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Motion in 1d

  • Thread starter ethinh
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  • #1
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Homework Statement



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?
 

Answers and Replies

  • #2
Simon Bridge
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Homework Statement



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.
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.
 
  • #3
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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!
 
  • #4
Simon Bridge
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OK. So what did you plot?
 

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