Mail Time: Linearizing a parabolic distance vs. time graph

An APSI participant writes:

I had a question regarding linearization.  We just did the free fall lab and a student graphed their data to get the parabolic curve of d vs t.  When asked what they would do to linearize, they stated to convert d to v and graph v vs. t.  

When they determined v, they did d/t for each point.  Then plotted that value for Vavg vs time.  Is this an acceptable way of linearizing?  

My first instinct is no, because the slope of the v-t graph, while still the acceleration, does not produce the value as it would if you did d vs t^2.  

The slope of their v-t graph was 5.15 m/s/s.  If they would have plotted d vs t^2, they would have a slope of 5.45 m/s/s with an acceleration of 10.9 m/s/s for g.

This is a common approach by early-in-the-year AP Physics students.  They did not do this right... 

By dividing just d/t for each point, they took the average velocity from the beginning - that's not how a velocity-time graph is made, and that quantity is rather meaningless in the experiment you describe.  Basically never divide the values of two data points!  :-)

What WOULD be a reasonable alternative approach to this experiment would be to make a legit velocity-time graph: by taking the slope of a tangent line to each point on the d vs. t curve.  Then plot those *instantaneous* speeds as a function of time.  That's a true velocity-time graph, for which the slope is acceleration.

Of course, that sounds way, way more complicated than just using lab linearization approaches of writing the relevant equation, then plotting data as it appears in the equation, d and t^2.  

Hope that helps!