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If the area under an acceleration-time graph denotes velocity and the area under a velocity-time graph denotes displacement, what exactly does the area under a displacement-time graph denote?

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A note on the "area under a graph" in teaching physics: I presume you are learning from a calculus-free source, in which case the "area under the graph" constructs that you're going to see several times are a way of representing integrals without having to teach you calculus. Or we could say that a definite integral is a notation for the area under a graph which is more in line with the historical development. In any case, there is no guarantee than an arbitrary integral has physical significance, we know which ones do empirically (i.e. they're useful in solving problems). – dmckee May 1 '13 at 15:12
Got it, thx! :) – Hele May 1 '13 at 15:35
In one problem that was presented in our text, the area under the positive half of the position time graph was the same as area in the negative half, and the displacement was zero. Perhaps, that helps students understand displacement of 0. – Donna Elias Nov 5 at 16:33

2 Answers 2

up vote 4 down vote accepted

what exactly does the area under a displacement-time graph denote?

I think it just represents what you said: the area under a displacement-time graph. I can't think of any other use for it. There are two main reasons for this:

  1. Your quantity, let's call it $f(t)$, retains a memory of where the object has been. That's because the area under the graph depends on the entire past history of the graph, and not just where the object is now. This flies in the face of the principle that physics is local in time: what happens next is determined entirely by the state of the system now and doesn't depend on what happened further back. This is called the Markov property of the laws of physics, and is a general feature of all the fundamental laws we know (relativity changes this in detail, but not in any essential way). There are examples of systems which are approximately non-Markovian because they interact with their environment in a way that preserves a memory, but this isn't the norm, and you can always get a Markovian description by including the environment as well. $f(t)$ might play some small role in the theory of such systems, but only in an approximation where you leave out the environment.

  2. Because of translation invariance (the laws of physics are the same here as anywhere else) there is no meaning to "absolute" position, only differences between positions matter. This means $f(t)$ is redundant: you can add an arbitrary linear function of time $f(t)\to f(t)+ u + w t$ and still get the same physics. This restricts the way $f(t)$ can appear in the laws of physics, leaving you with nothing new that isn't already recorded by position, velocity and acceleration.

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To add to the answer, OTOH, you might encounter systems (not Newtonian mechanics, but like say.. some electrical circuits) where the system might have memory. In such cases, I think $f(t)$ or analogous quantities might be useful. – Siva May 1 '13 at 4:49

I don't see this quantity computed very often so I don't think it is that useful, but here is one thing that can be said: If $A$ is the area under the graph and $T$ is the duration of time over which the integral was preformed then $A/T$ is the average displacement from the origin during that interval of time.

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protected by Qmechanic Nov 5 at 19:00

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