Question

A man can swim in still water with a speed of $2m{s}^{-1}$. If he wants to cross a river of water current speed $\sqrt{3}m{s}^{-1}$ along shortest possible path, then in which direction should he swim?

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

Answer 1

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

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,

$v_m\sin \theta =v_r$

$2\times \sin \theta =\sqrt{3}$

$\theta ={60}^{\circ }$

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.