Question

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

Answer 1

Here is the link: http://www.physics.arizona.edu/~leone/ur/phy141/eqns_motion_derivation.pdf
page 3,

Answer 2

Step 3 here

Answer 3

Looks tough; University of Arizona is fierce. This is incredible

Answer 4

Haha, it's actually very beautiful, if you do calculus you will realize it!

Answer 5

Are you majoring for physics to engineer? Astronomy? Aerodynamics? Taking calculus in August

Answer 6

Not decided yet, but I do enjoy learning all of the above :)

Answer 7

Commutative property of multiplication is what the instructor synthesizing, utilize her empowerment. Why do you feel she is expressing that it works so well?

Answer 8

\[\huge \vec a (x) = \frac{ d \left[ \frac{ (\vec v(x))^2 }{ 2 } \right] }{ dx }\]

Answer 9

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:

\[\frac{d}{dx} \left( \frac{1}{2} ( f(x))^2 \right) = \frac{1}{2} 2 (f(x))^{2-1} * f'(x)=f(x)*\frac{d f (x)}{dx}\]

Answer 10

Wow, I don't know how I missed that, that's pretty clever.

Answer 11

Thanks a lot! Now I can sleep in peace haha.

Answer 12

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 \[W = \int\limits \vec F \cdot d \vec r\]

Answer 13

I should specify by saying work energy theorem

Answer 14

Actually instead of typing it out, here is a nice one just from google images