Question:
A swimmer wishes to cross $500 \mathrm{~m}$ wide river flowing at $5 \mathrm{~km} / \mathrm{h}$. His speed with respect to water is 3 $\mathrm{km} / \mathrm{h}$.
(a) If he heads in a direction making an angle $\theta$ with the flow, find the time he takes to cross the river.
(b) Find the shortest possible time to cross the river.
Solution:
(a) Velocity responsible for crossing $=3 \sin \theta \mathrm{kmph}$
$=3 \times \frac{5}{18} \sin \theta$
Time to cross= distance/speed
$=\frac{500 \times 18}{3 \times 5 \sin \theta}=\frac{600}{\sin \theta}$
$=\frac{10}{\sin \theta}$ minutes
(b) For $t_{\min } ; \sin \theta=1$
When $\theta=90^{\circ}$
$\mathrm{t}_{\min }=10$ minutes