Acceleration day four

This is the first time I felt that the explanation of the ½ in d = ½at² + v₀t + d₀ ran in a smooth progression.

I started at the upper left with 2 m/s linear velocity on a time versus distance graph. Below this I sketched the time versus velocity. 

Then I demonstrated that the area under the curve is the distance. 

Moving directly to the right I sketched a continuously increasing velocity at 2 m/s per second. Then I worked out the areas for one, two, and three seconds. This approach avoids working with mean speed over a time interval on the time versus distance graph, a prior approach that is fraught with confusing details. 

After calculating the distances from the velocity graph, the distances were transposed up to a time versus distance graph. The result is a parabola. As promised. At this point a hand waving numeric argument is made that the ½ is from the area of the triangles rather than the mean speed over a time segment on the time versus distance graph. 

A presentation was prepared that follows these basic steps, but was not used in class. 

The wrap was "Where is the slope and intercept in d = ½at² + v₀t + d₀?"