Question

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

Answer 1

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.

Answer 2

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