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Math Help - Finding acceleration from graphs (s vs t^2)

  1. #1
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    Finding acceleration from graphs (s vs t^2)

    This is a bit of a physics question but I think you guys would know it, it's for a lab, anyways:

    If I知 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知 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



    This is the table I'll use
    Average Displacement (cm)---------->Average Time squared (tイ) s
    Distance 1 152.90 ア 0.05 cm--------->9.686
    Distance 2 143.66 ア 0.05 cm--------->9.321
    Distance 3 133.30 ア 0.04 cm--------->8.588
    Distance 4 123.56 ア 0.02 cm--------->7.896
    Distance 5 113.76 ア 0.02 cm--------->7.132
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by SportfreundeKeaneKent View Post
    This is a bit of a physics question but I think you guys would know it, it's for a lab, anyways:

    If I知 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知 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



    This is the table I'll use
    Average Displacement (cm)---------->Average Time squared (tイ) s
    Distance 1 152.90 ア 0.05 cm--------->9.686
    Distance 2 143.66 ア 0.05 cm--------->9.321
    Distance 3 133.30 ア 0.04 cm--------->8.588
    Distance 4 123.56 ア 0.02 cm--------->7.896
    Distance 5 113.76 ア 0.02 cm--------->7.132
    s will be the dependent variable and t^2 will be the independent variable.

    So the usual form for a line graph is y = mx + b

    In your case y = s and x = t^2. So your equation should be of the form
    s = mt^2 + b

    and you correctly predicted that the correct Physics equation is
    s = \frac{1}{2}at^2

    So when you graph (ie. do the regression analysis) the slope m will be equal to \frac{1}{2}a and your intercept b should be equal to (ie. close to) 0.

    -Dan
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    I was wondering if it would still work if I switched my numbers around so instead of going from 9.896 s, 9.321 s---> 7.132 s, I go from 7.132 s to 9.896 s on the graph for the x axis. So basically, I'd be switching this table going from bottom to top:

    152.9---------9.686
    143.66--------9.321
    133.3---------8.588
    123.56--------7.896
    113.76--------7.132
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by SportfreundeKeaneKent View Post
    I was wondering if it would still work if I switched my numbers around so instead of going from 9.896 s, 9.321 s---> 7.132 s, I go from 7.132 s to 9.896 s on the graph for the x axis. So basically, I'd be switching this table going from bottom to top:

    152.9---------9.686
    143.66--------9.321
    133.3---------8.588
    123.56--------7.896
    113.76--------7.132
    The order in which you graph the points is immaterial. The graph still looks the same.

    -Dan
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