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,
$t=\frac{d}{v} \ldots$ (i)
By shortest distance method, time taken,
$t'=\frac{d}{v \cos \left(90^{\circ}-\theta\right)}$
$t'=\frac{d}{v \sin \theta} \ldots \ldots \ldots$ (ii)
$\therefore \frac{t}{t}=\frac{d / v}{d / v \sin \theta}$
$=\sin \theta$