Finding the Average acceleration from a table of time vs distance

[H2Owned]

Homework Statement

http://img30.imageshack.us/img30/3598/physicsproblem.png

The Attempt at a Solution

I don't know how to approach it with the given data set.

 

[Stonebridge]

Welcome to Physics Forums.

Do you not know how to draw the graph (part 1)? Or is it the constructing of the tangents that is the problem?
You need to give us more detail about the exact difficulty you are having, and to show what you have done so far.

 

[H2Owned]

Drawing the graph isn't difficult, you plot the points and connect the line. The part that i got stuck at was getting the instantaneous velocity. How can you get the derivative of a function at specific points when you don't know what the funcrion is?

 

[Stonebridge]

Drawing the tangent to the curve at a number of points requires a bit of practice.
There is some info here that explains a lot better than I can.
http://cstl.syr.edu/fipse/graphb/Unit8/Unit8a.html
The tangent at a point on the curve is the instantaneous value of the rate of change of the function at that point, and is numerically equal to the derivative at that point.
In this case dx/dt. Taking 4 values at different places will give you 4 values for velocity at different times. These values can then be plotted on a v-t graph. The gradient of that graph is acceleration.

 

[H2Owned]

That link didnt help as i know how to find a tangent at a given point. Now what did help is solving for the quadratic equation that is formed by those dots. I realized that by calculating the quadratic regression, i can get a quadratic formula that holds all of those points.
We are allowed calculators in this class so this website adequately explained how to do it quite clearly.
http://calculator.maconstate.edu/quad_regression/index.html

 

[Stonebridge]

I'm glad you solved your problem; and thanks for the interesting link.
Just to point out:
The advice I gave relates to the question as presented in the original post, where you are asked to plot the graph of instantaneous tangent (=velocity) against time; and then find the gradient of the resulting graph in order to determine acceleration.
What you have done is fine, but the question didn't ask for the regression to be performed on the displacement data.
The question is about knowing that dx/dt = v (gradient at a point on first graph) and dv/dt = acceleration (gradient of 2nd graph for constant acceleration).

 

[H2Owned]

i completely agree with you but you still didnt provide a precise enough way in order to get that second graph. By calculating the quadratic regression, and getting the quadratic formula, you achieve a clear graph of which you can calculate the regression and therefor properly answer the question. Drawing by hand, you would lose a LARGE amount of accuracy with the amount of points that are given.