Question

A river flows $3km/h$ and a man is capable of swimming at the rate of $2km/h$. He wishes to cross it such that the displacement parallel to river is minimum. In which direction should he swim?

• A. ${\sin }^{-1}\left(\frac{2}{3}\right)$
• B. ${\cos }^{-1}\left(\frac{2}{3}\right)$
• C. ${\tan }^{-1}\left(\frac{2}{3}\right)$
• D. ${\cot }^{-1}\left(\frac{2}{3}\right)$

Answer 1

The correct option is A ${\sin }^{-1}\left(\frac{2}{3}\right)$

Let us assume he swims at an angle $\theta$ with the perpendicular as shown in figure.
If the river is $lm$ wide, time taken to cross it,
$t=\frac{l}{2\cos \theta }$,
$v_x=3-2\sin \theta$
horizontal distance covered along $x$ direction during this period $x=v_x\times t$.
$x=(3-2\sin \theta )\frac{l}{2\cos \theta }$
for $x$ to be minimum $\frac{dx}{d\theta }=0$, or
$l[\frac{3}{2}\sec \theta \tan \theta -{\sec }^2\theta ]=0$ or
$\sin \theta =\frac{2}{3}$