# Basics: Making graphs with kinematics stuff

**pre reqs:** [kinematics](http://blog.dotphys.net/2008/09/basics-kinematics/)

Suppose there is some experiment in which you throw a ball up and collect position and time data (with video analysis). What do you do with this data? Your instructor told you to make a graph, but how do you do that?

Here is the fictional data you (or I) collected:

Here is the text file with the data if you want to reproduce the graphs I make here [kinematics data](http://blog.dotphys.net/kinematics_data.txt)

**First idea: Use graph paper and plot what you see**

That makes sense, doesn’t it? Well, where do you get graph paper? Surprisingly, there are many places online that offer free graph paper that you can print out. Here is the one (this is the first hit from [google](http://www.google.com/search?source=ig&hl=en&rlz=&=&q=print+graph+paper&btnG=Google+Search) [http://www.printfreegraphpaper.com/](http://www.printfreegraphpaper.com/)).

So I have my paper now. This is going to be a weird graph though – look, the position data is all between 2 and 3 meters. Won’t this graph have a ton of wasted space? Actually, when you make a graph, the axes do NOT have to start at zero, they can start where ever you want. The other mistake to NOT make is to try and force each square to be one division of data points. Let me explain: The graph paper I printed out is 30 boxes by 39 boxes. If you choose time to be on the side with 30 boxes, do not make each square represent 0.1 seconds. This would only use 6 of the available boxes. You want at LEAST use half of the graph paper. If your data takes up less than half the paper, you can always let each TWO squares be what one square was before. In this case, I can let each 4 squares equal 0.1 seconds.

For the vertical data, my smallest value is 2.06 meters and my largest is 2.69 for a difference of 0.63 meters. For this data I can let each square (division) be 0.02 meters. Here is what I get:

A couple of things to note:

• data takes up more than half of the paper
• The axes are labeled WITH units.
• It has a title, which is just a good idea.

But what now? Do you connect the dots? Well, remember the purpose of a graph is not to make a graph (although many students think the purpose of a graph is because the instructor said to do it). There has to be some reason for making a graph. In this case, you would probably want to find the acceleration of this object. To do that, you could describe this data as a mathematical function (like *y(t)*). The data *looks* like a parabola, but how do you fit that? Truthfully, with graph paper that is quite difficult.

Ok, let us (me) think about exactly *what* we want. I want to show that this is constant acceleration motion. In this case, the object should follow the function: ([kinematics refresher](http://blog.dotphys.net/2008/09/basics-kinematics/))

But with graph paper, it is not trivial to fit a quadratic function. Can we cheat? Not really. If this had been a situation where the ball was dropped and the initial velocity was zero, then the function could be written as:

In this case the variable ?y is linear with respect to *t2*. But alas, the initial velocity is not zero. Then what is a student without a computer to do? There is one thing.

**Plot velocity versus time**

Velocity *is* linear with respect to time:

(is it confusing if I write v as a function of t? Maybe I should write it different). I could write:

Why is a linear relationship so important? Well, with a graph on graph paper, one CAN estimate a best fit straight line. Fine, well then how do I get velocity data? There is a way. Some may not like it, but there is a way. If you look at two consecutive positions, you can use them to get a velocity.

So, here is the plan. If I look at the first two times and positions, the ball (or whatever it was) went from 2.3 meters to 2.57 meters in 0.1 seconds. Yes, this was not constant motion. But I can get the average velocity. This is approximately:

But what time does this go with? How about I say it was the velocity at 0.05 seconds (halfway between 0.0 and 0.1 seconds). This is sort of cheating, but if the time interval is small it really doesn’t matter.

Doing this for the posted data, I have the new data:

Notice that before I had 7 data points, now I just have 6. I will go ahead and plot the data on a different sheet of graph paper (using the same ideas as before). Here is the finished graph (with the added best fit line):

A couple of things to point out:

• You probably see my mistake. If I were turning this in for a grade, I would probably redo this (or not do it in pen).
• I added my “best fit” line. This is really just a guess. There IS a way to determine the actual best fit line, but I will save that for another day. Notice that I did not draw a line from the first point to the last. If I had done that, what would be the point of all the data in between those?
• I picked two points from which to calculate the slope. The points chosen were as far apart as possible, not one of the points used to create the graph, and for ease I chose points that were on division lines.

Now I can calculate the slope – rise over run or:

The slope has units of acceleration and indeed it is the acceleration. Of course this is not -9.8 m/s^2 because I used realistic data. Of course you may get a better value of the acceleration if you had more data or if you had fit a quadratic function, but you work with what you have.

I think it is important that students understand the basics of graphing without using a spreadsheet or some other computer program. All too often students just feed numbers to a program and it spits out a picture. How do you know you can trust that program? You DO know that one day computers (and robots) will rule the world? You should prepare for that day now and understand how to do graphs by hand.

What if you want to use a program? How do you do that? I will save that for a part II. (this is longer than I expected)

**Here is [Part II](http://blog.dotphys.net/2008/09/basics-making-graphs-with-kinematics-stuff-part-ii/)

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