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Finding the Velocity from Acceleration as a function of position

  1. Aug 9, 2008 #1
    Hey, I've just found this forum and it looks like a great place for science-minded people. :smile:

    I'm having trouble finding the velocity from a graph of acceleration as a function of position.

    1. The problem statement, all variables and given/known data

    [​IMG]

    3. The attempt at a solution

    I've assumed that you can find the integral of each triangle and that this would be the velocity as a function of position. I used A = 0.5*x*a to find these areas. I calculated area 1 to be 2.925 m/s and area 2 to be -2.0475 m/s. I then summed the areas with the initial velocity given.

    My final answer was 4.8775 m/s but when I try to submit this answer, its wrong. Could someone please point me in the right direction so I can solve this?

    Thanks!
     
  2. jcsd
  3. Aug 9, 2008 #2

    tiny-tim

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    Welcome to PF!

    Hi astronutties! Welcome to PF! :smile:

    I stared at this for several minutes before I noticed what had happened …

    your method is correct :smile: … but only if the horizontal axis is time! :wink:
     
  4. Aug 9, 2008 #3
    Thanks for the welcome tiny-tim!

    I thought the same process could be used for acceleration vs position as for an acceleration vs time graph. I did manage to solve it in the end but I wasn't entirely sure what I did...just a lot crunching through different methods until one worked. :wink:

    In case anyone was interested, the answer was 4.198 m/s.

    Thanks again for the help.
     
  5. Aug 9, 2008 #4

    HallsofIvy

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    a= dv/dt= (dv/dx)(dx/dt)= v dv/dx, by the chain rule. Knowing the acceleration as a function of time means you have v dv/dx= a(x) or v dv= a(x)dx. You can integrate both sides to get v as a function of x.
     
  6. Aug 9, 2008 #5
    Hi HallsofIvy,

    That's what I thought I was doing. By finding the area under the graph, I thought I was finding the integral of a(x) which would be v(x) when rearranged as you have just said.
     
  7. Aug 9, 2008 #6

    D H

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    That is not what Halls said. He arrived at [itex]v dv= a(x)dx[/itex], which when integrated from position [itex]x=0[/itex] to position [itex]x=3[/itex] yields

    [tex]\frac 1 2 (v(x=3)^2 - v(x=0)^2) = \int_{x=0}^{x=3} a(x) dx [/tex]

    You can solve for the velocity at position [itex]x=3[/itex] using the above.

    You seem to have missed Tiny Tim's point: The graph is a function of distance, not time. While the integral of acceleration with respect to time is velocity, the integral of acceleration with respect to distance is something different. Just look at the units: Acceleration times time has units of velocity, but acceleration times distance has units of length2/time2, which is the same units as velocity squared.
     
  8. Aug 9, 2008 #7

    tiny-tim

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    Hi astronutties! :smile:
    No … the area under the curve is ∫vdv, which is … ? :smile:
     
  9. Aug 9, 2008 #8
    Oh I think I see where you are all getting at now. So the area under the graph is equivalent to the displacement? That puts everything into perspective now.

    Thanks for your help!
     
  10. Aug 9, 2008 #9

    Redbelly98

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    No, it's not displacement. The area under the curve is the same as

    [tex]
    \int v \ dv
    [/tex]

    Do that integral, and you'll be nearly done.
     
    Last edited: Aug 9, 2008
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