How is power proportional to acceleration?

1. Aug 24, 2005

hola

The above scenario is when distance and mass are held constant. ~ denotes proportional to
My work:
P= m*a*d/t
P~ a/t

a~1/t^2 when m and d are held constant

so P~ a^3/2

Is this right?

2. Aug 24, 2005

quasar987

It is important to say that you assumed a constant force. (what does 'd held constant' mean anyway?)

Personally, I don't see a flaw in your reasoning.

But wait 'til it gets approved or demolished by someone more knowledgable before you start expecting your nobel prize.

3. Aug 25, 2005

Crosson

Absolutely not. If you think that you can "hold d constant" and still have a non-zero energy transfer rate (power) then you don't even understand work. The error in your mathematics is the point when you say:

P= m*a*d/t

It should be:

P= m*a*(d(t)-d0)/(t-t0)

In other words, d should be "delta d" or "the change in d". But more importantly, on a physical level you are talking about transfering energy by accelerating in place, which is absurd.

"a~1/t^2 when m and d are held constant" This is mathematically laughable. Acceleration is the second derivative of position. If position is a constant, acceleration is zero.

4. Aug 25, 2005

lightgrav

Crosson: most people use "d" for distance, which is (x_f - x_i),
and it was pretty clear from the context.

Hola: your condition shows power proportional to a/t ,
(with fixed distance and mass). This is NOT proportional to a,
since the time to travel distance "d" is NOT the same for different P.
This is obvious since higher power run will have quicker acceleration
(your formulas presume CONTANT-ACCELERATION during each run)
By the way, this is *Average* power!

5. Aug 26, 2005

quasar987

I'm not sure I see your point lightgrav. I'll rewrite hola's work with more precision. Asume an object of mass m at rest at t_0 = 0 acted on my a constant force F, from which results an acceleration a. Then, after a time t', the object has traveled a distance d, and the rate at which work as been done by force (the power) for that period is P = Fd/t' = mad/t'. But since the force is constant, the equation of position of the object is x(t) = 0.5at², for which we know a solution to be d = 0.5at'² <==> $t' = \sqrt{2d/a}$. Hence we can rewrite the power for that particular trajectory of the object as

$$P = \frac{mad}{\sqrt{2d/a}} = \frac{ma\sqrt{ad}}{\sqrt{2}} = \sqrt{\frac{d}{2}}ma^{3/2}$$

6. Aug 27, 2005

lightgrav

Point #1 : the title prase "Power proportional to acceleration" does not apply
to P~a/t nor to P~a^3/2 . (Does Hola know what proportional means?)
Point #2 : "average Power over a given distance" is not the same thing as P.
Unmodified, unqualified Power seems to misleadingly imply constant P,
or that it is a relationship that could be used in most any situation.
But knowing that P ~ F.v ~ a.v makes it obvious that here (a=const) P ~ t .