Acceleration from Position vs Time^2 ?

[amd123]

Homework Statement

http://img824.imageshack.us/img824/6466/31231242.png

http://img835.imageshack.us/img835/5220/76022234.png

It's a free fall problem and I have to calculate a = g

Homework Equations

g = (2*(d-(vi * t)))/t^2
g = (2d)/t^2

The Attempt at a Solution

I've made a graph and I've obtained a best fit polynomial equation.
I know that m/t^2 = a, however if I just divide the position by time squared will that alone give me acceleration?

I've tried finding initial velocity by subtracting (p2-p1)/(t2-t1) and the change in time with (t2-t1) and delta d (d2-d1) and I used this as vi for the first equation I gave...but this calculation gives me 0. Basically all the calculations cancel each other out.

I've tried taking the position * 2 and dividing by t^2 and I'm still getting an acceleration value that keeps changing as the position and time increases.

What am I doing wrong?

 

[SammyS]

First of all, it's important to know what your data represent.

I'm guessing that your timer was at 1.0470 seconds when your falling object was at position 0.142 meters . It makes no sense to take the square of the time.

At 1.0973 s -- 0.0503 s later -- the object was at 0.222 m -- that's 0.080 from the earlier position.

What was the average velocity during that first 0.0503 s time interval? Also, since you are assuming that the acceleration is approximately a constant, that average velocity should be the instantaneous velocity at the mid-time of the first time interval, i.e. at time (1.0470 + 0.0503/2) s .

Do the same for each successive time interval.

Find average velocity and mid-time for 1.0973 s to 1.1477 s, 1.1477 s to 1.1983 s, etc.

Graph these velocities vs the "mid-time". The acceleration is the slope of this graph.​

Alternative to using the slope: Find the average acceleration from one "mid-time" to the next, for each of the mid-time intervals. Average these accelerations.

Additional comment: For a graph such as the one you show, notice that except for the last data point, the data very nearly fall on a straight line. Most experimenters would suspect that the last data point is an "outlier" and thus would tend to ignore it.

Since t² as calculated for this graph is meaningless, in this case the straight line is merely a coincidence.​