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Get acceleration from a distance-time graph

  1. Feb 11, 2014 #1
    1. The problem statement, all variables and given/known data

    I got a graph with time on the x-axis, and distance on the y-axis. (see attached picture)
    I have to calculate the acceleration using only the distance - time graph.

    2. Relevant equations

    I used S(t)=S(0) + V(0) * t + .5 * a * t2

    3. The attempt at a solution

    That gives 1.30=0+0*3.09 + .5 * a * 3.092
    1.30/(.5/9.5481)=a
    a=0.272 m/s2

    But I know the acceleration is 0.42171 m/s2, because that's the slope of the V,t graph. What am I doing wrong here?

    Thanks.

    http://alecm.nl/user/ruurd/test/pfhelp.jpg [Broken]
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Feb 11, 2014 #2
    The slope of the velocity time graph looks much less than 0.42. It looks to me (by eye) much more like 0.3.

    Chet
     
  4. Feb 11, 2014 #3

    CWatters

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    The slope of the velocity I made roughly..

    0.5/(3-1.8) = 0.42

    so that seems ok.

    The time axis doesn't appear to start at t=0 but at t=0.85.

    At t=0.85 the position still appears to be approx 0 but the velocity appears to be 0.1m/s.

    At t=2.8 the displacement appears to be 1m so..

    1 = 0.1(2.8-0.85) + 0.5 a (2.8-0.85)^2

    Gives a = 0.423
     
  5. Feb 11, 2014 #4

    BvU

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    Chet [strike]is[/strike] was missing the fact that the x-axis doesn't start at zero.

    I am impressed by the fancy teaching materials they have in NL. Now let's try to come up with a proper interpretation...

    Can you indicate how you found s(0) and v(0) ?

    Third time tonight quick replies cross. Starts to annoy me.
    Isn't it bedtime for you Rute ? 0:30 AM ?!

    Anyway, polynoom fit shows a3 = 0.21 which is 0.5 * 0.42, so data looks good.
    Not only that, polynoom a2 is lineair a1, so that looks almost too good !

    Don't know why the chi squareds are so low, but I suppose you don't have to worry about those in this stage of your physics education.

    Back to the question how you found s(0) and v (0).

    Since your mission is somewhat clearly formulated: "I have to calculate the acceleration using only the distance - time graph" I would be inclined to consider include the fit results as part of the Positie(Tijd) graph. But I am a lazy person. You Judge if that is allowed or not. If it is, life is easy: a1 = s(0), a2 = v(0), a3 = acceleration.
    If it is not, there is some more work to do...

    Did you really record this beautiful plot at 03:48 in the morning ? Wow....!
     
    Last edited: Feb 11, 2014
  6. Feb 11, 2014 #5
    You're right. Mia culpa.

    Chet
     
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