Question

Consider a river $500m$ wide. Speed of river water is $5km/hr$. Swimmers speed in still water is $3km/hr$. If swimmer wants to reach a point directly opposite the point from where he started to swim, find minimum distance he has to walk.

Answer 1

Velocity of man, $v_m=3$ km/h
BD = horizontal velocity of resultant R
$R_x=5+3\cos \theta$
$t=\frac{\left(0.5\right)}{3\sin \theta }$
Now $H_x=BD=(5+3\cos \theta )\left(\frac{0.5}{3\sin \theta }\right)$
$=\frac{5+3\cos \theta }{6\sin \theta }$
H to be minimum
$\frac{dH}{d\theta }=0$
$\frac{d}{d\theta }\left(\frac{5+3\cos \theta }{6\sin \theta }\right)=0$
$6\sin \theta \left(3\right(-\sin \theta \left)\right)-(5+3\cos \theta )\left(6\cos \theta \right)=0$
$-18{\sin }^2\theta -30\cos \theta -18{\cos }^2{\theta }^2\theta =0$
$-18({\sin }^2\theta +{\cos }^2\theta )-30\cos \theta =0$
$-18-30\cos \theta =0$
$\cos \theta =\frac{-18}{30}$
$=\frac{-3}{5}$
$\sin \theta =\sqrt{1-{\cos }^2\theta }$
$=\frac{4}{5}$
$H=\frac{5+3\cos \theta }{6\sin \theta }$
$=\frac{5+3\left(\frac{-3}{5}\right)}{6\times \frac{4}{5}}$
$=\frac{16}{24}=\frac{2}{3}$ km