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Standard XII

Physics

Relative Velocity

An express tr...

Question

An express train is moving with a velocity $v_{1}$ . Its driver finds another train is moving on the same track in the same direction with velocity $v_{2}$ $(v_{2} < v_{1})$ . To escape collision, driver applies a retardation a on the train. The minimum time to escape the collision is

\frac{(v_{1} - v_{2})}{a}

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\frac{(v_{1}^{2} - v_{2}^{2})}{2}

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\frac{(v_{2})}{a}

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\frac{(v_{1} + v_{2})}{a}

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Solution

The correct option is A $\frac{(v_{1} - v_{2})}{a}$
As the trains moving in the same direction so the initial relative speed
$(v_{1} - v_{2})$ and by applying retardation final relative speed become zero.
Initial relative velocity, $u_{r e l} = v_{1} - v_{2}$
Final relative velocity, $v_{r e l} = 0$
Relative acceleration, $a_{r e l} = - a - 0 = - a$
Using first equation of motion,
$v_{r e l} = u_{r e l} + a_{r e l} t$
$0 = (v_{1} - v_{2}) - a t$
$\Rightarrow t = \frac{(v_{1} - v_{2})}{a}$

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An express train is moving with a velocity v1. Its driver finds another train is moving on the same track in the same direction with velocity v2 (v2<v1). To escape collision, driver applies a retardation a on the train. The minimum time to escape the collision is

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