A river ` 800 m` wide flows at the rate of ` 6 km h^(_1)` in still water, wishes to cross the tiver straight. ltbRgt (a) Along what direction must he strike ?
(b) What should be his resultant velocity ?
(c ) How much time he would take ?

Explanation

Width of river ` = 800 m= 0.8 km , ` velocity of water flowing in river, ` v_r = 6 km h^(-1)`, velocity of swimmer, ` v_s= 10 km//h`. The swimmer will cross the river straight if the resultant velocity of river flow and swimmer acts perpendicular to the direction fo river flow. It will be so if swimmer moves making an angle ` theta` with upstream of river flow. Fig. 2 (c ) . 83 (a).
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Then, ` cos theta = 6/10 =06 ` or ` theta = 53^@ 8'`
Resultant velocity of swimmer, ltbRgt ` v = sqrt ( v_s^2 - v_r^2) = ( 10^2 - 6 0 ^(1//2) = 8 km h^(-1)`
Tiem taken to cross the river, ltbRgt ` t= S/v = (08)/8 = 0.1 h=0.1 xx 60 min = 6 min`.