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

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

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due north

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(A) To cross a river in the shortest time,the swimmer must maximize their velocity component perpendicular to the river flow.Let the velocity of the man in still water be $v_m = 20\,m/min$ and the velocity of the river be $v_r = 8\,m/min$.The time taken to cross the river of width $d$ is given by $t = \frac{d}{v_m \cos 	heta}$,where $	heta$ is the angle with the normal to the river bank.To minimize $t$,the denominator $v_m \cos 	heta$ must be maximized.This occurs when $\cos 	heta = 1$,which means $	heta = 0^{\circ}$.Therefore,the man should swim perpendicular to the river flow,which is due north.

$A$ river is flowing from west to east at a speed of $8\,m/min$. $A$ man on the south bank of the river,capable of swimming at $20\,m/min$ in still water,wants to swim across the river in the shortest time. He should swim in a direction

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