Question

Two swimmers start from point $P$ on one bank of the river to reach $Q$ on the opposite bank. Velocity of each swimmer in still waters is $2.5km{h}^{-1}$. 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 $2km{h}^{-1}$. If both the swimmers reach point $Q$ simultaneously, the velocity of walking of second swimmer is

• A. $3km{h}^{-1}$
• B. $4km{h}^{-1}$
• C. $2km{h}^{-1}$
• D. $3.5km{h}^{-1}$

Answer 1

The correct option is A $3km{h}^{-1}$

Resultant velocity of swimmer 1 is
$v_{PQ}=\sqrt{{2.5}^2-2^2}=\sqrt{2.25}$
$=1.5km{h}^{-1}$
Let width of the river be $W$ then time taken,
$t_1=\frac{W}{1.5}h$.
Time taken by swimmer 2 is
$t_2=\frac{W}{2.5}h$
Distance $Q{Q}^{\prime }=$velocity $\times$ time$=2\times \frac{W}{2.5}$
Time taken by swimmer 2 to move to distance $Q{Q}^{\prime }$ is
$t=t_1-t_2$
$=\frac{W}{1.5}-\frac{W}{2.5}=\frac{W}{1.5\times 2.5}$
Desired velocity$=\frac{Q{Q}^{\prime }}{t}=\frac{\frac{2W}{2.5}}{\frac{W}{1.5\times 2.5}}$
$=3km{h}^{-1}$