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Here is the last piece we needed tying Power to Energy as proven by DieselMaxPower and mentioned way back when by Mort:
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"I'll add some info, even though you've basically got it. So a simplified equation for your last post is
Power = M*A*V
It took several relationships, and if you actually did it through torque you would have several factors of 2*pi and the wheel radius to carry around and eventually cancel. Now I know most people don't know calculus, so you will have to trust me on this:
E = 1/2 M V^2
d(E)/dt = M V (d(V)/dt)
d(E)/dt is power, and d(V)/dt is acceleration. So check this out
P = M V A
You get the EXACT same equation using calculus. And acceleration is:
A = P/(MV),
so as you go faster, your acceleration decreases (assuming constant P). This doesn't really add too much to your current understanding, but if you want to more advanced stuff it will really help. For instance, we may want to take into account the force it takes to spin the wheels. This gets so ridiculously tedious looking at forces. Instead, add it to the energy equation
E = 1/2 M*V^2 + 1/2 I omega^2
d(E)/dt = M*V*(d(V)/dt) + I omega (d(omega)/dt)
I is the mass moment of inertia and omega is the angular velocity of the wheel, and now d(omega)/dt is the angular acceleration of the wheel. In the simplest case (tires aren't slipping on the road, or slip by a constant percentage), d(omega)/dt is proportional to A:
P = M*V*A + I*omega*c*A
P = A*(M*V+I*omega*c)
A = P/(M*V+I*omega*c)
A = P/(M*V+I*V*d)
where c and d are constants. We can change omega to V by the same argument that we could change d(omega)/dt to A. So acceleration decreases with increased velocity (assuming constant power), but now we get a more realistic estimate. Some of the power produced goes into spinning the wheels.
Rockets get into a whole new mess, which I would actually say makes more sense by looking at momentum instead of energy. If you're interested, the Reynolds transport theorem for momentum should give you some good information. Most simple problems don't need calculus, even though the theorem in its purest form needs it. As a bonus, looking at model rockets is one of the most common simple problems."
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