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Posted (edited)

So I know the normal derivative and anti-derivative relationships, velocity, position, acceleration, but what do you get when you take the anti-derivative of a position tie graph? What does that represent? normally it goes m/s^2 -> m/2 -> m -> ? What does the "un-slope-ing" of a position time graph do?

Edited by SamBridge
Posted (edited)

[math]\int dt \: x(t)[/math] ?

 

Over some finite interval, this would be the average position multiplied by the time taken. As elfmotat has said, I don't think it has any useful purpose.

Edited by ajb
Posted

Although you could only do it numerically - rather than algebraically - it might make a rather nice single figure to sum up position on field and time of possession in a rugby game

Posted

There seems to be no information on it from a formal source, but there is some original research: Derivatives of Displacement (antiderivatives a little down the page)

 

I agree though that it seems to hold little significance in application. What do you think? Tentative at best?

Posted (edited)

There seems to be no information on it from a formal source, but there is some original research: Derivatives of Displacement (antiderivatives a little down the page)

 

I've heard of the names of derivatives up to snap, crackle, and pop before. The others (including "absement") I've never come across, though pop, lock, and drop made me smile.

Edited by elfmotat
Posted

I agree though that it seems to hold little significance in application. What do you think? Tentative at best?

I have never come across any applications of this notion. It does not look like a particularly nice notion, other than possibly only in one dimension.

 

I've heard of the names of derivatives up to snap, crackle, and pop before.

They come up in engineering, but I have never come across a physics discussion of them. They don't seem particularly fundamental I guess.

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