Two trains A and B which are 100 km apart are travelling towards each other on different tracks with the initial speed of 50 km/h for both. The train A accelerates at $20 k m h^{- 2}$ and the train B retards at the rate $20 k m h^{- 2}$ The distance covered by the train A when they cross each other is approximately :

45 km

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55 km

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150 km

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60 km

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Solution

The correct option is C 60 km
Suppose , the train A and B meet at a common point A after time t
from the start .
Common initial velocity is 50 km/h.
Distance travelled by A is, $S_{A} = x$
Distance travelled by is Bis , $S_{B} = 100 - x$
Now, A is accelerating with $\frac{20 k m}{h^{2}}$
So,using equation:
$S_{A} = u t + \frac{1}{2} a t^{2}$
$x = 50 t + \frac{1}{2} 20 t^{2} = 50 t + 10 t^{2}$
and B is retarding with $\frac{20 k m}{h^{2}}$ So using equation
$S_{B} = u t + \frac{1}{2} a t^{2}$
$100 - x = 50 t - 10 t^{2}$
i.e. adding both equation
$100 = 100 t$
$t = 1 h$
Distance travelled by A = $50 t + 10 t^{2} = 50 + 10 = 60 k m$

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