A train is travelling at a speed of $90 k m p h$ . Brakes are applied so as to produce a uniform acceleration of $- 0.5 m s^{- 2}$ . Find how far the train will go before it is brought to rest.

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Solution

Given:
Initial speed of the train, $u = 90 k m p h = 90 \times \frac{5}{18} = 25 m s^{- 1}$
Final speed of the train, $v = 0$ (as the train finally comes to rest)
Acceleration of the train, $a$ $= - 0.5 m s^{- 2}$

Solving for the distance covered by the train:
Let $s$ be the distance covered.
According to the third equation of motion, $v^{2} = u^{2} + 2 a s$
$\Rightarrow$ $(0)^{2} = (25)^{2} + (2 \times - 0.5 \times s)$
$\Rightarrow$ $s = \frac{- 25^{2}}{2 \times - 0.5} = 625 m$

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