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