A swimmer can swim in still water with speed | vedclass

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(C) Let $d$ be the width of the river. The swimmer swims with speed $v$ in still water.For the shortest time,the swimmer must swim perpendicular to th

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$\sin 	heta$

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(C) Let $d$ be the width of the river. The swimmer swims with speed $v$ in still water.For the shortest time,the swimmer must swim perpendicular to the river flow. The time taken is $t = \frac{d}{v}$.For the shortest distance,the swimmer must swim at an angle such that the resultant velocity is perpendicular to the river bank. Let the angle with the normal to the stream be $\alpha$. Then $\sin \alpha = \frac{v_{river}}{v} = \frac{v/2}{v} = \frac{1}{2}$,so $\alpha = 30^{\circ}$.The angle with the upstream is $	heta = 90^{\circ} + \alpha = 120^{\circ}$.The time taken for the shortest distance is $t^{\prime} = \frac{d}{v \cos \alpha} = \frac{d}{v \sin 	heta}$.The ratio of the time taken is $\frac{t}{t^{\prime}} = \frac{d/v}{d/(v \sin 	heta)} = \sin 	heta$.

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