Question

A boat which has a speed of $5 \, km \, h^{- 1}$ in still water crosses a river of width $1 \, km$ along the shortest possible path in $15$ minutes. The velocity of the river water in $km \, h^{- 1}$ is

• A. $1kmh^{- 1}$
• B. $3 \, km \, h^{- 1}$
• C. $4 \, km \, h^{- 1}$
• D. $\sqrt{41} km h^{-1}$

Answer 1

The Correct Option is B

Vertical displacement = $1km$ $t=\frac{15}{60}=\frac{1}{4}h$ $\therefore \, \, \, V_{b}cos\theta =\frac{ \, \, 1 \, \, }{\left(\right. \frac{1}{4} \left.\right)}=4kmh^{- 1}$ $\Rightarrow cos\theta =\frac{4}{V_{b}}=\frac{4}{5} \\ \Rightarrow sin\theta =\frac{3}{5}$ By velocity triangle ABC, $ sin \theta = \frac{V_{r}}{V_{b}}$ $\therefore \, \, \, \frac{V_{r}}{V_{b}}=\frac{3}{5} \\ \Rightarrow \frac{V_{r}}{5}=\frac{3}{5} \\ \Rightarrow V_{r}=3kmh^{- 1}$