If a train travelling at 72 kmph is to be brought to rest in a distance of 200 metres, then its retardation should be

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20 ms^-2

10 ms^-2

2 ms^-2

1 ms^-2

answer is D.

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Detailed Solution

$u = 72 kmph = 20 m / s, v = 0$
By using
$\begin{array}{l} v^{2} = u^{2} - 2 as \Rightarrow \\ a = \frac{u^{2}}{2 s} = \frac{(20)^{2}}{2 \times 200} = 1 m / s^{2} \end{array}$

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