Can someone provide to me an intuitive explanation of the second integral of position with respect to time?

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.

• I don't think there is something more deep to it than the relation of "being second integral of" . – AoZora Apr 2 '19 at 22:13
• 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). – jacob1729 Apr 2 '19 at 23:18
• An example of absity is the answer provided by @Steeven. Typically, it's consider to be a worthless measure. – Cinaed Simson Apr 3 '19 at 20:20

• 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 $$q_{pedal}$$: $$\int\int a_{car}\;\mathrm dt \;\mathrm dt=k\int\int s_{pedal}\;\mathrm dt\;\mathrm dt\quad\Leftrightarrow\quad s_{car}=k\,q_{pedal}$$