Question

A swimmer is seen crossing the river along the line making an angle of ${45}^{\circ }$ with the direction of flow. Velocity of the river is $5\text{m/s}$. Swimmer takes $6\text{seconds}$ to cross the river of width $60\text{m}$. The velocity of the swimmer with respect to water (in $m{s}^{-1}$)will be

Answer 1

Lets assume that the swimmer swims at an angle $\alpha$ from perpendicular to river flow direction. As river is flowing with velocity of $5\text{m/s}$, resulting velocity of man is ${\overrightarrow{V}}_{m,g}$ at ${45}^{\circ }$ to the river flow.


${\overrightarrow{V}}_{m,g}={\overrightarrow{V}}_{m,r}+{\overrightarrow{V}}_{r,g}$

Resolving velocity vectors into components along horizontal and vertical directions, we get
i.e., $V_{r,g}-V_{m,r}\sin \alpha =V_{m,g}\cos {45}^{\circ }$
$5-V_{m,r}\sin \alpha =V_{m,g}\cos {45}^{\circ }$ ... (1)
and
$V_{m,r}\cos \alpha =V_{m,g}\sin {45}^{\circ }$... (2)

From (1) and (2)
$5-V_{m,r}\sin \alpha =V_{m,r}\cos \alpha$....(3)
Time taken for swimmer to cross the river
$\frac{60}{V_{m,r}\cos \alpha }=6\text{sec}$
$V_{m,r}\cos \alpha =10$....(4)
Substituting in (3)
$5-V_{m,r}\sin \alpha =10$
$V_{m,r}\sin \alpha =-5$....(5)
Squaring and adding (4) and (5), we get
$V_{m,r}^2=100+25$
$V_{m,r}=5\sqrt{5}\text{m/s}$