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Position vs. Time graph equation

  1. May 11, 2010 #1
    I just completed a lab on the acceleration due to gravity and I have to answer a few questions about my data. Before I share what I am confused about I will give a quick run through of the lab procedures:

    --First I used a spark timer to mark a strip of paper with a clamp attached to it and let the strip free fall. Then I measured the distance of each mark made from the spark timer with a 2m stick and made a distance vs time graph from the data. I made a best fit polynomial line of degree 2 and found the equation of the fit line that turned out to have the form ___x^2+___x+___, where the ___ is where my coefficients are. I multiplied the x^2 coefficient by 2 to get acceleration.

    One of the lab follow up questions asked me why I multiply that coefficient by 2 to get acceleration and I am totally stumped.

    Can someone explain why multiplying by 2 gives accel?
  2. jcsd
  3. May 11, 2010 #2


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    In your polynomial fit, x represents time (independent variable) and position is y (dependent variable). You fitted coefficients a, b and c to the equation

    y(t) = at2 + bt + c.

    What kinematic equation gives y(t)? Compare the coefficients between what you know to be the case and what you fitted.
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