A swimmer can swim in still water with speed \(v\) and the river flowing with velocity \(\frac{v}{2}\). To…
A swimmer can swim in still water with speed \(v\) and the river flowing with velocity \(\frac{v}{2}\).
To cross the river in shortest time, he should swim making angel \(\theta\) 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?
\(\cos \theta\)
\(\cot \theta\)
\(\sin \theta\)
\(\tan \theta\)
Solution
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,
\(\text {time taken, } t=\frac{d}{v}\) ...(i)
By shortest distance method,
\(\begin{aligned}
\text {time taken, } t^{\prime} & =\frac{d}{v \cos \left(90^{\circ}-\theta\right)} \\
t^{\prime} & =\frac{d}{v \sin \theta} \quad \ldots (ii) \\
\therefore \quad \frac{t}{t^{\prime}} & =\frac{d / v}{d / v \sin \theta}=\sin \theta
\end{aligned}\)