Algebra in Constant Motion Derivation.

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x = x_o + v_0\Deltat + \frac{1}{2} a ( \Delta t )^2

If \frac{v - v_0}{a} = \Delta t

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

\Rightarrow 2 a (x - x_0) = 2v_0v - v_0^2 + v^2 - 2v_0v v_0^2

\Rightarrow 2 a (x - x_0) = v^2 - v_0^2

\Rightarrow v^2 = v_0^2 + 2a(x - x_0)





My Question is about this part;

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

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

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

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

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

And, during this part;

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

\Rightarrow 2 a (x - x_0) = 2v_0v - v_0^2 + v^2 - 2v_0v v_0^2

To clarify, the 2v_0v - v_0^2 comes from the fact every part of the equation is multiplied by 2a and the a cancels that fraction leaving just the two, right?

 

[w3390]

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.