If a train travelling at 72 km/h is to be brought to rest in a distance of 200m, then its retardation should be:-

Options

A. `20ms^(-2)`
B. `10ms^(-2)`
C. `2ms^(-2)`
D. `1ms^(-2)`

Answer

D

Explanation

Stoping distance = (u^2)/(2a)`
`rArr a = ((72 xx 5/18)^2)/(2xx200) = ((20)^2)/(400) = 1m//s^2`