A river is flowing due east with a speed ` 3 m//s` (Fig. 2 (EP) .31 ).
. (a) If swimmer starts swimming due borth, what will be his resultant velocity (magnitude and direction) ? (b) If he wants to start from point (A) on South bank and reach opposite point (B) on North bank,
(i) Which direction should he swim? (ii) What will be his resultant speed ? (c ) From two differenrent casses as mentioned in (a) and 9b) above, in which casse will he reach opposite bank in shorter time ?

Explanation

(a) Refer Fig. 2 )EP) 32 (a) , ` v_r =3 m//s along AD`
Resultant velocity ` v= sqrt (v_r^2 +v_s^2) = sqrt (3^2 + 4^2 = 5 m//s`
` tan theta = v_r/v_s = 3/2 = 0. 75 = tan 36^2 54' ~~tan 37^@`
` theta = 37^@` with borth towars the direction of river flow.
.
(b) Refer Fig, 2 (EF) . 32 (b) , the swimmer will go along (AB) if he goes along (AC), so that reaultant of river velocity and swimmer veoocuitymay act along (AB).
` Velocity along (AB), v = sqrt( v_s^2 = v_r^2 ) = sqrt ( 4^2 - 3^2 ) = sqrt 7 m//s`
Let ` /_BAC = theta , ehen tan theta = (v_r / v = 3/(sqrt 7) ` or ` tan ^(-1) ( 3/( sqrt 7)) West of Noth
(C) In case (a) , time taken to ceross then river , `t_1 = d/ v = d/ (sqrt 7)`
As ` d/ 4 lt d/ (sqrt 7) , so ` t_1 lt t_2`, therefore the swimmer will cross the river in showrte time in care (a).