A river is flowing from east to west at a spe | vedclass

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(A) To cross the river in the shortest time,the swimmer must maximize the component of their velocity perpendicular to the river flow.Let the width of

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Due north

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(A) To cross the river in the shortest time,the swimmer must maximize the component of their velocity perpendicular to the river flow.Let the width of the river be $d$ and the velocity of the swimmer in still water be $v_s = 10\ m/min$.The time taken to cross the river is given by $t = \frac{d}{v_s \cos 	heta}$,where $	heta$ is the angle with the direction perpendicular to the river flow.To minimize $t$,$\cos 	heta$ must be maximum,which occurs at $	heta = 0^\circ$.Therefore,the swimmer should swim perpendicular to the river flow,i.e.,due north.In this case,the swimmer will reach point $C$ on the opposite bank due to the river's velocity,but the time taken will be the minimum possible.

$A$ river is flowing from east to west at a speed of $5\ m/min$. $A$ man on the south bank of the river,capable of swimming $10\ m/min$ in still water,wants to swim across the river in the shortest time. He should swim:

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