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Motion in 1d

  1. Sep 26, 2013 #1
    1. The problem statement, all variables and given/known data

    when graphing a velocity squared vs. time graph how do I find the acceleration. I realize the slope gives me units of acceleration but this is not correct because the theoretical acceleration is no where near this and I didnt botch this lab up.

    Theoretical
    a = gsin(3.03) +/- 0.085 m/s^2

    when i graph my data from the photogate the linear trend line gives me a slope of 1.0481 which is the acceleration in this case but like I said that is far beyond the theoretical.

    also i did this experiment by using a stopwatch and I calculated an acceleration using the equation
    a=2s/t^2 and got an acceleration of 0.58 +/- 0.10 m/s^2

    So again how do I calculate the correct acceleration using a v^2 vs. x graph?
     
  2. jcsd
  3. Sep 26, 2013 #2

    Simon Bridge

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    The slope of the velocity-time graph is the acceleration.

    But you wrote: "velocity-squared vs time" ... the slope of that graph will not give you the acceleration.
    If you means "displacement-squared vs time" then that is also wrong ... as you can see by rearranging your own equations from later: a=2s/t^2 means s=(a/2)t^2 ... i.e. a plot of s vs t^2 gives a slope of a/2.

    But it may be a typo - in which case there is probably something else going on that you have not taken into account with the theory.
     
  4. Sep 26, 2013 #3
    Thanks Simon but I figured out that I had to use a kinematics equation V^2=2a(xf-xi) to find acceleration and it matched the theoretical value pretty well!
     
  5. Sep 26, 2013 #4

    Simon Bridge

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    OK. So what did you plot?
     
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