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Kinematics particle velocity problem

  1. Sep 13, 2008 #1
    1. The problem statement, all variables and given/known data

    This is from the 1983 ap test. I'm using it to study the concepts but so far it only confused me more.

    It goes:

    A particle moves so that the x-component of its velocity has the constant value vx = C; that is, x = Ct

    1. Determine the y-component of the particle's velocity as a function of x.

    2. Determine the y-cpmponent of the particle's acceleration.

    Part b.

    Suppose, instead, that the particle moves along the same parabola with a velocity whose x-component is given by vx = C/(a+x2)1/2

    3. Show that the particle's speed is constant in this case.

    3. The attempt at a solution

    I have the solution, but it doesn't make any sense to me. For the first question they showed dy/dt = (dy/dx)(dx/dt) They said it's the chain rule, but where did that come from? I thought I knew the chain rule until I saw that. Where did t come from?

    I can't figure out where the answer to the 2nd question is from either. They put ay = (dvx/dt) = (d/dt)(C2t) = C2
    I can't figure out where that came from either.

    Number 3 confuses me even more than the previous 2. This is what they did:

    v = sqrt(vx2 + vy2)

    vy = dx/dt = (dy/dx)(dx/dt) = xvx

    v = sqrt((vx2)(1 + x2)) = sqrt( (C2/ 1+x2)(1 + x2)) = C

    If you have access to 1983 mech #1 everything looks much better than what I typed.

    Can someone please explain all this to me? I'm trying to prepare for the ap test by looking at old problems, but so this one is written in hieroglyphics and I could really use some help.

    And please explain how the chain rule makes dy/dx = (dy/dx)(dx/dt)

    Any help is greatly appreciated.
    Last edited: Sep 13, 2008
  2. jcsd
  3. Sep 13, 2008 #2
    Comments on the chain rule:

    [tex]\frac{dy}{dt}[/tex] may be written as [tex]\frac{dy}{dt}\frac{dx}{dx}[/tex]. Rearranging we get [tex]\frac{dy}{dx}\frac{dx}{dt}[/tex]. Does this make a bit more sense by writing in the middle step?

    For acceleration [tex]a = \frac{dv}{dt}[/tex] and velocity is [tex]v \frac{dx}{dt}[/tex]. We put these together by substituting v to get [tex] a = \frac{d2x}{dt2}[/tex].
  4. Sep 13, 2008 #3
    That last formula was meant to be

    [tex] a = \frac{d^{2}x}{dt^{2}}[/tex]
  5. Sep 13, 2008 #4
    Thanks for the chain rule explanation. It makes a lot more sense now.
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