Help with Displacement vs. time and acceleration

AI Thread Summary
The discussion focuses on analyzing a displacement vs. time graph of a ball dropped from rest, which shows a parabolic trend. The quadratic regression equation derived is -4.03x^2-2.06x+1.20, with a high R^2 value of 0.99857. It is clarified that the quadratic regression should align with the kinematic equation h(t)= (1/2)At^2+V0t+h0, where the initial velocity (V0) must be zero since the ball was dropped from rest. Guidance is provided to redo the regression by constraining the linear term's coefficient to zero for accuracy. The discussion emphasizes the importance of matching regression parameters with expected values from the kinematic equation.
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Homework Statement



I have the following points that were gathered from a displacement vs. time graph (the graph is of a ball being dropped from rest). The line is parabolic. The quadratic regression for the points is -4.03x^2-2.06x+1.20 where R^2=0.99857.

Correct me if I'm wrong but the quadratic regression is the same as this: h(t)= \frac {1}{2}At^2+V_{0}t+h_{0}

I need to make a velocity vs. time graph with this information but I need a little guidance.

(time (s), displacement (m))
(0.000, 1.194)
(0.097, 0.979)
(0.129, 0.881)
(0.161, 0.774)
(0.193, 0.654)
(0.258, 0.399)
(0.290, 0.254)
(0.322, 0.102)
(0.354, 0.000)


Homework Equations


h(t)= \frac {1}{2}At^2+V_{0}t+h_{0}


The Attempt at a Solution


I performed the quadratic regression but now I'm stuck
 
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Think of x as representing time t, then match whatever constants you found from your quadratic regression to the constants in the kinematic equation that you quoted. But (and this is a big "but") if the ball was dropped from rest, then the linear term in t must have its coefficient constrained to be zero for a meaningful fit (i.e. v0=0). I think you need to redo your regression with that in mind. You should be able to predict from the kinematic equation what the fitted parameters ought to be and then verify that they are what you expected.
 
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