Lets assume a function a = |sin t| for acceleration over time. If we integrate it, we get instantaneous velocity. Now i have taken a limit for time. How will this graph look like?

I have been told that integrating a function provides me with the area under the curve. If i represent this in a linear way (without area), do i get a function which has a periodic increase in rate of change? Now if i want the displacement, I would integrate the function for velocity. As i now have the function for displacement, how will this graph look like?

I understand that it will simply be a graph which increases just like velocity does, but since the graph a = |sin t| after integration, gives us the area under it for velocity, is it possible to represent the displacement in the same graph? Just like velocity was in an a-t graph? If i look at it in terms of dimensions, Its obvious why this happens, its simply because velocity = m/s. And assuming t is in seconds and acceleration in m/s^2, area would naturally give us velocity. But to obtain displacement, i would need to multiply the square of time with the acceleration.

What implications does this have on the graph?

Thank you.


It would be the area under the curve which represents the area under the acceleration curve. There isn't really any particularly easy way to represent this, as far as I know. You could numerically integrate it twice to get a crude approximation, but that's about it.

  • $\begingroup$ Would it be apt to extent the numerical value dimension-ally to represent it in a 3d graph, even if the idea is not mathematically sound? $\endgroup$ – Mathbreaker Jun 30 '14 at 10:24
  • $\begingroup$ Interesting idea, I hadn't thought of doing it that way, but that would definitely work, you would get a line (Acceleration), an area on the t-acceleration plane (velocity), and a solid extruded from this plane (displacement). Actually, that's fascinating, good idea! :) $\endgroup$ – FundThmCalculus Jun 30 '14 at 12:07
  • $\begingroup$ I have asked this question on a different post as i have a few more doubts. math.stackexchange.com/questions/852331/… Thank you. $\endgroup$ – Mathbreaker Jun 30 '14 at 15:09

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