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Prove Velocity is constant

  1. Sep 8, 2007 #1
    1. The problem statement, all variables and given/known data
    A particle is moving in the xy plane with velocity [itex]\vec{v}(t)=v_x(t)\vec{i}+v_y(t)\vec{j}[/itex] and acceleration [itex]\vec{a}(t)=a_x(t)\vec{i}+a_y(t)\vec{j}[/itex]. By taking the appropiate derivative show that the magnitude of v can be constant only if [itex]a_xv_x+a_yv_y = 0[/itex]

    2. Relevant equations



    3. The attempt at a solution
    So I know that in order for velocity to have a constant magnitude then acceleration must = 0.

    Since acceleration is the derivative of velocity and dv/dt=0 iff v=some constant.
    expanding the LHS:

    [tex]\frac{d\vec{v}(t)}{dt}=0[/tex]

    [tex]\frac{dv_x(t)}{dt}\vec{i}+\frac{dv_y(t)}{dt}\vec{j} = 0[/tex]

    [tex]a_x(t)\vec{i}+a_y(t)\vec{j} = 0[/tex]

    since i,j are unit vectors and do not equal 0 the components of acceleration must = 0.

    [tex]a_x(t)=a_y(t)=0[/tex]

    But it's also possible for [itex]a_x=-a_y[/itex] in which case the acceleration would still = 0 so is this a proof really correct?
     
  2. jcsd
  3. Sep 8, 2007 #2

    learningphysics

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    Homework Helper

    The velocity need not be constant. The magnitude of the velocity is constant. Get the magnitude of the velocity, and use the fact that it is constant... go from there..
     
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