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(A) To cross the river in the shortest time,the component of the swimmer's velocity perpendicular to the river flow must be maximized.Let the width of

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Due north

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(A) To cross the river in the shortest time,the component of the swimmer's velocity perpendicular to the river flow must be maximized.Let the width of the river be $d$ and the swimmer's velocity 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 the swimmer makes with the direction perpendicular to the river flow.To minimize $t$,$\cos 	heta$ must be maximum,which occurs when $	heta = 0^\circ$.Therefore,the swimmer should swim perpendicular to the river flow,i.e.,due north.

$A$ 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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