mam i dont know how to do this.plz answer.

Here main thing is to note that the 1 one is trying to swim at an angle theta and reached directly at B. The 2 one try to swim straight and reach at C then he walk toB.

 

let the width of river =d=AB

 

and let BC=x.

 

let the time taken by first person to reach at B =t.

 

let the time traken by 2 person to reach at C= t1

 

and to reach from Cto B=t2

 

so t=t1+t2--------(1)

 

For the first person-----

 

from the figure--------

 

U COS(?)=d/t----(2)

 

U sin(?)=2

 

Sin(?)=2/U=2/2.5

 

means cos (?)=(?2.25)/2----(3)

 

from (2) and (3)

 

t=d/?2.25----(4)

 

For the 2one--------

 

t1=d/2.5-----(5)

 

( because for the 2 one the the drift perpendicular to the river is only due to his own velocity 2.5, as the river velocity is perpendicular to his velocity)

 

t1=x/2---(6)

( because the displacement of the 2 one along the river is only due to the river velovity as , his velocity is perpendicular to the river velocity)

 

from (5) and(6)

 

x=2d/2.5----(7)

 

now he cover some distance by walking

 

t2=x/v=2d/2.5v----(8)

 

from 1, 4, 5 and 8

 

d/?2.25=d/2.5+2d/2.5v

 

solve above expresiuon and find v.