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Velocity squared and position graph

  1. Sep 7, 2008 #1
    I asked this question a couple days ago but I am still confused. I have a velocity squared vs position graph and I am confused on what the slope means. I'm guessing that the slope is the displacement? I'm am also suppose to find the acceleration using this slope. The slope I calculated right now is 2.0x10^3 cm^3/s^2 and I can't really connect that into finding an acceleration which would be close to 9.80m/s^2
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  3. Sep 7, 2008 #2


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    Hi azn4lyf89,

    What is the formula for the slope? Displacement would be change in position, so I don't think that would match the slope formula.

    How did you calculate this? It does not appear to be correct to me.

    There's not many details here, but is it supposed to be close to 9.8m/s[itex]^2[/itex]? Is the object in free fall?
  4. Sep 7, 2008 #3
    Yes, its from an experiment I did in class. It was a free falling plummet ignoring air resistance. I basically got the slope from my line of best fit. So the y-value would be the velocity squared and the position would b the x-values. Dividing y/x would get the slope and since the y-value is in cm^2/s^2 and the x-value is in cm, the slope would be cm^3/s^2, correct me if I'm wrong. The 2 points I got from my line are (21cm,1.55x10^5 cm^2/s^2) and (49cm, 2.10x10^5 cm^2/s^2). I'm suppose to determine how this slope is related to the acceleration and from this slope calculate the acceleration of the plummet.
  5. Sep 7, 2008 #4


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    Because you're dividing by the x value, a cm in the units denominator will cancel one in the units numerator. Since you're trying to find acceleration, that's definitely a good sign.

    To find the relationship between the slope and the acceleration, you should use symbols only (not your experimental numbers). Let your two points be [tex](x_i,v_i^2)[/tex] and [tex](x_f,v_f^2)[/tex]. Using those symbols, write down the slope. In other words, if [itex]m[/itex] is the slope, use those two points and write out:


    Once you have that written out, see if you can rearrange it to match a formula that you are very familiar with. Do you get the answer?
  6. Sep 7, 2008 #5
    Ooo I think I get it. That means the slope should be twice the acceleration? So that means my acceleration from the slope would be 1000cm/s^2 which converts into 10m/s^2 which seems pretty close to 9.8m/s^2?
  7. Sep 7, 2008 #6


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    That sounds right to me. (You got that answer because you rounded off the slope to the number 2000. If you don't round off until the end, I think you'll get even closer to the given value.)
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