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 a
screeching 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-1
Speed started from to slow down = (acceleration-2) (Time-2) = 1 time-2
The speeds are equal.
.2 time-1 = 1 time-2
time-1 = 5 x time-2
It 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 = 2,700
T-squared = 2700/3 = 900
T = 30 seconds
The 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.
