1800-212-7858 (Toll Free)
9:00am - 8:00pm IST all days

or

Thanks, You will receive a call shortly.
Customer Support

You are very important to us

022-62211530

Mon to Sat - 11 AM to 8 PM

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

Asked by 31st December 2011, 11:48 AM
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.

Answered by Expert 13th January 2012, 5:05 PM
• 1
• 2
• 3
• 4
• 5
• 6
• 7
• 8
• 9
• 10

You have rated this answer /10