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This question already has an answer here:

We are all used to the mechanical definitions of velocity, as the variation in time of the distance, the acceleration as the variation in time of the velocity, and even more other quantities like the jerk, the pop and so on (which are derivatives of further order of the acceleration with respect to time).

Now my question is this: could there exist a quantity, let's call it $b$, such that ?

$$\mathbf{\dot b} = \mathbf{r}$$

I mean something, the derivative of which, with respect to time, gives us the distance.

If it's a meaningless question, please just don't down vote. Explain me why end eventually I'll delete it...

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marked as duplicate by sammy gerbil, Qmechanic Apr 19 '18 at 14:31

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Such a quantity is just the time integral of position. There have been a few questions about it, and as far as I know there's no particularly useful physical interpretation. However, I can provide a contrived and unrealistic example in which it appears.

We know that unlike some physical objects, our muscles have to continuously expend energy to exert a force. Let's make a simple guess that this work per unit time $P$ is just proportional to the force: $P = \beta F$ with $\beta$ a constant of proportionality. Let's say we are using our muscles to move around a mass $m$ attached to a spring, so that the equation of motion of the mass is

$$m \ddot{x} = -kx + F,$$

where $F$ is the force we are doing. The work per unit time is then

$$P = \beta F = \beta m \ddot{x} + \beta k x,$$

and the energy used between $t=0$ and now is

$$\begin{align} E &= \beta m \int_0^t \ddot{x}\, dt + \beta k \int_0^t x\, dt + \int_{x(0)}^{x(t)} kx\, dx\\ &= \beta m (v(t)-v(0)) + \beta k \int_0^t x\, dt + \Delta U \end{align}$$

where I have included the work done to actually give energy to the spring. You can see the $x\, dt$ term there in the middle. Whether this is a useful result is left as an exercise to the reader.

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  • $\begingroup$ Left as an exercise to the reader!?!? You have become everything you swore to destroy! Don't make me read "left as an exercise to the reader" ever again on this site. $\endgroup$ – Dominik Car Apr 19 '18 at 14:07

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