# Help with Displacement vs. time and acceleration

1. Sep 12, 2009

### lmn

1. The problem statement, all variables and given/known data

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)

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

3. The attempt at a solution
I performed the quadratic regression but now I'm stuck

2. Sep 12, 2009

### kuruman

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.