Question

A river is flowing at a speed of $10\text{m/s}$. A man standing on the bank of the river wants to reach the opposite side of the river through the shortest path. Find the shortest angle with the bank of the river, at which the man should swim in order to reach the opposite side. Speed of man with respect to river water is $20\text{m/s}$.

• A. ${75}^{\circ }$
• B. ${60}^{\circ }$
• C. ${45}^{\circ }$
• D. ${30}^{\circ }$

Answer 1

The correct option is B ${60}^{\circ }$

Let us take the river velocity along $+vex-$ direction as shown in the figure.


In order to reach the opposite end through the shortest possible path, the man should head in such a way that the net velocity of man w.r.t ground $($i.e ${\overrightarrow{v}}_{MG})$ should be perpendicular to the river flow.

${\overrightarrow{v}}_{MR}={\overrightarrow{v}}_{MG}-{\overrightarrow{v}}_{RG}$

$\Rightarrow {\overrightarrow{v}}_{MG}={\overrightarrow{v}}_{MR}+{\overrightarrow{v}}_{RG}$

Here, ${\overrightarrow{v}}_{RG}=10\hat{i}\text{m/s}$

${\overrightarrow{v}}_{MR}=(-v_{MR}\cos \theta \hat{i}+v_{MR}\sin \theta \hat{j})\text{m/s}$

$\therefore {\overrightarrow{v}}_{MG}=(10-v_{MR}\cos \theta )\hat{i}+\left(v_{MR}\sin \theta \right)\hat{j}\text{m/s}$

Now, applying the condition for shortest path,

$(\overrightarrow{v}{}_{MG}{)}_x=0$

$\Rightarrow 10-v_{MR}\cos \theta =0$

$\Rightarrow \cos \theta =\frac{10}{v_{MR}}=\frac{10}{20}$

$\therefore \theta ={60}^{\circ }$ with river bank.

Hence, option $\left(\text{D}\right)$ is the right choice.