Question

A swimmer can swim with a speed $v$ in still water


If the swimmer crosses a swimming pool of width ${}^{\prime }{d}^{\prime }$ from $A$ to directly opposite point $B$ on other side in time $t_1$ as shown in figure and in a flowing river (river velocity $u$) of same width ${}^{\prime }{d}^{\prime }$ from $A^{\prime }$ to directly opposite point $B^{\prime }$ on other bank in time $t_2$, then $\left(\frac{t_1}{t_2}\right)$ is equal to: (Assume $v>u$)

• A. $\sqrt{1-\frac{v^2}{u^2}}$
• B. $\sqrt{1+\frac{v^2}{u^2}}$
• C. $\sqrt{1-\frac{u^2}{v^2}}$
• D. $1$

Answer 1

The correct option is C $\sqrt{1-\frac{u^2}{v^2}}$

Case 1- In still water

$t_1=\frac{d}{v_{sw}}=\frac{d}{v}$ (where $v_{sw}$ is the velocity of swimmer in still water)

Case 2- In flowing water, the swimmer will be swimming the shortest distance across, so his velocity $v_{sw}=\sqrt{v^2-u^2}$

$t_2=\frac{d}{v_{sw}}=\frac{d}{\sqrt{v^2-u^2}}$
$\Rightarrow \frac{t_1}{t_2}=\frac{\frac{d}{v}}{\frac{d}{\sqrt{v^2-u^2}}}=(\frac{\sqrt{v^2-u^2}}{v})=\sqrt{1-\frac{u^2}{v^2}}$