Accepted Answer
The train accelerates at the rate of 20 for some time, until it's just exactly time to put on the brakes, decelerate at the rate of 100, and come to ascreeching stop after a total distance of exactly 2.7 km.The speed it reaches while accelerating is exactly the speed it starts decelerating from.Speed reached while accelerating = (acceleration-1) (Time-1) = .2 time-1Speed started from to slow down = (acceleration-2) (Time-2) = 1 time-2 The speeds are equal..2 time-1 = 1 time-2time-1 = 5 x time-2It spends 5 times as long speeding up as it spends slowing down.The distance it covers speeding up = 1/2 A (5T)-squared = 0.1 x 25 T-squared = 2.5 T-squared.The distance it covers slowing down = 1/2 A (T-squared) = 0.5 T-squared.Total distance = 2,700 meters.(2.5 + .5) T-squared = 2,700T-squared = 2700/3 = 900 T = 30 secondsThe train speeds up for 150 seconds, reaching a speed of 30 meters per sec and covering 2,250 meters. It then slows down for 30 seconds, covering 450 meters.Total time = 180 sec = 3 minutes, minimum. Observation:This solution is worth more than 5 points.
Suggested Answer
This can be done by hit and trial easily.Considering train accelerate for 13000s and deaccelerate for 1000sso total distance covered is = 20cm/s * 13000s + 100cm/s*100s= (260000 + 10000 )cm= 270000cm= 2.7kmSo total time taken is - 13100 seconds
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