$ A $ train is travelling at a speed of $ 90 \mathrm{km} \mathrm{h}^{-1} . $ Brakes are applied so as to produce a uniform acceleration of $ -0.5 \mathrm{m} \mathrm{s}^{-2} $. Find
how far the train will go before it is brought to rest.

Solution:

Initial velocity of train, u = 90 kmph = $90 \times \frac{5}{18}$

                                         $u = 5 \times 5 m/s = 25 m/s$

Final velocity of train, $v = 0$

Distance travelled before coming to stop = S

acceleration, $a = - 0.5 m / s^2$

Using formula $v^2 - u^2 = 2aS$

$S = \frac{(v^2 - u^2)}{2a} = \frac{(0^2 - 25^2)}{2 \times (- 0.5)}$

= $\frac{(- 25 \times 25)}{ - 1} = 625 m$

So the train travels a distance of 625 m before coming to stop.  

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Updated on: 10-Oct-2022

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