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.


1 Answer 1


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? $\endgroup$ Jun 30, 2014 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! :) $\endgroup$ Jun 30, 2014 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. $\endgroup$ Jun 30, 2014 at 15:09

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