Question

A man crosses a $ 320\,\,m $ wide river perpendicular to the current in $4\, \min.$ If in still water he can swim with a speed $ 5/3 $ times that of the current, then the speed of the current, in $ mm^{-1} $ in is

• A. 30
• B. 40
• C. 50
• D. 60

Answer 1

Answer: (D)

$v_{r}^{2}=v_{m}^{2}-v^{2}$
$v=\frac{320}{4} \,m / \min =80\, m / \min$
$v_{m}=\frac{5}{3} v_{r}$
$v_{r}^{2}=t \frac{5}{3}\left(v_{r}\right)^{2}-(80)^{2}$
$\frac{16}{9} v_{r}^{2}=(80)^{2}$
$v_{r}=60 \,m / \min$
$u^{2}, u_{0}$ form canours linear