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

asked by Anonymous
  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

    posted by Damon
  2. or
    Vi = a t
    a = Vi/t

    posted by Damon
  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.

    posted by Henry

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