1.

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»

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.



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