Question

A swimmer wishes to cross a $500$ m wide river flowing at $5$ km/h. His speed with respect to water is $3$ km/h. He reaches the other shore in shortest time. The man has to reach the other shore at the point directly opposite to his starting point. If he reaches the other shore somewhere else, he has to walk down to this point. Find the minimum distance that he has to walk.

• A. $2/3$ km
• B. $4$ km
• C. $1$ km
• D. $3$ km

Answer 1

The correct option is A $2/3$ km

Given that :

Velocity of man $V_m=3km/h$

$BD$= Horizontal distance for resultant velcoity $R$

$X$= component of resultant $R=5+3cos\theta$

$t=\frac{0.5}{3}sin\theta$

Which is same for horizontal component of velcoity

$H=BD=(5+3cos\theta )\left(\frac{0.5}{3}\right)sin\theta$

$H=5+3cos\theta /6sin\theta$

For $H$ to be minimum $\frac{dH}{d\theta }=0$

$\frac{d}{d\theta }\frac{5+3cos\theta }{6sin\theta }=0$

$18(sin²\theta +cos²\theta )-30cos\theta =0$

$30Cos\theta =18$

$cos\theta =\frac{-18}{30}=\frac{-3}{5}$

$sin\theta =\sqrt{1-Cos²\theta }=\frac{4}{5}$

$H=\frac{5+3cos\theta }{6sin\theta }$

$H=\frac{5+3\times \frac{-3}{5}}{6\times \frac{4}{5}}$

$H=\frac{25-9}{24}$

$H=\frac{16}{24}$

$H=\frac{2}{3}km$

The minimum distance is $\frac{2}{3}km$.

Hence, Option A is correct.