A man can swim at the rate of `5 kmh^(-1)` in still water A. One km wide river flows at the rate of ` 3 kmh^(-1)` . The man wishes to swim across the river directly opposite to the starting point. How much time will be take to cross the river

Explanation

Given, width of rivet ` =1 km`.
Velocity of swimmer, ` v_s = 5 km//h , Velocity of water flowing in river, ` v_r =3 km h^(-1)` along ` OA`.
(a) The swimmer will cross straight if the resultant velocity of river flow and swimmer acts perpendicular to the dierction of river flow i.e. along ` OC`. It will be so if swimmer moves making an angle ` alpha` with upstream i.e. goes along ` OB` Refer to Fig. 2 (c ) . 48. Here, theta _ alpha= 90^@ or theta = (90^@ -alpha) `
.
In ` Delta OBC, sin theta = sin ( 90^@-alpha)`
`= cos alpha = (BC)/(OB) = v_r/v_s = 3/5 = 0.6`
or ` cos alpha = cos 53^@ 8'`
or ` alpha =53^@ 8'` upstream.
(b) Let (v) be the resultant velcity along ` OC`, then
` v = sqrt 9 v-s ^2 -v_r ^20 = sqrt 5^2-3^2) = 4 km//h`
(c ) Time taken by swimmer to cross the river is
` t = (1 km)/( 4 km //h) = 1/4 h =15 minutes.