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A particle starts with an initial velocity of 9 m/s due east and has a constant acceleration of 2 m/s² due west. Find the total distance covered by the particle during the fifth second of its motion.
- 0 m
- 0.5 m
- 2 m
- None of these
Correct answer: 0.5 m
Solution
Taking east as positive: u = +9 m/s, a = -2 m/s². The particle stops at t = 9/2 = 4.5 s. In the 5th second (t = 4 s to t = 5 s), the particle first travels east from t=4 to t=4.5 s, then travels west from t=4.5 to t=5 s. Distance in each segment = (1/2)(1 m/s)(0.5 s) = 0.25 m. Total distance = 0.25 + 0.25 = 0.5 m.
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