Day one acceleration

Once again I began the week plus on acceleration by asking the students what will happen on a time versus distance graph if the RipStik starts from rest and goes faster and faster.

The layout was as in the prior term. Timing marks were set at Fibonacci distances of 1, 2, 3, 5, and 8. Once again this worked well in terms of my ability to sustain an acceleration of nearly 0.5 meters per second squared. Maintaining a linear increase in velocity beyond eight meters seems like it would not be possible for me. I started on a push off from the usual post. Once again hindsight has suggested that the first meter is being covered too quickly.

Athina Viola between the five and eight meter mark

Desmos

The speed into the first meter mark is a tad high against the downstream data points. The velocity, however, is still nicely linear in its rate of increase. 

Desmos has powerful value reference notations as seen in the use of t₁[5] to pull the value of the fifth element of t₁ in the table seen earlier above.

In theory the two accelerations should be equal. That they are not is a result of timing errors and, more significantly, non-linearity in the increase of the velocity.

I decided not to return to the classroom and tried to sketch out an argument for the parabolic nature of the distance versus time curve. I started with the equation from last week that distance equals the velocity × time, d = vt. I then made a table where the velocity increases by one metre per second each second. 

Using a table on the poster pad that was arranged as above, I argued the if d = vt and v increases by one each second, then the resulting distances covered are the produce of v × t. This of course is more than problematic because at four seconds the RipStik has not been traveling at 4 m/s for 4 seconds. And that holds true for each of the earlier calculations. I did not mention this issue which is resolved in part by looking at the area under the segments and using ½vt (the area under the velocity line is the distance and is ½ base × height for the triangle. And that cuts each value above in half. But then the squared nature disappears for the students. Seeing that 0.5, 2,  4.5, and 8 have an underlying quadratic nature is tougher. Not obvious. 

Each term I vacillate between a more rigorous mathematical approach and a more hand waving argument for the quadratic nature of these systems. A year ago I started in the classroom with a more rigorous approach and used data recorded every meter for seven meters. Perhaps that worked better than what I did today on the sidewalk. Last term I tried to make this argument clearer, again hoping to retain the ½ that comes from the triangular area and will appear in the d = ½at² formula. This term I abandoned that approached and ran an argument straight from d = vt which puts the focus on the t² nature of the relationship. This term the difficulty was building the time versus velocity table in order to show that this was linear. On a tablet. While down on the sidewalk. But I had decided not to go back to the classroom. Maybe Wednesday. Maybe.