Two swimmers leave point A on the 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 of his walking if both swimmers reached the destination simultaneously? The stream velocity `v_0=2.0 km//hour` and the velocity `v^'` of each swimmer with respect to water equals to `2.5km` per hour.

Answer

`u=(v_0)/((1-v_0^2//v^('^2))^(-1//2)-1)=3.0km` per hour.

Explanation

Let one of the swimmer (say 1) cross the river along AB, which is obviously the shortest path. Time taken to cross the river by the swimmer 1.
`t_1=(d)/(sqrt(v^('^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.


In the time `t_2`, drifting of the swimmer 2, becomes
`x=v_0t_2=(v_0)/(v^')d`, (using Eq. 2) (3)
If `t_3` be the time for swimmer 2 to walk the distance x to come from C to B(fig), then
`t_3=x/u=(v_0d)/(v^'u)` (using Eq. 3) (4)
According to the problem `t_1=t_2+t_3`
or, `(d)/(sqrt(v^('^(2))-v_0^2))=d/v^'+(v_0d)/(v^'u)`
On solving we get
`u=(v_0)/(((1-v_0^2)/(v^('^2)))^(-1/2)-1)=3km//hr`.