A swimmer’s speed in still water is double the speed of the river. He crossses the river by the shortest path in time t. For another swimmer whose speed in still water is equal to the speed of the river, find the shortest time to cross the river.

Question

Arun · Answer

Let distance = d, angle made with the upstream = theta Therefore, 90- theta = angle made with the normal stream Therefore time taken = t’ = d/v cos (90-theta) = d/v sin theta Also t = d/v Therefore, required ratio = t/t’= (d/v)/ (d/v sin theta) = sin theta

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A swimmer’s speed in still water is double the speed of the river. He crossses the river by the shortest path in time t. For another swimmer whose speed in still water is equal to the speed of the river, find the shortest time to cross the river.

A swimmer’s speed in still water is double the speed of the river. He crossses the river by the shortest path in time t. For another swimmer whose speed in still water is equal to the speed of the river, find the shortest time to cross the river.

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A swimmer’s speed in still water is double the speed of the river. He crossses the river by the shortest path in time t. For another swimmer whose speed in still water is equal to the speed of the river, find the shortest time to cross the river.

A swimmer’s speed in still water is double the speed of the river. He crossses the river by the shortest path in time t. For another swimmer whose speed in still water is equal to the speed of the river, find the shortest time to cross the river.

1 Answers

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