- #1

Kandy

- 27

- 0

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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- Thread starter Kandy
- Start date

- #1

Kandy

- 27

- 0

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.)

- #2

EnumaElish

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- #3

andrevdh

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[tex]a=grad\ F[/tex]

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.

- #4

whozum

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Kandy said:

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.

- #5

HallsofIvy

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On the other hand, I notice that all of your points except (40, 25)

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

- #6

Ouabache

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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.

You can double check your answer by plugging your data into a spreadsheet program such as

I believe both these techniques use the "least squares" method of finding the

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- #7

EnumaElish

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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).Ouabache said:Since this is supposed recorded data, my first choice would belinear regression analysisespecially for data that doesn't fall perfectly on a straight line.

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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- #8

Ouabache

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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).

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

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

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- #9

Integral

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

- #10

Diane_

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- #11

Ouabache

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HallsofIvy already covered your sentiment:Diane_ said:

We don't know what level Kandy is at. So we gave several equally useful methods. If I were Kandy, I would try more than one method and use the one I felt most comfortable with.HallsofIvy said: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.

- #12

Diane_

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- #13

Ouabache

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