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A train is travelling with a speed of 72 kmph. The driver applies brakes so that a uniform acceleration of $- {0.2 ms}^{- 2}$ is produced. The distance travelled by the train before it comes to rest is

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0.01 km

1 km

0.1 km

10 km

answer is A.

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Detailed Solution

$u = 72 \times \frac{5}{18} = 20m/s$

$a = - 0. {2m/s}^{2}$

$v^{2} - u^{2} = 2 as$

$0 - {(20)}^{2} = 2 (- 0.2) s$

$20 \times 20= \frac{2 \times 2}{10} . s$

$\frac{20 \times 20 \times 10}{2 \times 2} = s$

$s = 1000 m$

$s = 1 km$

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