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Two swimmers start from point P on one bank of the river to reach Q on the opposite bank. Velocity of each swimmer instill waters is 2.5 ${kmh}^{-1}$ . One of the swimmers crosses the river along the straight route PQ and the other swims right angles to the stream and then walks the distance which he has been carried away by the river to get to point Q. Stream velocity is 2 ${kmh}^{- 1}$ .If both the swimmers reach point Q simultaneously, the velocity of walking of second swimmer is

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$3 {kmh}^{-1}$

$4 {kmh}^{-1}$

$2 {kmh}^{-1}$

$3.5 {kmh}^{-1}$

answer is A.

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Detailed Solution

Resultant velocity of swimmer 1 is

$v_{P Q} = \sqrt{{2.5}^{2}} = \sqrt{2.25}$

Let width of river be W then time taken,

$t_{1} = \frac{W}{1.5} h .$ Time taken by swimmer 2 is \

$t_{2} = \frac{W}{2.5}$

$∴$ Distance $Q Q'$ = velocity $\times$ time $= 2 \times \frac{W}{2.5}$

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Time taken by swimmer 2 to move to distance QQ’ is

$t = t_{1} - t_{2}$

$= \frac{W}{1.5} - \frac{W}{2.5} = \frac{W}{1.5 \times 2.5}$

Desired velocity

$= \frac{Q Q'}{t} = \frac{2 W}{2.5} = \frac{W}{1.5 \times 2.5}$

$= 3 {kmh}^{- 1}$

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