Question
Sat December 31, 2011 By:

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

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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