Question

A train starting from rest attains a velocity of 72km/h in 5 metres. assuming the acceleration uniform find the acceleration and distance tavelled by the train

Answer 1

u = 0 m/s

v = 72 km/h = 72*5/18 m/s

= 20 m/s

t = 5 s

As v = u + a t

so, 20 = 0 + a*5

so. a = 20/5 = 4 m/s^2

Also, s = u t + 1/2 a t^2

so, s = 0 + 1/2 * 4 * 5^2

so, s = 50 m

Hence, the acceleration is 4 m/s^2 and

distance tavelled by the train is 50 m.