Question

A man wants to reach point $\text{B}$ on the opposite bank of a river flowing at a speed $u$ as shown in figure. what minimum speed relative to water should the man have so that he can reach point $\text{B}$?

• A.
  $\sqrt{2}u$
• B.
  $\sqrt{3}u$
• C.
  $\frac{u}{\sqrt{3}}$
• D.
  $\frac{u}{\sqrt{2}}$

Answer 1

The correct option is D

$\frac{u}{\sqrt{2}}$
Let $v$ be the speed of the man in still water.

Resultant of $v$ and $u$ should be along $\text{AB}$.
Components of $\overrightarrow{v_b}$ (absolute velocity of boatman) along $x$ and $y$ direction are,

$v_x=u-v\sin \theta$ and $v_y=v\cos \theta$

Further,
$\tan {45}^{\circ }=\frac{v_y}{v_x}$

Substituting the values,
$\Rightarrow 1=\frac{v\cos \theta }{u-v\sin \theta }$

$\Rightarrow v=\frac{u}{\sin \theta +\cos \theta }$

$\Rightarrow v=\frac{u}{\sqrt{2}(\frac{1}{\sqrt{2}}\sin \theta +\frac{1}{\sqrt{2}}\cos \theta )}$

$\Rightarrow v=\frac{u}{\sqrt{2}(\cos {45}^{\circ }\sin \theta +\sin {45}^{\circ }\cos \theta )}$

$\Rightarrow v=\frac{u}{\sqrt{2}\sin (\theta +{45}^o)}...\left(1\right)$

$v$ is minimum at,

$\theta +{45}^o={90}^o$

$\therefore \theta ={45}^o$

So, from equation $\left(1\right)$, we get
$v_{min}=\frac{u}{\sqrt{2}}$

Why this question? This question checks the application of motion of $\text{2-D}$ in river-boat problems.