Does the integral of distance have any physical meaning?

Question

the derivative of distance is velocity, but does the integral of distance have any physical meaning at all?

My guess is that it would have the units of distance*time, is this right and are there any physical measures with the units distance*time?

yes i mean with respect to time

in maple command format
int(distance,x)

Answer 1

epidavros - What's wrong with saying distance from something, instead of position? You're confusing distance with distance travelled, which isn't what's being asked. Derivative of distance certainly is velocity.

falzoon - Why did you do all that... It doesn't lead anywhere... All you need to figure out what q is, is to think about what an integral means.

For the integral, obviously the units will come out as distance*time, if we're integrating with respect to time, which I'll assume. So the physical meaning is that we add up individual bits of distance at every moment of time, creating a sort of wacky version of the total of distance from your reference point.

Is this integral really useful? Meh, maybe for some task I cannot conceive. Here's a poorly conceived example though:

Let's say we want to see which NBA players have the most dunk-friendly hops. Suppose, a lot of them can jump 40 inches, but some of them can't hang in the air long enough to have cool dunks. On the other hand, some have huge hangtime, but can't those high-flying acts others do. What is the best way to take into account both hangtime and jump height? We want some sort of composite height/time thing.

So, we take the integral of the distance with respect to time (not saying this is how they should do it). We basically judge a player's jumps based on adding their heights at each moment in time. This gives that weird Distance*Time unit. In mathematical terms, we plot the jump height against time, and take the area under the curve.

Answer 2

If distance is a function of time D(t), then while the derivative of
D(t) has the physical dimensions of L/T, or velocity, the integral of D(t) would have the physical dimensions of LT. There are very few published physical quantities in use that have physical dimensions of the form M^a L^b T^c, where c is positive nonzero, and there isn't any at all with physical dimensions of LT. That doesn't necessarily mean that it has no physical meaning, it just means that there's no commonly used physical quantity with dimensions LT.

What could a quantity with dimensions LT physically mean? Good question. Let me think on it and maybe I'll get back to you.

I've given your question a star because while maybe to others this question may seem silly, I think it's a non-trivial question. Why is that almost everything in physics that involves time only have time as an inverse quantity? And, by the way, arguing that distance isn't location is pendantic and missing the point, even if correct.

Addendum: I liked Jeffrey's suggestion for an use of the term LT. Give the man his 10 points.

Answer 3

Sorry, I have no answer, just a re-iteration of the question.
I wanted to see what I'm dealing with. Hope it's all correct!
Starting at the end, with acceleration, we have :
dv = a dt
Integrating both sides :
∫ dv = ∫ a dt
Let acceleration be constant, so:
∫ dv = a ∫ dt
Therefore, v = at + C.
At t = 0, let v = v0, so C = v0.
Final equation is : v = v0 + at.

Continuing on backwards, we have :
dx = v dt
Integrating both sides :
∫ dx = ∫ v dt
Allowing v to vary, we have :
∫ dx = ∫ (v0 + at) dt
Therefore, x = v0*t + (½)at^2 + C’
At t = 0, let x = x0, so C’ = x0.
Final equation is : x = x0 + v0*t + (½)at^2.

OK! Now let’s postulate an unknown
quantity, q, with the following property :
dq = x dt
Integrating both sides :
∫ dq = ∫ x dt
Allowing x to vary, we have :
∫ dq = ∫ [x0 + v0*t + (½)at^2] dt
Therefore, q = x0*t + (½)v0*t^2 + (1/6)at^3 + C”
At t = 0, let q = q0, so C” = q0.
Final equation is : q = q0 + x0*t + (½)v0*t^2 + (1/6)at^3.

Now I can see it for what it is, but I still don’t understand it.
Maybe if I substitute, by letting :
x0 = x – v0*t - (½)at^2 and v0 = v – at.
so, x0 = x – (v – at)t - (½)at^2 = x – vt + (½)at^2
This gives :
q = q0 + [x – vt + (½)at^2]t + (½)(v – at)t^2 + (1/6)at^3
or absolutely finally :
q = q0 + xt - (½)vt^2 + (1/6)at^3.

We now have q in terms of q0, t, x, v and a.
This is something that some bright people
can play around with and try to figure it out,
because I still haven’t got a clue.

Example : let q0, x0 and v0 be zero.
Let a = 1 m/s2 and let t = 1 s.
After 1 second, x = 0 + 0 + (½)(1)(1^2) = ½ metre.
v = 0 + (1)(1) = 1 m/s.
Therefore, q = 0 + (½)(1) - (½)(1)(1^2) + (1/6)(1)(1^3)
which is 1/6. What is it, that is 1/6???

The q-value seems to have something to do with
where you are, how fast you are going and what
your acceleration is at a certain point in time, so
it doesn't seem to be a simple matter.
Calling all geniuses!

Answer 4

distance with respect to what. Using your derivitive of distance is velocity is with respect to time. If it is just a constant distance you can pull it out of your integral. Is there any other variables involved? Like time for instance the integral of distance (x) with respect to time (t) will come up with the total distance for any time ie INT(x)dt = x*INT dt or xt. (the area under the curve). So it does have meaning with respect to the right variable time, temperature, bananas etc (d?)

Answer 5

Velocity is NOT the derivative of distance, it is the derivative of POSITION. The integral of position with respect to time is DISTANCE travelled.

You are confusing POSITION (or location) with distance from a point.

Answer 6

I don't recollect that there is a physical meaning to INT (location)dt