 # 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 time for answering the question is over
52 cents 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 = <u>180 sec</u> = 3 minutes, minimum.

Observation:
This solution is worth more than 5 points.
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