Question

A swimmer wishes to cross a $1\text{km}$ wide river flowing at a speed of $5{\text{kmh}}^{-1}$. His speed in still water is $3{\text{kmh}}^{-1}$. He has to reach the directly opposite point in minimum possible time. If he does not reach the directly opposite point by swimming, he has to walk the extra distance at $5{\text{kmh}}^{-1}$. Find the minimum time taken by the swimmer to reach the desired point. Use the following data if needed:
$\sqrt{91}=9.5,\frac{1}{2.85}\simeq 0.35$

• A. $0.64\text{hr}$
• B. $1\text{hr}$
• C. $0.33\text{hr}$
• D. Not possible

Answer 1

The correct option is A $0.64\text{hr}$

Since speed of swimmer in still water $\left(3{\text{km h}}^{-1}\right)$ is less than the speed of river flow $\left(5{\text{km h}}^{-1}\right)$, he won't be able to reach the opposite point directly.
Velocity of swimmer w.r.t ground is:
$v_{s,g}=v_{s.r}+v_{r,g}...\left(i\right)$


where velocity of swimmmer w.r.t river, $v_{s,r}=v_s=3{\text{km h}}^{-1}$
& velocity of river w.r.t ground,$v_{r,g}=v_r=5{\text{km h}}^{-1}$
$\Rightarrow v_{s,g}=v_s\cos \theta \hat{j}+(v_r-v_s\sin \theta )\hat{i}$
$\therefore v_{s,g}=3\cos \theta \hat{j}+(5-3\sin \theta )\hat{i}$

Time $t_1$ to cross river $t_1=\frac{d}{v_s\cos \theta }=\frac{1}{3\cos \theta }...\left(ii\right)$
The distance $\text{AP}$ after drifting from point $\text{O}$,
$\text{AP}=(v_r-v_s\sin \theta ).t_1$
$\Rightarrow \text{AP}=(5-3\sin \theta )\times \frac{1}{3\cos \theta }$
Time $t_2$ to cover distance $\text{AP}$ by walking is given by,
$t_2=\frac{\text{AP}}{5}=\frac{5-3\sin \theta }{15\cos \theta }$

For total time taken to be minimum :
$\frac{d(t_1+t_2)}{d\theta }=0$
$\Rightarrow \frac{d}{d\theta }(\frac{1}{3\cos \theta }+\frac{5-3\sin \theta }{15\cos \theta })=0$
Or, $\frac{2}{3}\sec \theta \tan \theta -\frac{1}{5}{\sec }^2\theta =0$
$\therefore \sin \theta =\frac{3}{10}$


From the triangle shown in figure, $\cos \theta =\frac{\sqrt{91}}{10}$
Putting the value of $\sin \theta ,\cos \theta$:
$t_{\text{min}}=t_1+t_2$
$\Rightarrow t_{\text{min}}=\frac{1}{3\cos \theta }+\frac{5-3\sin \theta }{15\cos \theta }$
$\Rightarrow t_{\text{min}}=\frac{1}{2.85}+\frac{4.1}{14.25}$
$\therefore t_{\text{min}}=0.35+0.29=0.64\text{hr}$