A $150 m$ long train accelerates uniformly from rest. If the front of the train passes a railway worker $50 m$ away from the station at a speed of $25 m/s$ , what will be the speed of the back part of the train as it passes the worker?

30 m/s

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40 m/s

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50 m/s

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60 m/s

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Solution

The correct option is C $50 m/s$
Applying $3^{r d}$ equation of motion
$v^{2} = u^{2} + 2 a s$
$25 \times 25 = 0 + 2 \times a \times 50$
$a = \frac{25}{4} {m/s}^{2}$
Now for speed of the back part of the train as it passes the worker
We have $v^{' 2} = v^{2} + 2 a l$
where $l$ is the length of the train or distance covered by the back part to pass the worker
$v^{' 2} = 25^{2} + 2 \times \frac{25}{4} \times 150$
$v^{' 2} = 625 + (25 \times 75) = 2500$
$v^{'} = 50 m/s$

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