Question

if a train at 72kmphr is to be brought to rest in a distance of 200 m then its retardation is A 20m/s^2 b 10m/s^2 c 2m/s^2 d 1m/s^2

Answer 1

72/3.6 = 20 meters/second
(helpful to remember 1000 meters /3600 seconds = 1/3.6)

0 = 20 + a t
t = -20/a

x = Xi + Vi t + (1/2) a t^2
200 = 0 + 20 t + .5 a t^2
200 = 0 -400/a + .5 a (400/a^2)

200 a = -400 + 200
a = -1
so d, retardation of 1 m/s^2
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alternatively
average speed during stop = 10 m/s^2
200 = 10 t
t = 20
a = change in velocity/ time = -20/20 = -1