# Volume moved through by an object at variable velocity

1. Jul 23, 2009

### Nabeshin

Sorry if the title is a bit cumbersome, but I don't know any better way to describe the problem. The essential question is as follows:

An object with a cross sectional area of A moves with some time-dependent speed v(t) through some medium. How much volume does it pass through at a given time, in m^3/s.

At first glance I thought it should move through a distance v(t)dt in time dt, but setting dt=1s is essentially assuming v(t) is constant in that one second. Of course, I could integrate v(t)dt for, say, a one second interval around the time t, in something similar to:
$$\int_{t-.5}^{t+.5}v(s)ds$$
Where s is a dummy variable, but I have some misgivings about this approach too. I thought this through a lot more earlier and I don't recall why but I remember thinking this wouldn't work. I think it had something to do with the arbitrary choice of midpoint method, and the arbitrary time interval. At any rate, I don't think that's it either.

I finally arrived at saying just v(t)*A, but again I don't know if this is correct. Dimensionally it gives m^3/s, which is what I want, but I find it hard to justify otherwise. I keep going back and forth in my mind between thinking it is correct and questioning it.

I want to be sure before I put this in something I'm working on, because a silly mistake like this would be embarrassing. Thanks for any input

2. Jul 24, 2009

### rl.bhat

If the object moves with some time-dependent velocity, it must be either accelerating or retarding. Using kinematic equation x = ut + 1/2*at^2, you can find the displacement of the object. Then the volume = A*x/t

3. Jul 24, 2009

### Nabeshin

Thanks for the response!

For one, that kinematic equation is only for constant acceleration.

Also, the problem is what do your t's in your equation represent? They represent some time interval. The first equation you provide will give the displacement during a time interval t, and the second the volume covered within the same time interval t. But the choice of time interval seems arbitrary... Basically your solution seems essential in form to the integral solution I mentioned in my first post, which I'm not comfortable with.

I think my problem is that my brain is too wrapped around this to even see the essential problem, because it seems this should be simple.

4. Jul 24, 2009

### Andy Resnick

Maybe I'm misunderstanding something, but I think the problem is straightforward:

The distance the object moves in a time interval dt is simply v(t)dt. I suppose if v(t) is constant over an interval you could write v(t)$$\Delta t$$, but the meaning is the same. The displaced volume is then A* v(t)*dt.

Then integrate as needed?

5. Jul 24, 2009

### Nabeshin

Andy Resnick,

I'm on board with you here, I'm just not sure if this is the right thing I need for the problem I'm solving. Ok, let me give a little bit of elaboration and you can tell me if I'm still missing the simplicity:

For the case of a body moving at constant speed, we can easily say that the amount of volume it moves through in one second at any given time t is v*A, right?

Now, extending to a non-constant speed, the amount of volume it moves through in any one second at any given time t should be...?

Perhaps my confusion arises from the fact that since the units are in seconds, I cannot see the justification in setting the time interval to be one second (that should work out on its own), but then the only logical course of action is to shrink dt -> 0, which of course yields volume=0.

Perhaps a better way to phrase this is, Andy Resnick, your solution is perfectly good finding the volume gone through for a large time interval, say, from t=0 to t=1 year, or something. But what I'm interested is more... the instantaneous amount of volume being traveled through at any given time.

P.S: Sorry for wording so poorly, I cannot seem to put this down in concise enough terms for someone to understand why my brain is short circuiting.

6. Jul 24, 2009

### Andy Resnick

Hmmm.. I wonder if you are getting hung up on the units concepts rather than the math.

For example, when you wrote "volume it moves through in one second at any given time t is v*A", that's not exactly true: I think you mean, using [] to identify the units,

V[m^3] = v[m/s]*A[m^2]*1. That's obviously different than V[m^3]=v[mi/hr]*A[m^2]*1. So the missing piece is how you specified the velocity.

The quantity "instantaneous volume" needs a unit specification, because I don't know what that means. Maybe you mean [m^3/s], or more generally [L^3/T], becasue that corresponds to v*A.

It's almost like you are re-discovering the concept of 'volume flux', which is used a lot in fluid mechanics.

Does that make sense?

7. Jul 25, 2009

### Nabeshin

Yeah, I think you're getting at part of what is getting me so confused about this whole thing. When I said instantaneous volume, I do mean [L^3/T], specifically in [m^3/s]. I don't really have any formal education of fluid mechanics, so it's possible what you say about the concept of volume flux. Let me just explicitly give my problem and perhaps you can sort out exactly the type of expression I'm looking for.

There's a spaceship with cross sectional area A traveling at a speed v(t) through a medium with density p. The goal is to get a function for [particles/sec] that the spaceship collides with for any given t. Then, knowing the change in energy imparted by each particle, know the change in energy per second at any given t, plug into differential equation and find expression for energy as a function of time. That's basically the nuts and bolts of it.

It seems I need my "instantaneous volume" with units [m^3/s] multiplied by density, multiplied by energy per particle gives [energy/s], which is the ultimate expression I'm going for here.

If I am over-complicating this please put my mind at ease.

8. Jul 25, 2009

### Staff: Mentor

This is correct. The differential volume is A dx so the differential volume rate is A dx/dt = A v

9. Jul 25, 2009

### Nabeshin

Thank you very much for the confirmation, DaleSpam. And thanks for the thoughts as to what's confusing me, Andy Resnick.