300/30 = 10. But that's option A 10 m/s. I need to check: Did I compute right? 200 m forward, then returns 100 m back, so total path length = 300 m. Time total = 30 s. Average speed = 10 m/s. Option A is 10 m/s, B is 6.67. That gives A. But to test concept, many confuse with velocity. I'll set correct answer A. But I'll adjust to make distinction: Maybe distance first 200 m at 10 m/s, return 100 m at 5 m/s? Already given times. So correct answer A. It's moderate. I'll keep.
Total distance = 200 m (forward) + 100 m (back) = 300 m. Total time = 20 s + 10 s = 30 s. Average speed = total distance / total time = 300/30 = 10 m/s. Average velocity would be displacement (200-100=100 m) / 30 s = 3.33 m/s. But average speed is 10 m/s. Wait, 300/30 = 10. But that's option A 10 m/s. I need to check: Did I compute right? 200 m forward, then returns 100 m back, so total path length = 300 m. Time total = 30 s. Average speed = 10 m/s. Option A is 10 m/s, B is 6.67. That gives A. But to test concept, many confuse with velocity. I'll set correct answer A. But I'll adjust to make distinction: Maybe distance first 200 m at 10 m/s, return 100 m at 5 m/s? Already given times. So correct answer A. It's moderate. I'll keep.
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