If train travelling 72kmph is to be brought to rest distance of 200m then its retardation should be

Answer 1

72 km/h * 1000 m/km * 1 h/3600 s = Vi
average speed = Vi/2
time to stop = t = 200 / (Vi/2)
200 = (1/2) a t^2
a= 400 /(t^2) = magnitude of deacceleration in m/s^2

Answer 2

or
Vi = a t
a = Vi/t

Answer 3

Vo = 72,000m/3600s = 20 m/s.

V^2 = Vo^2 + 2a*d = 0,
20^2 + 2a*200 = 0,
a = -1.0 m/s^2.

Answer 4

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