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When lab data are plotted, a best fit line is drawn through the data points. If the line is straight, then it has a constant slope value. The slope value can be calculated as the rise per run ratio; that is, the slope is calculated as the change in the y-axis variable per change in the x-axis variable.
 

There are three versions of this question. Each version requires you to calculate the slope of the line on a graph and to determine the units of the slope. Here is one of the versions:

Version 1:

The graph shows the mass of samples of a metal as a function of its volume. Determine the slope of the line and indicate the units on the slope.
 

 

The slope is the "rise per run" ratio for a line on a graph. To determine this ratio, one must first know the coordinates of any two points on the graph. The points you select do not need to be data points; they merely need to be points on the line. Pick points for which at least one (preferrably two) coordinates are clearly known. Write down the coordinates in (X, Y) form for each of the two points. It wouldn't hurt to label them (X1, Y1) and (X2, Y2).

Once the coordinates are known and written down, you are ready to compute the slope. To do so, calculate the change in y-coordinate values and the change in x-coordinate values. A change means to subtract the value for the first coordinate from the value of the second coordinate. Write down the result and label your work so that you know what you've done. (I know that this "write it down" idea may seem like a lot of work ... but if you miss the question you will have to get it correct two more times and that's guaranteed more work. Sometimes the "long way is the short cut".)

Finally, use the change in y coordinates (∆Y) and the change in x-coordinates (∆X) to calculate the slope. The slope is the ∆Y value (rise) divided by the ∆X value (run).

What about the slope units? Easy - it's just the unit on the y-coordinate divided by the x-coordinate; that would be y-uit per x-unit. And that makes sense since the slope is determined by dividing ∆Y by ∆X.

One final remark: because you are estimating coordinate values for your two selected points, the whole procedure is subject to measurement error. But fear not as you allotted a little more than 10% error in your answer. 

Unfortunately there are currently no resources at The Physics Classroom Tutorial that focus exclusively on using graphs.
 


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