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 (\theta\) with the flow, find the time he takes to cross the river.

(b) Find the shortest possible time to cross the river.

Answer 1

(a) The vertical compoent 3 sin $\theta$ take him to opposite side.

Distance = 0.5 km

Velocity = 3 sin \(\theta) km/h

Time = $\frac{\text{Distance}}{\text{Velocity}}=\frac{0.5}{3Sin\theta }hr$

= $\frac{0.5}{3Sin\theta }\times 60\frac{1}{Sin\theta }min$

(1 hr=60 min.)

(b) Here vertical component of velocity, i,e. 3 km/hr takes him to opposite.

$\therefore t=\frac{d}{v^{\prime }}=\frac{\frac{500}{3\times 1000}}{60}$

= 10 minutes