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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 that another train is moving on the same track in the same direction with velocity $v_{2} (v_{1} > v_{2})$ . To escape collision, driver applies break to produce retardation $‘ a^{'}$ on the train. The minimum time to avoid collision will be

t = \frac{v_{1} - v_{2}}{a}

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t = \frac{v_{1}^{2} - v_{2}^{2}}{a}

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

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Solution

The correct option is C $t = \frac{2 (v_{1} - v_{2})}{a}$

Assuming the frame of reference at 2, we can say:
Relative velocity of 1 with respect to 2: $v_{12} = v_{1} - v_{2}$

We know,
$d = u t - \frac{1}{2} a t^{2}$
At limiting condition, $d = 0$
$0 = (v_{1} - v_{2}) t - \frac{1}{2} a t^{2}$
$\frac{1}{2} a t^{2} = (v_{1} - v_{2}) t$

$t = \frac{2 (v_{1} - v_{2})}{a}$

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An express train is moving with a velocity v1. Its driver finds that another train is moving on the same track in the same direction with velocity v2(v1>v2). To escape collision, driver applies break to produce retardation ‘a′ on the train. The minimum time to avoid collision will be

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