Question

A river is flowing from west to east at a speed of $5$ metre per minutes. A man on the south bank of the river, capable of swimming at $10$ metre per minute in still water, wants to swim across the river in the shortest time. He should swim in a direction.

Answer 1

According to diagram

Velocity of river with respect to ground $=V_{RG}$

Velocity of man with respect to river $=V_{MR}$

Velocity of man with respect to ground $=V_{MG}$

Now,

${\overrightarrow{V}}_{MR}+{\overrightarrow{V}}_{RG}={\overrightarrow{V}}_{MG}.....\left(I\right)$

${\overrightarrow{V}}_{MG}$ is resultant of ${\overrightarrow{V}}_{MR}$ ${\overrightarrow{V}}_{RG}$

Now, the magnitude of $V_{RG}$

$=\sqrt{5^2+{10}^2+2\times 5\times 10\cos {45}^0}$

$=\sqrt{25+100+50\sqrt{2}}$

$=\sqrt{195.7}$

$=13.9m/min$

For the angle b/w $V_{MR}$ and $V_{mg}$

$\cos \alpha =\frac{V_{MR}^2+V_{MG}^2-V_{RG}^2}{2\times V_{MR}\times V_{MG}}$

$\cos \alpha =\frac{268.21+100-25}{278}$

$\cos \alpha =0.96$

$\alpha ={16.26}^0$

So, the heading angle will be

${16.26}^0+{45}^0={61.26}^0$ from east to north