The speed of a swimmer in still water is 20 ; | vedclass

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(A) Let the speed of the swimmer in still water be $v = 20 \; m/s$ and the speed of the river be $u = 10 \; m/s$.To cross the river along the shortest

Vedclass · Accepted Answer

$30$

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(A) Let the speed of the swimmer in still water be $v = 20 \; m/s$ and the speed of the river be $u = 10 \; m/s$.To cross the river along the shortest path (perpendicular to the river flow),the swimmer must swim at an angle $	heta$ with respect to the north (perpendicular to the bank) such that the horizontal component of his velocity cancels the river's velocity.From the geometry of the velocity vector triangle,we have:$\sin 	heta = \frac{u}{v}$Substituting the given values:$\sin 	heta = \frac{10}{20} = \frac{1}{2}$Therefore,$	heta = \arcsin(1/2) = 30^{\circ}$.Thus,the swimmer should make his strokes at an angle of $30^{\circ}$ west of north.

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