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Graphing Questions

  1. Aug 29, 2007 #1
    1. The problem statement, all variables and given/known data
    Why did we choose to linearize our data by plotting velocity versus time rather than by distance versus time-squared. Use d-v-a-t (distance=velocity*time+0.5*acceleartion*time-squared) equations to present a convincing answer.

    2. Relevant equations
    1. d=vt+1/2at^2
    2. (v2)=(v1)+at
    3. (v2)^2-(v1)^2=2ad
    4. d=1/2(v1+v2)*t
    V1 is initial velocity
    v2 is final velocity
    t is time
    a is acceleration
    d is distance

    3. The attempt at a solution
    I must say that I am perplexed since they are essential the same graphic shape and equivalent slopes (which equals accleration if you take derivative). I guess that it could have something to do with my 3rd equation above.What has got me is that v vs. t and d vs. t^2 are both linear graphs! Is it because acceleration is defined as the change in velocity divided by the change in time and not change in distance divided by change in time squared. But that doesn't use a d-v-a-t equation. Help please.
    Thanks in advance for your time and thoughts.
  2. jcsd
  3. Aug 29, 2007 #2


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    Homework Helper

    Can you give some more background on the question?

    "Why did we choose to linearize our data by plotting velocity versus time rather than by distance versus time-squared."

    This question makes no sense to me unless I know what the goal of the experiment was, or what you're trying to obtain...
  4. Aug 30, 2007 #3
    Alright, this is a physics lab in which the motion of a glider on an airtrack is being used to calculate gravity. The lab asks us to find gravity in two ways: one-algebraically by using the fact that (v2)^2-(v1)^2=2ad and a=g sin theta where theta is the angle of the airtrack with the ground. I did the graphs and the algebra and got an algebraic g of 945 and a graphic g of 825 (yes, there is some error in this, but the teacher actually admitted to introducing error into the lab=the sparker we used ass a timing interval was broken. Maybe she is checking for honesty in our first lab?). Anyway one of the lab asks me to find g by graphing v versus t and then asks why I did not graph d versus t^2
  5. Aug 30, 2007 #4
    could it be that rearranging the equation (v1)=(v2)+at yields a = delta v/t
    while rearranging d=(v1)t + 0.5at^2 d/t^2= (v1)/t^2 + 0.5a
  6. Aug 30, 2007 #5


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    Homework Helper

    you mean d/t^2 = v1/t + 0.5a

    Yes, the graph of d vs. t^2 will not be linear unless v1 = 0.

    Suppose you want to plot d vs. t^2... so let s = t^2

    [tex]d = v1*\sqrt{s} + 0.5s[/tex]

    So you can see that this is not the equation of a line unless v1 is 0. And when it equals 0... the slope of the graph is 0.5a... so you have to multiply the slope by 2 to get the acceleration...
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