Recent content by phasacs
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Undergrad Looking for a proof that u(x) du(x)/dx = 0.5 d(u(x)^2)/dx
let u*du/dx=y <=> 2(u*du/dx)=2y <=> u*du/dx+u*du/dx=2y (use inverse product rule a'b+ab'= (ab)' ) <=> d/dx[u2]=2y <=> y=1/2 d/dx[u2] -
Undergrad A limit as t-->0 of log(t) / SQRT(t)
I don't see why not, you can write -∞ as a product of ∞ and -1 so you get -(∞2)=-∞ -
Force/time graph acceleration problem
By using calculus it's easy, we know that the rate of change of acceleration is constant (from shape) => let da/dt=j => a=jt (no initial rate of change of acceleration) integrate with respect to time => v=1/2jt2 but a=jt => j=a/t so v=1/2at. But I don't think he is supposed to do this way...- phasacs
- Post #7
- Forum: Introductory Physics Homework Help
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Speed at Takeoff and Average Acceleration
True, I forgot to multiply velocity with time: x=vi*t+1/2at^2- phasacs
- Post #9
- Forum: Introductory Physics Homework Help
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Speed at Takeoff and Average Acceleration
distance: x=vi+1/2at^2 is also a relevant equation, and since there is acceleration v=x/t (velocity=distance/time) is Not true- phasacs
- Post #5
- Forum: Introductory Physics Homework Help
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Undergrad Why isn't Kinetic Energy always equal to Potential Energy?
K= ∫mvdv = ∫m dx/dt dv = ∫m dx/dv dv/dt dv = ∫m dv/dt dx = ∫Fdx = U => K=U, why isn't this true? If it is, wouldn't that mean that Kinetic Energy is always equal to Potential Energy?