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The Forum > Math & Science > Integration of Distance
While pondering the value of integrals and derivatives, I stumbled on to a question.

What are you left with if you take the integral of distance? In case I'm wording this wrong (it's been awhile since math, and I feel rusty) when I say distance I refer to what you get when you take the integral of velocity, given that you know C.
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I see that there is no usage of the KB anymore.:(
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Sadly, yes. It's entirely too many clicks to try to discover the same amount of information, people get to it slower, and it's being removed soon. Plus, the forum has the same outcome.
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Put a question on the KB, it takes a week to get answered. Post one on the forum, the answer arrives tomorrow.
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It doesn't really have a physical definition. If you have an equation for the position and you integrate, you get another equation. It is mathematically valid, but it doesn't represent anything. At least, not so far as I know.
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In Dimensional Analysis (using L for Length and T for Time):

Acceleration is represented by L1T-2
Velocity is represented by L1T-1
Distance is represented by L1T0

So integrating Distance would produce something represented by L1T1

As far as I know there are no real measurements that use this representation (or a similar Dimension, which could easily be changed)

The closest I could find to "ms" (Metre-Second) Would be a Coulomb-Metre per Amp, which breaks down to this, but would be completely fictitious:
(Including A for Amperage)

Coulomb = A1L0T1
Metre = A0L1T0
Coulomb-Metre = A1L1T1
Amp = A1L0T0
Coulomb-Metre per Amp = A0L1T1

I have no idea how this measurement would be implemented in reality
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The Forum > Math & Science > Integration of Distance
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