Question

A man wishes to cross a river and come back to original position by swimming minimum distance. If the width of the river is 50 m, speed of river is 2 m/s and velocity of man with respect to river is 4 m/s, then time taken by man to come back to original position is

• A.
  $\frac{50}{\sqrt{3}}$ sec
• B.
  50$\sqrt{2}$ sec
• C.
  25 sec
• D.
  25$\sqrt{3}$ sec

Answer 1

The correct option is A

$\frac{50}{\sqrt{3}}$ sec
For minimum distance
$v_r=v_{m}sin\theta \Rightarrow 2=4sin\theta$
$\therefore v_{m}cos\theta =4cos30=2\sqrt{3}$
so time $t=\frac{2w}{v_{m}cos\theta }=\frac{2\times 50}{2\sqrt{3}}=\frac{50}{\sqrt{3}}$