Question

A man can swim in still water with a speed of $2m/s$. If he wants to cross a river of water current speed $\sqrt{3}m/s$ along shortest possible path, then in which direction should he swim?

• A. At an angle ${90}^{\circ }$ to the water current
  
• B. None of these
• C. At an angle ${120}^{\circ }$ to the water current
  
• D. At an angle ${150}^{\circ }$ to the water current
  

Answer 1

The correct option is D At an angle ${150}^{\circ }$ to the water current


Draw a labelled diagram.

Find velocity component acting opposite to that of the velocity of water current.

Given,
The speed of the man is $2m/s$ and the speed of the water current is $\sqrt{3}m/s$.

The velocity component acting opposite to that of the velocity of water current is given as,

when, $v_w<v_{BW}$
$v_m\sin \theta =v_r$
$2\times \sin \theta =\sqrt{3}$
$\theta ={60}^{\circ }$

The angle between the speed of man and the water current is given as,

${\theta }_1={60}^{\circ }+{90}^{\circ }$
$={150}^{\circ }$

Thus, the direction of swimming is at an angle ${150}^{\circ }$ to the water current.