Question
Sat December 31, 2011 By:

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

Expert Reply
Fri January 13, 2012
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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