Question

A man can swim at the rate of $5km{h}^{-1}$ in still water. A 1-km wide river flows at the rate of $3km{h}^{-1}$. The man wishes to swim across the river directly opposite to the starting point
a. Along what direction must the man swim?
b. What should be his resultant velocity?
c. How much time will he take to cross the river?

Answer 1

a. Velocity of man with respect to river water, $v=5km{h}^{-1}$. This is greater than the river flow velocity. Therefore, he can cross the river directly (along the shortest path or no drift condition from flow velocity). The angle of swim,
$\theta =\frac{\pi }{2}+si{n}^{-1}\left(\frac{u}{v}\right)$
$={90}^o+si{n}^{-1}\left(\frac{u}{v}\right)$
$={90}^o+si{n}^{-1}\left(\frac{3}{5}\right)={90}^o+{37}^o={127}^o$ w.r.t. the river flow or ${37}^o$ w.r.t. the perpendicular in upstream direction
b. Resultant velocity or velocity of mass will be
$v_m=\sqrt{v^2-u^2}=\sqrt{5^2-3^2}=4km{h}^{-1}$
In the direction perpendicular to the river flow.
c. time taken to cross the river
$t=\frac{d}{\sqrt{v^2-u^2}}=\frac{1km}{4km{h}^{-1}}=\frac{1}{4}h=15min$