### Integral of Position

Here's something to think about...

Let's consider a rocket. It's accelerating at a constant rate (F, for FAST).

a = F

It's a fact that the integral of acceleration, when considering time, is velocity. The constant of integration, in this case, is replaced with S, for SPEED, or the initial velocity of the rocket.

v = Ft + S

...and of course, the integral of velocity is position, and the constant again is replaced with another known letter (B, for beginning or butt or something, this time).

x = (1/2)Ft^2 + St + B

So, suppose I integrate this. What could the unit of this thing be?

? = (1/6)Ft^3 + (1/2)St^2 + Bt + ?

It is something that changes as a result of change in position. What would it be?

acceleration = x/t/t

velocity = x/t

distance = x

? = ?

Consider the pattern. The degree of "t" begins at -2, goes to -1, then to 0. The pattern, therefore implies the development of a time unit with a degree of 1.

? = xt

This wouldn't be meters per second, it would be meterseconds: distance * time.

Does this mysterious unit have any use? Any meaning what-so-ever? Can it be applied practically to any problems? I'll sleep on it.

Let's consider a rocket. It's accelerating at a constant rate (F, for FAST).

a = F

It's a fact that the integral of acceleration, when considering time, is velocity. The constant of integration, in this case, is replaced with S, for SPEED, or the initial velocity of the rocket.

v = Ft + S

...and of course, the integral of velocity is position, and the constant again is replaced with another known letter (B, for beginning or butt or something, this time).

x = (1/2)Ft^2 + St + B

So, suppose I integrate this. What could the unit of this thing be?

? = (1/6)Ft^3 + (1/2)St^2 + Bt + ?

It is something that changes as a result of change in position. What would it be?

acceleration = x/t/t

velocity = x/t

distance = x

? = ?

Consider the pattern. The degree of "t" begins at -2, goes to -1, then to 0. The pattern, therefore implies the development of a time unit with a degree of 1.

? = xt

This wouldn't be meters per second, it would be meterseconds: distance * time.

Does this mysterious unit have any use? Any meaning what-so-ever? Can it be applied practically to any problems? I'll sleep on it.

## 6 Comments:

I'm a high school Physics teacher, and a student asked me this last week. We came up with:

The integral of position over a time period varies proportionately with both the value of position and the length of the time integral, so we need a meaning that does the same.

Also, since position is determined by the reference frame chosen, the value of the integral for the position of an object over a time period will depend on the reference frame. For example, if I set my origin at my teacher's desk, then Tyler's integral of p vs t will be non-zero, but in his reference frame (with him at the origin), the integral will be zero. So we need a meaning whose value depends on the reference frame.

So, in short, we need an interpretation that accumulates with distance and time, and which depends on what you choose as your reference frame. We chose "Loneliness," with the origin chosen as whatever you consider to be "home."

Cheers!

I am an Electrical Engineering student at University right now. I was considering this very idea just the other day. I got just about as far as you did in my reasoning.

Although I am unsure what this unit of distance*time is, I am reasonably certain it must exist. Allow me to show you my logic here. In electronics, an inductor does not permit instantaneous changes in current. The equation that describes an inductor is V(t) = L * di(t)/dt. I notice that the current that cannot change instantly is a derivative. A similar relationship occurs with capacitors and their voltage, with a similar equation (dv(t)/dt).

Now, I examine space. We cannot make instantaneous changes in location. We may be able to make wormholes to make a path shorter, but we cannot instantly go from point A to point B (outside the quantum scale). Therefore, following the pattern of the electronics components, position MUST be the derivative of something (this distance*time).

There you have it. It might not be a rock-solid proof, but it is reasonably convincing. Whatever this distance*time thing it, it would seem that it DOES exist.

MS Engineering Physics student's 2 cents. Try relating it to a field's pull. Think magnet near bar or maglev train over rail, spaceship near planet, etc. Distance here might relate to amount of pull or push needed to hover. It could also relate to orbital speed needed to maintain an altitude. In the pull or push needed case, the integral of position might relate to fuel consumption. There's the minor detail of a power of two associated with most field effects...

Integral of position or displacement is called absement.

Here's a practical example: Consider a water faucet that is linear in flow versus openness. For simplicity consider a linear slider that opens and closes the valve (e.g. a gate-valve that slides open or closed along a linear path).

The amount of water in a bucket held under the faucet is the absement (i.e. the area under the displacement-time curve).

Just as there are derivatives of displacement (velocity, acceleration, jerk, jounce, etc.), there are integrals of displacement: absement, absity, abseleration, abserk, absounce, etc..

A practical application of the second derivative is a 2-stage valve or 2-stage hydraulophone, where the amount of water in the bucket is the absity.

See http://wearcam.org/absement

For some illustrative examples of absement, etc., see also

http://wearcam.org/absement/examples.htm

I've contemplated this question before and I came up with the only possible solution:

The integral of position is the end of the universe.

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