Question

a train takes 5 minutes to cover a distance of 3 km between 2 stations P and Q. Starting from rest at P, it accelerates at a constant rate of 40 km/h and maintains it's speed until it is brought uniformly to rest at Q. If the train takes 3 times as long to decelerate as it does to accelerate, find the time taken by the train to decelerate.
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Answer 1

Answer:

HERE IS YOUR ANSWER

Explanation:

40 km/h = 40,000 m/ 3600 s = 11.1 m/s

5 min = 300 s

time to accelerate = t

d1 = (1/2) a t^2

11.1 = a t

so

d1 = (1/2)(11.1) t

d1 = 5.56 t

runs at 11.1 m/s for t2 seconds

distance = d2 = 11.1 t2

deaccelerate time = 3 t

0 = 11.1 + 3 a2 t

a2 t = -11.1/3 = - 3.7 t

a2 = -3.7/t

d3 = 11.1 (3t) + (1/2) (-3.7 /t )(3t)^2

d3 = 33.3 t - 16.7 t

d3 = 16.7 t

so

5.56 t + 11.1 t2 + 16.7 t = 3000 meters

t + t2 + 3 t = 300 seconds

so t2 = (300 -4t)

solve for t

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