Two swimmers leave point A on the bank of the river to reach point B lying right across on the other bank. One of them crosses the river along the straight line AB while the other swims at right angles to the stream and then walks the distance that he has been carried away by the stream to get to point B. What was the velocity u of his walking if both swimmers reached the destination simultaneously? The stream velocity `v_0=2.0 km//hour` and the velocity `v^'` of each swimmer with respect to water equals to `2.5km` per hour.

Options

A. `3km//hr`
B. `3.5km//hr`
C. `4km//hr`
D. `5km//hr`

Answer

A

Explanation

According to problem,
`t_(A)=t_(B)+` time taken in walking from C to B.
`t_(A)=t_(B)+(CB)/"walking speed"=t_(B)+(v_(r)t_(B))/u`
`d/(v_(A)cos theta)=d/v_(B)+(v_(r)t_(B))/u=d/v_(B)+(v_(r)d)/(u v_(B))`
`d/(v_(A)xxsqrt(v_(A)^(2)-v_(r)^(2))/v_(A))=d/v_(B)+(v_(r)d)/(u v_(B))`
`u=v_(0)/((1-v_(0)^(2)/v^(2))^(1/2)-1)=3.0km//hr`