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Algebra in Constant Motion Derivation.

  1. Sep 22, 2009 #1
    [tex] x = x_o + v_0\Deltat + \frac{1}{2} a ( \Delta t )^2 [/tex]

    [tex] If [/tex] [tex] \frac{v - v_0}{a} = \Delta t [/tex]

    [tex] x = x_0 + v_0 (\frac{v - v_0}{a} ) + \frac{1}{2}a(\frac{v - v_0}{a})^2[/tex]

    [tex] \Rightarrow 2 a (x - x_0) = 2v_0v - v_0^2 + v^2 - 2v_0v v_0^2 [/tex]

    [tex] \Rightarrow 2 a (x - x_0) = v^2 - v_0^2 [/tex]

    [tex] \Rightarrow v^2 = v_0^2 + 2a(x - x_0) [/tex]






    My Question is about this part;

    [tex] x = x_0 + v_0 (\frac{v - v_0}{a} ) + \frac{1}{2}a(\frac{v - v_0}{a})^2[/tex]

    [tex] \Rightarrow x - x_0 = (\frac{v_0v - v_0^2}{a}) + (\frac{v^2 - 2v_0v + v_0^2}{2a}) [/tex]

    Where did the [tex] a [/tex] in the [tex] \frac{1}{2}a [/tex] from the part [tex] \frac{1}{2}a(\frac{v - v_0}{a})^2[/tex]

    go to? I don't see how it works. I imagine the [tex]1[/tex] numerator multiply's the [tex] {v - v_0}[/tex] and the

    denominator [tex]2[/tex] multiples the [tex]a[/tex] from the fraction [tex] (\frac{v - v_0}{a})^2[/tex]
    but I can't see what happens to the [tex]a[/tex] outside the brackets...

    And, during this part;

    [tex] x = x_0 + v_0 (\frac{v - v_0}{a} ) + \frac{1}{2}a(\frac{v - v_0}{a})^2[/tex]

    [tex] \Rightarrow 2 a (x - x_0) = 2v_0v - v_0^2 + v^2 - 2v_0v v_0^2 [/tex]

    To clarify, the [tex] 2v_0v - v_0^2 [/tex] comes from the fact every part of the equation is multiplied by [tex] 2a[/tex] and the [tex]a[/tex] cancels that fraction leaving just the two, right?
     
    Last edited: Sep 22, 2009
  2. jcsd
  3. Sep 22, 2009 #2
    To answer your first question, when you square the a that is in the denominator is cancels the a from the (1/2*a) leaving only one a in the denominator. For your second equation, not every part of the equation is multiplied by 2a, it is just to get common denominators. When you do this, you see one term cancels and you can see the rest.
     
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