Consider a train which can acceleration of 20cm/sec and slow down with acceleration of 100 cm/sec find the minimum time for train to travel between the stations 2.7 km per a part

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.

This can be done by hit and trial easily.

Considering train accelerate for 13000s and deaccelerate for 1000s
so total distance covered is = 20cm/s * 13000s + 100cm/s*100s
= (260000 + 10000 )cm
= 270000cm
= 2.7km

So total time taken is - 13100 seconds

Iwould choose the same momentum because if they throw a baseball very light and not hard at all then doing it with a bowling ball won’t hurt that worse

The answer is a i think