Question

A swimmer wishes to cross a 500 m wide river flowing at 5 km/h. His speed with respect to water is 3 km/h.
(a) If he heads in a direction making an angle θ with the flow, find the time he takes to cross the river.
(b) Find the shortest possible time to cross the river.

Answer 1

Given:
Width of the river = 500 m
Rate of flow of the river = 5 km/h
Swimmer's speed with respect to water = 3 km/h
As per the question, the swimmer heads in a direction making an angle θ with the flow.
We know that the vertical component of velocity 3sinθ takes him to the opposite side of the river.
Distance to be travelled = 0.5 km
Vertical component of velocity = 3sinθ km/h
Thus, we have:
$\mathrm{Time}=\frac{\mathrm{Distance}}{\mathrm{Velocity}}=\frac{0.5}{3\sin \theta }\mathrm{h}=\frac{500\times 6}{5\sin \theta }=\frac{600}{\sin \theta }\mathrm{s}$
∴ Required time = $\frac{10\mathrm{minutes}}{\sin \theta }$


(b) Shortest possible time to cover the river:
Take θ = 90^∘
$\mathrm{Time}=\frac{0.5}{3\sin \theta }\mathrm{h}=\frac{500\times 6}{5\sin \theta }=\frac{600}{\sin 90°}\mathrm{s}=600\mathrm{s}$
Hence, the required time is 10 minutes.