Question

A river flows $3{\text{km h}}^{-1}$ and a man is capable of swimming $2{\text{km h}}^{-1}$. He wishes to cross it such that displacement parallel to river is minimum. In which direction will 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 that he swims at an angle $\theta$ with the perpendicular as shown. If 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.
$x=v_{x}t=(3-2\sin \theta )\frac{l}{2\cos \theta }$
for $x$ to be min $\frac{dx}{d\theta }=0$, or
$l[\frac{3}{2}\sec \theta \tan \theta -{\sec }^2\theta ]=0\text{or}\sin \theta =\frac{2}{3}$