A passenger train is moving with speed $v_{1}$ , on rails. The driver of this train observes another goods train moving in the same direction with speed $v_{2} (v_{1} > v_{2})$ . If on applying brakes, the retardation produced is $a$ , then the minimum distance covered by the passenger train to avoid collision is

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

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

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

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Solution

The correct option is A $\frac{(v_{1} - v_{2})^{2}}{2 a}$
For minimum distance $s$ , velocity of the passenger $v_{1}$ should be reduced to $v_{2}$ i.e. the relative velocity should be zero so that collision is avoided and both train moves with same velocity $v_{2}$
For passenger train, $0 = (v_{1} - v_{2})^{2} + 2 (- a) s o r s = \frac{(v_{1} - v_{2})^{2}}{2 a}$

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