Area under a displacement graph

Question

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?

Answer 1

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.

Answer 2

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.