# Interpreting Graphs

Kandy
Suppose you recorded the following data below during your study of the relationship of forces to acceleration. Plot the graph.

Force (N) | Acceleration (m/s^2)

10 | 6.0
20 | 12.5
30 | 19.0
40 | 25.0

This is where I'm having problems: Write an equation for the line. (Recall that slope = change of y over change of x.)

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Presumably F=0 imples a=0 so you have 5 points including the origin. I'd have written a "step" (parts) function for each segment. E.g., a = g0(F) for 0 < F < 10, a = g1(F) for 10 < F < 20, etc.

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The data gives a directly proportional graph which means that the relationship between the acceleration and the force is given by
$$a=grad\ F$$
where grad is the gradient of the graph, which, according to Newton's second law, is equal to the inverse of the mass of the object that was used in the experiment. One would normally plot the force on the x-axis, since this is what was changed directly by the experimenter.

whozum
Kandy said:
Suppose you recorded the following data below during your study of the relationship of forces to acceleration. Plot the graph.

Force (N) | Acceleration (m/s^2)

10 | 6.0
20 | 12.5
30 | 19.0
40 | 25.0

This is where I'm having problems: Write an equation for the line. (Recall that slope = change of y over change of x.)

Are these measured from experiment? Because that isn't a direct proportion.

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You can't write the equation of THE line because these points do not all lie on a line. They are close to lieing on a line but that means that there are an infinite number of lines you could use. I assume you don't want to go to all the work of finding the "least squares line", the line that best fits all the points. What I would do is take the two points that are farthest apart, (10,60) and (40, 25), find the slope of that to find the equation of the line. If you are sure that, as EnumaElish said, 0 force gives 0 acceleration, the (0,0), (40, 25) would do. What is the slope of that? Once you know the slope m, y= mx.

On the other hand, I notice that all of your points except (40, 25) do lie on a straigt line! Perhaps there was an error measuring that and it would be better to give the equation for that line.

In any case, an experiment is never perfect. You can only expect an approximate answer and any of these would be good.

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Since this is supposed recorded data, my first choice would be linear regression analysis especially for data that doesn't fall perfectly on a straight line. A very handy tutorial is given http://phoenix.phys.clemson.edu/tutorials/excel/regression.html.

You can double check your answer by plugging your data into a spreadsheet program such as MS Excel, If you have that program, enter your data in two columns, just like you have them. Under Insert>Function use the category Statistical, then use INTERCEPT function for y-intercept and SLOPE function for slope. Just be sure to select an empty cell before you calculate either parameter.

I believe both these techniques use the "least squares" method of finding the best fit line that HallsofIvy mentioned.

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Ouabache said:
Since this is supposed recorded data, my first choice would be linear regression analysis especially for data that doesn't fall perfectly on a straight line.
Technically that is correct, you can run a simple regression on 5 data points. But don't expect to see any statistical significance (that's not the reason you are running the regression anyways so it's okay).

A slightly more troublesome issue is the intercept. The theory tells you that intercept = 0. You have 2 choices:

1. include the origin as a 5th data point and run the regression in the standard way (unconstrained). Then see how close your intercept ends up being to zero (you may also look at whether it is statistically significantly different from zero; my guess is it won't be). Never the less, you will end up with a nonzero constant term as your least squares intercept.

2. Do not include the origin in your data but constrain (restrict) the intercept to be identically zero. That will change the estimation procedure; it will also change the interpretation of the regression results. For example, the R squared statistic can not be interpreted as a standard measure of model fit as it would in a standard (unconstrained) model. On the upside, your intercept will be exactly zero.

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All those who replied here, have given some excellent suggestions! It is also fun to observe, from how many different directions we approach this question.

My earlier response was made from an empirical perspective.. In other words, suppose you didn't know that force is directly proportional to acceleration and you conducted this experiment. You could plot the data, observe that it seems to be linear and draw by eyeball your best line through the points and read the y-intercept directly off the graph. You could take all the (change-y / change-x) differences and average them for the slope. Since no criteria for precision was given, this may be accurate enough. Or you can apply statistics and find a line using linear regression method.

If you repeated the experiment a few more times, you may converge to a better representation of y-values (acceleration) or if there is imprecision in the experiment, you may find more variation. But if you knew ahead of time (from theory), that there was a perfectly linear relationship and that 0 force would yield 0 acceleration (giving you the y-intercept); then I would make some assumptions as have been suggested, maybe even exclude outlier data points, before calculating the slope. But then I would be biasing the data in a direction I thought it should be heading (If you're practising science, it's best not to fudge your results to fit what you believe, even Einstein made that mistake).

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Staff Emeritus
Gold Member
What am I missing? When I plot this data I see an excellent line, Excel agrees. There is little difference if I force it to pass thorough the origin or let it compute the intercept, the answer is the same, within errors. This looks like just about to good a data set to be real.

Plot the points, eye ball a good line through them, find the slope. Should be very straight forward.

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Folks, just a thought: if Kandy is in high school, the chances are that linear regression analysis and least-squares fits are not in her repertory.

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Diane_ said:
Folks, just a thought: if Kandy is in high school, the chances are that linear regression analysis and least-squares fits are not in her repertory.
I've always found it advantageous to give a person the benefit of the doubt, and offer equally valid methods to approach a problem. I was taught linear regression in high school. It's a basic function on my calculator. It came in quite handy for h.s. science labs.. 