@Empty like my heart and @ganeshie8 of course...hey can you help me figure this "trick" out.

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Here is the link: http://www.physics.arizona.edu/~leone/ur/phy141/eqns_motion_derivation.pdf page 3, |dw:1435816385662:dw|
Step 3 here
Looks tough; University of Arizona is fierce. This is incredible

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Haha, it's actually very beautiful, if you do calculus you will realize it!
Are you majoring for physics to engineer? Astronomy? Aerodynamics? Taking calculus in August
Not decided yet, but I do enjoy learning all of the above :)
Commutative property of multiplication is what the instructor synthesizing, utilize her empowerment. Why do you feel she is expressing that it works so well?
\[\huge \vec a (x) = \frac{ d \left[ \frac{ (\vec v(x))^2 }{ 2 } \right] }{ dx }\]
I remember the first time I saw a trick like this I was surprised, but it's essentially just taking the derivative of the integral, which are inverses of each other. Work it backwards, it's basically the reverse of the chain rule:\[\large \frac{ d \left[ \frac{ (\vec v(x))^2 }{ 2 } \right] }{ dx }=\vec v(x)\frac{ \vec v(x) }{ dx }\] To make it more clear, suppose we have: \[\frac{d}{dx} \left( \frac{1}{2} ( f(x))^2 \right) = \frac{1}{2} 2 (f(x))^{2-1} * f'(x)=f(x)*\frac{d f (x)}{dx}\]
Wow, I don't know how I missed that, that's pretty clever.
Thanks a lot! Now I can sleep in peace haha.
Hey, if you want you can derive the kinetic energy here, I know we can do it using the work definition, I remember deriving it a while back, all we need is F = ma and \[W = \int\limits \vec F \cdot d \vec r\]
I should specify by saying work energy theorem
Actually instead of typing it out, here is a nice one just from google images |dw:1435824066157:dw|

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