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A man can swim with a certain speed, in still water to cross a river to a point directly opposite to the starting point in time t0. When the river flows, he crosses the river directly along the same path in a time t1. If ‘d’ be the river width, then the velocity of water current is

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$\frac{d (t_{1}^{} - t_{0}^{})}{t_{1} t_{0}}$

$\frac{d (t_{1}^{} + t_{0}^{})}{t_{1} t_{0}}$

$\frac{d \sqrt{t_{1}^{2} + t_{0}^{2}}}{t_{1} t_{0}}$

$\frac{d \sqrt{t_{1}^{2} {-t}_{0}^{2}}}{t_{1} t_{0}}$

answer is A.

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Detailed Solution

$t_{0} = \frac{d}{Vm} {and t}_{1} = \frac{d}{\sqrt{{Vm}^{2} {-Vr}^{2}}}$

$\frac{t_{0}}{t_{1}} = \frac{\sqrt{{Vm}^{2} {-Vr}^{2}}}{Vm} \Rightarrow \frac{t_{0}^{2}}{t_{1}^{2}} =1- \frac{{Vr}^{2}}{{Vm}^{2}}$

$\frac{{Vr}^{2}}{{Vm}^{2}} = \frac{t_{1}^{2} {-t}_{0}^{2}}{t_{1}^{2}}; vr = \frac{Vm}{t_{1}} \sqrt{t_{1}^{2} {-t}_{0}^{2}} = \frac{d}{t_{0} t_{1}} \sqrt{t_{1}^{2} {-t}_{0}^{2}}$

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