Question

A boat which has a speed of $5\text{km/h}$ in still water crosses a river of width $1\text{km}$ along the shortest possible path in $15\text{min}$. The magnitude of velocity of the river water in $\text{km/h}$ is

• A. $1$
• B. $3$
• C. $4$
• D. $\frac{4}{3}$

Answer 1

The correct option is B $3$


Width of the river $=1\text{km}$,
Velocity of boat w.r.t river $V_{Br}=5\text{km/h}$
The boat crosses the river in time $t=15\text{min}=1/4\text{h}$
To cross the river along shortest possible path, the velocity of boat w.r.t ground must be perpendicular to the river.
So, the net horizontal velocity must be zero.
$\Rightarrow V_{Br}\sin \theta -V_r=0$
$\Rightarrow 5\sin \theta =V_r.........\left(1\right)$
We can also write,
$V_{Br}\cos \theta =\frac{1}{\frac{1}{4}}=4\text{km/h}$
$\Rightarrow 5\cos \theta =4$
$\Rightarrow \cos \theta =\frac{4}{5}$
$\Rightarrow \sin \theta =\frac{3}{5}.........\left(2\right)$
Using $\left(1\right)$ and $\left(2\right)$, we can say
$V_r=3\text{km/h}$