Question 153: Motion in 2D & Vectors

Question

The speed of a swimmer is $$4 \mathrm{~km} \mathrm{~h}^{-1}$$ in still water. If the swimmer makes his strokes normal to the flow of river of width $$1 \mathrm{~km}$$, he reaches a point $$750 \mathrm{~m}$$ down the stream on the opposite bank.
The speed of the river water is ___________ $$\mathrm{km} ~\mathrm{h}^{-1}$$

Accepted Answer

Time to cross the River width $\omega=1000 \mathrm{~m}$ is $=\frac{1 \mathrm{~km}}{4 \mathrm{~km} / \mathrm{h}}$
Drift $\mathrm{x}=\mathrm{Vm} / \mathrm{g} 	imes \mathrm{t}$
Where $\mathrm{Vm} / \mathrm{g}$ is velocity of River w.r. to ground.
$$
\begin{aligned}
& \mathrm{x}=\mathrm{Vm} / \mathrm{g} 	imes \frac{1}{4}=750 \mathrm{~m}=\frac{3}{4} \mathrm{~km} \\
& \mathrm{Vm} / \mathrm{g}=3 \mathrm{~km} / \mathrm{hr}
\end{aligned}
$$

Motion in 2D & Vectors

Integer / Numeric Answer Type