- #1

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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?

- Thread starter hola
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- #1

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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

quasar987

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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

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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:Is this right?

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

lightgrav

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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

quasar987

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

- #6

lightgrav

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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 .

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