Exams › JEE Advanced › Physics
A train travelling at 120 km/h is slowed uniformly to 30 km/h, travels at 30 km/h for some time, then accelerates uniformly back to 120 km/h. The distances covered during deceleration, uniform motion, and acceleration are 3 km, 5 km, and 2 km respectively. Find the total time lost compared to travelling the entire 10 km at 120 km/h.
- 4 min
- 5 min
- 7 min
- 9 min
Correct answer: 9 min
Solution
During deceleration: avg speed = (120+30)/2 = 75 km/h; time = 3/75 h = 0.04 h = 2.4 min. During uniform motion at 30 km/h: time = 5/30 h = 1/6 h = 10 min. During acceleration: avg speed = 75 km/h; time = 2/75 h = 1.6 min. Total actual time = 2.4 + 10 + 1.6 = 14 min. Time at 120 km/h for 10 km = (10/120)*60 = 5 min. Time lost = 14 - 5 = 9 min.
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