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Vector Problem involving Cartesian plane

  1. Sep 5, 2010 #1
    My main issue is trying to understand the question being asked (is it asking for the magnitude of the resultant vector for speed?)
    1. The problem statement, all variables and given/known data
    Particle 1 is moving on the x-axis with an
    acceleration of 5.00 m/s^2 in the positive x-
    direction. Particle 2 is moving on the y-
    axis with an acceleration of 6.00 m/s^2 in the
    negative y-direction. Both particles were at
    rest at the origin at t = 0 s.
    Find the speed of particle 1 with respect to particle 2 at 4.00 s. Answer in m/s.


    2. Relevant equations
    I know that this is a problem involving constant acceleration so:
    [tex]v = v_0 + a t[/tex] One for the x direction and one for the y direction.




    3. The attempt at a solution
    So V0x and V0y are both 0. Then, Vxf=(5m/s^2)(4s)=20 m/s and Vyf=(-6m/s^2)(4s)=-24 m/s. Now the problem is whether the problem wants me to find the resultant speed (the wording is somewhat confusing for me). If so, then sqrt(20^2+(-24)^2)) or about 31.2 m/s. If not, I have no idea. I'd appreciate if someone could let me know if I'm on the right track. Thanks!
     
  2. jcsd
  3. Sep 6, 2010 #2

    rl.bhat

    User Avatar
    Homework Helper

    Hi proster, welcome to PF.

    If x is the distance traveled by particle 1 along x-axis and y is the distance traveled by the particle 2 along the y-axis, then

    x^2 + y^2 = R^2.

    By taking the derivative you get

    2x*dx/dt + 2y*dy/dt = 2R*dR/dt..........(1)

    x = 1/2*a(x)t^2 and dx/dt = V(x) = at.

    y = 1/2*a(y)t^2 and dy/dt = V(y) = at.

    R^2 = x^2 + y^2 So R = sqrt(x^2 + y^2)

    Substitute these values in eq.(1) and find dR/dt.
     
  4. Sep 6, 2010 #3
    Thanks a lot! I got the same answer as the way I did it the first time, but it's definitely an interesting way of thinking about the problem.
     
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