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Uniform acceleration kinematics lab

  1. Sep 12, 2011 #1
    1. The problem statement, all variables and given/known data

    This is more of a general question but it is part of my physics course so i decided to post here, im sure many of you have done this lab but what it is is a cart with tickertape attached to it going down a ramp, once we plot a d-t graph there are lots of application questions. One of them says plot a d-t^2 graph and a v-t graph, it says to calculate the slope of each graph and compare them.

    2. Relevant equations

    None really just slope, (y2-y1)/(x2-x1)

    3. The attempt at a solution

    By looking at the units i thought that the slope of each graph should both represent acceleration as cm/s^2 and (cm/s)/s are the same thing, since they came from the same data set they should have the same slope but I get two different slopes, they are exactly off by a factor or 2. Am I making an arithmatic error or is it supposed to be like this, ( I get a slope of 61 for the d-t^2 graph and 120 for the v-t
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 12, 2011 #2

    PhanthomJay

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    What is the equation that relates velocity with acceleration and time? What is the equation that relates distance with acceleration and time?
     
  4. Sep 12, 2011 #3
    v2^2=v1^2 + 2aD?
    D= v1T + 1/2aT^2?

    im sorry but im not sure what exactly you are getting at? why are my accelerations different if the units worked out correctly and they were both taken from the same data set
     
  5. Sep 12, 2011 #4

    PhanthomJay

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    Your 2nd equation relates distance with acceleration and time,

    D = 1/2at^2

    you didn't express the equation which relates velocity with acceleration and time,

    v = at

    Now when you plot v vs t, the slope of the line is a , correct?

    Now plot D vs. t^2. The slope of the line is ??????
     
  6. Sep 12, 2011 #5
    D vs t^2 should also be acceleration no! m=rise/run = D/t^2= cm/s^2
    Or are you implying I use your D=1/2at^2 and t^2 for it and do (1/2at^2)/(t^2) which leaves me with 1/2 a. Which makes sense slightly but what is this the acceleration for? And howcome it doesnt work when i just use the units on the axis, only when i use an equivelant relationship to D
     
  7. Sep 12, 2011 #6

    PhanthomJay

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    When you plot D vs. t^2, you should get more or less a straight line sloping up and to the right. Since distance = 1/2 at^2 (d = mt^2, where m is the slope)then the slope of the line is ??????. Be sure to plot distance on the y axis, and time squared (not time) on the x axis.
     
  8. Sep 12, 2011 #7
    1/2a. Thanks for all the help I just dont understand what this acceleration represents. Is it just the graph is set up like the kinematic equation so "solving for the slope" is really just isolating the 1/2a, being 1/2 of the acceleration of the cart. Is doing D over t^2 times 2 just the graphical equivelent of the kinematic equation. I think thats all been cleared up thanks alot. What is the the actual graph, ie what does it really mean, when i read it what information can i extract without differntiating or integrating it.
     
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