Question

A swimmer can swim in still water with speed ' $v$ ' and the river flowing with velocity $v/2$, To cross the river in shortest time, he should swim making angle ' $\theta$ ' with the upstream. What is the ratio of the time taken to swim across in the shortest time to that in swimming across over shortest distance?

• A. $\cos \theta$
• B. $\cot \theta$
• C. $\sin \theta$
• D. $\tan \theta$

Answer 1

Answer: (C)

If $d$ be the width of river and $\theta$ be the angle made with the upstream.
Therefore, angle made with normal stream
$=90^{\circ }-\theta$
Hence, by shortest time method, time taken,
$t=\frac{d}{v}\dots$ (i)
By shortest distance method, time taken,
$t^{\prime }=\frac{d}{v\cos \left(90^{\circ }-\theta \right)}$
$t^{\prime }=\frac{d}{v\sin \theta }\dots \dots \dots$ (ii)
$\therefore \frac{t}{t}=\frac{d/v}{d/v\sin \theta }$
$=\sin \theta$