Question

A river is flowing from West to East at a speed of $5\text{m/min}$. A man on the south bank of the river, capable of swimming at speed $10\text{m/min}$ in still water, wants to swim across the river in the shortest time. What angle with the bank of the river should he swim at? (in degree) .

Answer 1

Velocity of river, $V_r=5\text{m/min}$
Velocity of man w.r.t river, $V_m=10\text{m/min}$

Let the width of the river be $d$ and assume the man heads in a direction making angle $\theta$ with the bank of the river.


On breaking component of $V_m$ along and perpendicular to river flow


We can see that component $V_m\sin \theta$ will be responsible for crossing the river.

Time taken to cross river,
$t=\frac{\text{width of river}}{\text{perpendicular component of velocity}}$

$\therefore t=\frac{d}{V_m\sin \theta }$

For $t$ to be minimum, $\sin \theta$ should be maximum, and maximum value of $\sin \theta$ is $1$ at $\theta =\frac{\pi }{2}={90}^{\circ }$.

$\therefore$ Man should head at angle of ${90}^{\circ }$ with the bank of river i.e. in North direction.