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Q.

For a train engine moving with speed of $20 m s^{- 1}$ , the driver must apply brakes at a distance of 500 m before the station for the train to come to rest at the station. If the brakes were applied at half of this distance, the train engine would cross the station with speed $\sqrt{x} m s^{- 1}$ . The value of x is ______.
(Assuming same retardation is produced by brakes)

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answer is 200.

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Detailed Solution

$u = 20 m s^{- 1}, s = 500 m, v = 0$

$a = \frac{v^{2} - u^{2}}{25} = \frac{0 - 400}{2 \times 500} = - 0.4 m s^{- 2}$

Now, $u = 20 m s^{- 1}, s = 250 m,$

$a = - 0.4 m s^{- 2}, v = 7$

$v^{2} - 400 = 2 (- 0.4) 250 \Rightarrow v v^{2} = 200$

$\Rightarrow v = \sqrt{200} m s^{- 1}; x = 200$

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For a train engine moving with speed of 20ms−1, the driver must apply brakes at a distance of 500 m before the station for the train to come to rest at the station. If the brakes were applied at half of this distance, the train engine would cross the station with speed x ms−1 . The value of x is ______.(Assuming same retardation is produced by brakes)