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Math Help - fit to quadradic

  1. #1
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    Exclamation fit to quadradic

    need help on a particular question of my assignment...

    Using strobe photography a badminton shuttle's path is followed. the strobe light takes photos every 0.2 seconds and lasts for a total of 3.2 seconds.

    The actual positions of the shuttle are below...

    Horizontal - 0.00 0.80 1.56 2.29 2.97 3.61 4.21 4.76 5.28 5.75 6.18 6.59 6.97 7.33 7.67 8.00 8.32

    Vertical - 0.00 2.24 4.08 5.56 6.71 7.54 8.09 8.36 8.38 8.16 7.68 6.97 6.05 4.93 3.64 2.19 0.59

    i plotted a graph of the horizontal position against time.

    i am then asked to find the horizontal velocity of the shuttle using the gradient of the previous graph. it is supposebly a linear function and i just need to differentiate it. i think it may be y=ax^2+bx+c...i need a table and a graph.

    not completely sure tho. help would be great. thanks.
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  2. #2
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    If we're talking about ONLY the horizontal position against time, we do get a function that is nicely approximated by a quadratic.

    I get x(t) = -0.4373t^2 + 3.9568t + 0.0490
    The derivative of this is x'(t) = -0.8746t + 3.9568
    This gives the first three pieces of velocity
    x'(0) = 3.9568
    x'(1) = 3.0822
    x'(2) = 2.2076

    On the other hand, the first three first differences of the given data suggest:

    (0.8-0)/0.2 = 4
    (1.56-0.8)/0.2 = 3.8
    (2.29-1.56)/0.2 = 3.65

    There is some discrepancy there. If you can come to terms with exactly when the two versions are measured, perhaps you can reconcile them.

    Hint: Derivative are instantaneous and differences are averages.
    Last edited by TKHunny; May 11th 2008 at 05:21 AM.
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  3. #3
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    Thumbs up Thanks!

    Thanks!
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