Question

A river $800m$ wide flows at the rate of $5km{h}^{-1}$ . A swimmer who can swim at $10km{h}^{-1}$ in still water, wishes to cross the river straight.
(i) Along what direction must he strike?
(ii) What should be his resultant velocity?
(iii) How much time he would take?

Answer 1

Let the angle between the swimmer and the perpendicular to the flow of river is $\theta$ with the perpendicular to the flow of river and time to cross the river is $t$

(i)

The direction is given as,

$\sin \theta =\frac{5}{10}$

$\theta ={30}^{\circ }$

(ii)

The resultant velocity is given as,

$V=V_{sg}\cos {30}^{\circ }$

$=10\cos {30}^{\circ }$

$=8.66km/h$

(iii)

Displacement in vertical direction is given as,

$\left(V_{sg}\cos \theta \right)t=800\times {10}^{-3}$

$\left(10\cos {30}^{\circ }\right)t=800\times {10}^{-3}$

$t=0.092s$

$t=5.5min$