Question

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

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

Answer 1

The correct option is B $\frac{u}{\sqrt{2}}$

Given,
Speed of river $=u$
Assuming,
Speed of man swimming $=v$ in direction '$\theta$' anticlock with line $AC$
On breaking speed of man $v$ into $x$ and $y$ component
$v_x=v\sin \theta$
$v_y=v\cos \theta$
Let us suppose $v_n$ is the net speed of the man, we know that to reach point $B$ from $A$, $v_n$ should be along $AB$.
Also, $x$ component of $v_n$, $v_{nx}=u-v\sin \theta$
$y$ component of $v_n$, $v_{ny}=v\cos \theta$
In $\Delta ACB$,
$\tan {45}^{\circ }=\frac{v_{nx}}{v_{ny}}$
$\Rightarrow 1=\frac{u-v\sin \theta }{v\cos \theta }$
$\Rightarrow v\cos \theta =u-v\sin \theta$
$\Rightarrow v\cos \theta +v\sin \theta =u$
$\Rightarrow v(\cos \theta +\sin \theta )=u$
$\Rightarrow v=\frac{u}{\cos \theta +\sin \theta }...\left(1\right)$

Here, to get minimum value of $v$, value of denominator should be maximum.
Assuming, $t=\cos \theta +\sin \theta$
$\therefore \frac{dt}{d\theta }=-\sin \theta +\cos \theta$ [Applying concept of maxima and minima]

For maxima or minima,
$\frac{dt}{d\theta }=0\Rightarrow -\sin \theta +\cos \theta =0$
$\Rightarrow \cos \theta =\sin \theta$
$\Rightarrow 1=\tan \theta$
$\Rightarrow \theta ={45}^{\circ }$

Now, $\frac{d^{2}t}{d{\theta }^2}=-\cos \theta -\sin \theta$
$=-(\cos \theta +\sin \theta )$
$=-(\cos {45}^{\circ }+\sin {45}^{\circ })=-\sqrt{2}$
= Negative value
$\frac{d^{2}t}{d{\theta }^2}<0$
$\therefore t$ has maximum value at $\theta ={45}^{\circ }$
On putting $\theta ={45}^{\circ }$ in equation $\left(1\right)$
$v_{\text{min}}=\frac{u}{\cos {45}^{\circ }+\sin {45}^{\circ }}=\frac{u}{\sqrt{2}}$
$v_{\text{min}}=\frac{u}{\sqrt{2}}$