A river is flowing due east with a speed ` 3 m//s` (Fig. 2 (EP) .31 ).
. (a) If swimmer starts swimming due north, 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

Given , Speed of the river `(v_(r))`=3 m/s (east)
Speed of swimmer `(v_(s))=4m//s`(east)
(a) When swimmer starts swimming due north then his resultant velocity
`v=sqrt(v_(r)^(2)+v_(s)^(2))=sqrt((3)^(2)+(4)^(2))`
`=sqrt(9+16)=sqrt(25)=5m//s`
`tan theta=(v_(r))/(v_(s))=(3)/(4)`
`=0.75=tan 36^(@)54'`
Hence, `" " theta = 36^(@)54'N`

(b) To reach opposite points B, the swimmer should swim at an angle `theta` of north.
Resultant speed of the swimmer
`v=sqrt(v_(s)^(2)-v_(r)^(2))=sqrt((4)^(2)-(3)^(2)) `
`=sqrt(16-9)=sqrt(7) m//s`
`tan theta =(v_(r))/(v)=(3)/(sqrt(7))`
`rArr " " theta = tan ^(-1) ((3)/(sqrt(7)))` of north

(c) In case (a),
Time taken by the swimmer to cross the river , `t_(1)=(d)/(v_(s))=(d)/(4)s`
In case (d) ,
Time taken by the swimmer to cross the river
`t_(1)=(d)/(v)=(d)/(sqrt(7))`
As `" " (d)/(4) lt (d)/(sqrt(7))`, therefore `t_(1) lt t_(2)`
Hence , the swimmer will cross the river in shorter time in case (a).