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

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Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.3Here is the link: http://www.physics.arizona.edu/~leone/ur/phy141/eqns_motion_derivation.pdf page 3, dw:1435816385662:dw

misssunshinexxoxo
 one year ago
Best ResponseYou've already chosen the best response.0Looks tough; University of Arizona is fierce. This is incredible

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.3Haha, it's actually very beautiful, if you do calculus you will realize it!

misssunshinexxoxo
 one year ago
Best ResponseYou've already chosen the best response.0Are you majoring for physics to engineer? Astronomy? Aerodynamics? Taking calculus in August

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.3Not decided yet, but I do enjoy learning all of the above :)

misssunshinexxoxo
 one year ago
Best ResponseYou've already chosen the best response.0Commutative property of multiplication is what the instructor synthesizing, utilize her empowerment. Why do you feel she is expressing that it works so well?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.3\[\huge \vec a (x) = \frac{ d \left[ \frac{ (\vec v(x))^2 }{ 2 } \right] }{ dx }\]

Empty
 one year ago
Best ResponseYou've already chosen the best response.3I 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: \[\frac{d}{dx} \left( \frac{1}{2} ( f(x))^2 \right) = \frac{1}{2} 2 (f(x))^{21} * f'(x)=f(x)*\frac{d f (x)}{dx}\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.3Wow, I don't know how I missed that, that's pretty clever.

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.3Thanks a lot! Now I can sleep in peace haha.

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.3Hey, 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 \[W = \int\limits \vec F \cdot d \vec r\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.3I should specify by saying work energy theorem

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.3Actually instead of typing it out, here is a nice one just from google images dw:1435824066157:dw
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