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Finding acceleration from graphs (s vs t^2)

  1. Nov 11, 2007 #1
    1. The problem statement, all variables and given/known data

    If I’m to graph a plot of s (displacement) vs t^2 (time squared) then what would be the independent and dependent variable, t or s? I’m supposed to supposed to use this graph to find a slope with error bars and use the appropriate kinematics equation [which I think is s(t) = 1/2 at^2 (the vi*t is 0)]. The slope I will attain from the t^2 versus s graph, what will it represent? And how could I possibly plug it back into the equation to find acceleration? Thanks.

    Also when I graph my results, should it be a positive or negative slope if the distance and time are decreasing every time? Because if it's a negative slope than that would mean that the x values for time squared would decrease going from 9.826 to 7.132

    2. Relevant equations

    s(t) = vi^t + 1/2a*t^2

    3. The attempt at a solution

    This is the table I'll use
    Average Displacement (cm)-------->Average Time squared (t²) s
    Distance 1 152.90 ± 0.05 cm--------->9.686
    Distance 2 143.66 ± 0.05 cm--------->9.321
    Distance 3 133.30 ± 0.04 cm--------->8.588
    Distance 4 123.56 ± 0.02 cm--------->7.896
    Distance 5 113.76 ± 0.02 cm--------->7.132
     
    Last edited: Nov 11, 2007
  2. jcsd
  3. Nov 11, 2007 #2
    You are trying to draw a linear graph, right? And, as your graph will pass through the origin, this will have the form y=mx, correct? I'm going to assume that time is your independent variable, as it is the one with with you don't have a recorded error. Now, rearrange the formula to find m. Does the result seem familiar?

    When drawing your graph, always start at the origin (0,0), and go towards infinity, for both x and y. Start with your lowest value of time, plot it, and continue up until you run out of points. In this way, you will not have a negative slope.
     
  4. Nov 11, 2007 #3
    Actually, the error on time is +/- 0.0005 s. Anyways, I don't know about starting from 0, I mean, wouldn't that throw off the graph because the four intervals for time are 9.686 s, 9.321, 8.588 etc.

    My other question now is should I have the time going from 9.686, 9.321, 8.588 or should I have the time increasins going from 7.132 s to 9.686?

    Reason is because I'm supposed to find gravity from the acceleration. I'm guessing that the slope will be the value of 1/2*a from the kinematics equation.

    Also, would anyone know how to show error bars on excel?
     
  5. Nov 11, 2007 #4
    Right click on one of your points.

    "Format data series"

    "X Error Bars"
    "Y Error Bars"


    Plot it as increasing
     
  6. Nov 12, 2007 #5

    andrevdh

    User Avatar
    Homework Helper

    Best to consider s as the dependent and t^2 as the independent variables.
    Assuming you are using Excel?

    If the object started out of rest the term [tex]ut[/tex] can be dropped from the formula so

    [tex]s = s_o + \frac{1}{2} at^2[/tex]

    which menas that the gradient of the graph should be half of the acceleration and the y-intercept the displacement of the object from the origin when it started to accelerate.

    This approach gives you a negative gradient. This fits the situation where the origin was below the point where the object was released and it fell towards the origin, giving you a negative acceleration.
     
    Last edited: Nov 12, 2007
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