Question

Two swimmers leave point $A$ on one bank of the river to reach point $B$ lying right across on the other bank. One of them crosses the river along the straight line $AB$ while the other swims at right angles to the stream and then walks the distance that he has been carried away by the stream to get to point $B$. What was the velocity $u$ in km/hr of his walking if both swimmers reached the destination simultaneously?The stream velocity $v_0=2.0km/h$ the velocity $v^{\prime }$ of each swimmer with respect to water equals $2.5km/h$

Answer 1

Let one of the swimmer (say 1) crosses the river along $AB$, which is obviously the shortest path. Time taken to cross the river by the swimmer 1.
$t_1=\frac{d}{\sqrt{{v^{\prime }}^2-v_0^2}}$, (where $AB=d$ is the width of the river) (1)
For the other swimmer (say 2), which follows the quickest-path, the time taken to cross the river.
$t_2=\frac{d}{v^{\prime }}$ (2)
In the time $t_2$, drifting of the swimmer 2, becomes
$x=v_0{t}_2=\frac{v_0}{v^{\prime }}d$, (using equation 2) (3)
If $t_3$ be the time for swimmer 2 to walk the distance $x$ to come from $C$ to $B$ (shown in figure above), then
$t_3=\frac{x}{u}=\frac{v_{0}d}{v^{\prime }u}$ (using equation 3) (4)
According to the problem $t_1=t_2+t_3$
or, $\frac{d}{\sqrt{{v^{\prime }}^2-v_0^2}}=\frac{d}{v^{\prime }}+\frac{v_{0}d}{v^{\prime }u}$
On solving, we get
$u=\frac{v_0}{{\left(\frac{1-v_0^2}{v^{\prime 2}}\right)}^{-\frac{1}{2}}-1}=3.0km/h$