# Homework Help: Acceleration from Position vs Time^2 ?

1. Sep 17, 2011

### amd123

1. The problem statement, all variables and given/known data
http://img824.imageshack.us/img824/6466/31231242.png [Broken]

http://img835.imageshack.us/img835/5220/76022234.png [Broken]

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

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

3. 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?

Last edited by a moderator: May 5, 2017
2. Sep 17, 2011

### SammyS

Staff Emeritus
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. ​