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I am aware of what the first integral of position, absement means (at least to a very superficial level). However, I can find nothing regarding the physical intuitive meaning of absity, the second integral of position.

If anyone is able to explain to me, that would be great.

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    $\begingroup$ I don't think there is something more deep to it than the relation of "being second integral of" . $\endgroup$ – AoZora Apr 2 '19 at 22:13
  • $\begingroup$ What do you want these quantities for? They seem somewhat useless (for instance, simply changing origin drastically changes them in a way that seems unphysical). $\endgroup$ – jacob1729 Apr 2 '19 at 23:18
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    $\begingroup$ An example of absity is the answer provided by @Steeven. Typically, it's consider to be a worthless measure. $\endgroup$ – Cinaed Simson Apr 3 '19 at 20:20
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Consider the gas pedal in your car. When you push the pedal down and hold it there, the car accelerates. Push it a bit further and the acceleration is larger. In other words, one's acceleration depends on the other one's position - which means that one's position (double integral) depends on the other one's absity (double integral).

  • Specifically, the car acceleration $a_{car}$ relates to the pedal position $s_{pedal}$: $$a_{car}=k\,s_{pedal}$$ $k$ is the proportionality constant (we assume a simple, linear and thus proportional relationship for simplicity). For a stiffer pedal (where less displacement is needed for the same acceleration) $k$ is larger.

  • Therefore, the car velocity $v_{car}$ (integral) relates to the pedal absement (or absition) $p_{pedal}$: $$\int a_{car}\;\mathrm dt=k\int s_{pedal}\;\mathrm dt\quad\Leftrightarrow\quad v_{car}=k\,p_{pedal}$$

  • and the car position $s_{car}$ (double integral) relates to the pedal absity $q_{pedal}$: $$\int\int a_{car}\;\mathrm dt \;\mathrm dt=k\int\int s_{pedal}\;\mathrm dt\;\mathrm dt\quad\Leftrightarrow\quad s_{car}=k\,q_{pedal}$$

If someone asks you, where the car is after a certain time, then you can answer it if you know the absity function of the gas pedal.

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You can say that the position is the second derivative of it, so that the relation between this quantity and the position is the same between the position and the acceleration. I've never seen this quantity be used for a physical purpose, it carries indeed very redundant information.

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