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Vector Kinematics Take 2

  1. Sep 18, 2007 #1
    A particle is moving in the xy plane with velocity v(t) = Vx(t)i + Vy(t)j
    and acceleration a(t) = ax(t)i + ay(t)j
    By taking the appropriate derivative, show that the magnitude of v can be constant only if axvx + ayvy = 0

    I did it in a similiar way to the other problem I had as it seems like a similiar problem,
    My thinking is:

    h = (Vx(t)i)^2 + (Vy(t)j)^2

    s(t) = sqrt(h)

    dh/dt = d(vx^2 + vy^2) / dt
    = d(vx)^2/dt + d(vy)^2/dt
    = (d(vx^2)/dvx)(dvx/dt) + (d(vy^2)/dvy)(dy/dt)
    = 2vx(dx/dt) + 2vy(dy/dt)

    ds/dt = (ds/dh)(dh/dt)
    = ((1/2)h^(-1/2))[(2vx(dvx/dt))+(2vy(dvy/dt))]

    ((1/2)h^(-1/2))[(2vx(dvx/dt))+(2vy(dvy/dt))] = 0

    ((1/2)h^(-1/2))[(2vx(ax))+(2vy(ax))] = 0

    2((1/2)h^(-1/2))[vxax+vyax] = 0

    axvx + ayvy = 0


    Thanks
     
  2. jcsd
  3. Sep 18, 2007 #2

    learningphysics

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    Homework Helper

    Looks good! Very nice!
     
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