To cross the river in shortest distance, a swimmer should swim making

1.	To cross the river in shortest distance, a swimmer should swim making angle theta with the upstream.What is the ratio of the time taken to swim across in the shortest time to that in swimming across overshortest distance. (Assume speed of swimmer in still water is greater than the speed of river flow](A) cos theta(B) sin theta(C) tan theta(D) cot theta
Answer» Answer: Explanation: Taking d to be the WIDTH of the RIVER and also the the shortest distance COVERED by the swimmer. If the swimmer makes an angle θ with the upstream, then angle (90-θ) will be normal to the stream. Thus, shortest time TAKEN will be the own distance i.e t=d/v. If speed of the river is v. Thus, making an angle (90-θ) will be cos(90-θ) and so speed will be VCOS(90-θ). The shortest distance to cover the river in time will be t1 = d/vcos(90-θ) t1= d/vsin(θ). The ratio will be t/t1 =(d/v)/(d/vsin(θ)) = sinθ.

To cross the river in shortest distance, a swimmer should swim making angle theta with the upstream.What is the ratio of the time taken to swim across in the shortest time to that in swimming across overshortest distance. (Assume speed of swimmer in still water is greater than the speed of river flow](A) cos theta(B) sin theta(C) tan theta(D) cot theta

Answer»

Answer:

Explanation:

Taking d to be the WIDTH of the RIVER and also the the shortest distance COVERED by the swimmer.

If the swimmer makes an angle θ with the upstream, then angle (90-θ) will be normal to the stream.

Thus, shortest time TAKEN will be the own distance i.e t=d/v.

If speed of the river is v.

Thus, making an angle (90-θ) will be cos(90-θ) and so speed will be VCOS(90-θ).

The shortest distance to cover the river in time will be

t1 = d/vcos(90-θ)

t1= d/vsin(θ).

The ratio will be t/t1

=(d/v)/(d/vsin(θ))

= sinθ.

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