Question

A boat which has a speed $5\text{km/h}$ in still water crosses a river $100\text{m}$ wide along the shortest possible path in $1.5$ minutes, the velocity of river water in $\text{km/h}$ is

Answer 1

The shortest path that the boat could cross is $OB$. For this, the boat will have to move against the direction of flow of river at some angle.
Assuming origin at point $O$, rightward direction be positive $x-$ axis and $OB$ be positive $y-$ axis.
Let velocity of the river be, $\overrightarrow{v_r}=v\hat{i}$ and velocity of boat w.r.t. the ground be $\overrightarrow{v_b}$.
Given that velocity of the boat w.r.t. river,
$\left|{\overrightarrow{v}}_{br}\right|=5\text{km/h}$
Now, width of the river, $OB=100\text{m}=0.1\text{km}$
Time taken by the boat to cross the river,
$t=1.5\text{min}=\frac{1.5}{60}\text{h}$
$\therefore \left|{\overrightarrow{v}}_b\right|=\frac{OB}{t}=\frac{0.1\text{km}}{\frac{1.5}{60}\text{h}}=\frac{60}{15}=4\text{km/h}$
$\Rightarrow \overrightarrow{v_b}=4\hat{j}\text{km/h}$
Velocity of boat w.r.t. river, $\overrightarrow{v_{br}}=\overrightarrow{v_b}-\overrightarrow{v_r}$
${\overrightarrow{v}}_{br}=4\hat{j}-v\hat{i}$
$\left|{\overrightarrow{v}}_{br}\right|=5=\sqrt{4^2+v^2}$
$v=\sqrt{25-16}=3\text{km/h}$
$\therefore$ Velocity of river $=3\text{km/h}$