Acceleration day one

After glancing at the approach taken last January, I decided to attempt a parallel approach this fall and tread along the edge of calculus.

I began with a graph of a medium speed velocity of 3.5 meters per second from the 11:00 class last Thursday. I then made a second graph of time versus velocity. I noted that the 2.88 second time recorded at 10 meters was only an approximate time. I then showed that the area under the velocity line in the second graph was the distance: 2.88 seconds horizontally × 3.5 meters per second vertically is 10.08 meters. In other words, the time × velocity is the distance - which is exactly what d = vt expresses. Distance is velocity × time. 

The third graph was another time versus velocity graph, but here I supposed that each second my velocity would be one meter per second faster. So I would start at 0 meters per second at time zero, be moving at one meter per second at one second, increase my speed again to two meters per second by the two second mark, and be moving at three meters per second after three seconds. 

Using the formula for a triangle where the area equals ½base × height to obtain the distance covered yields distance of 0.5 meters, 2 meters, 4.5 meters, and 8 meters at 1, 2, 3, and 4 seconds respectively. Plotting these values back onto a time versus distance graph yields a curve. Arguably, but not proven to be, a parabola. For now, a curve that appears that it could be parabolic.  But is this what will happen?

At 12:26 the class returned to the sidewalk. The red lines are at Fibonacci numbers: 1, 2, 3, 5, and 8 meters. Use of the Fibonacci sequence has worked well in recording temperatures for cooling curves. The ever increasing distances work against "timing rhythms" developing.

A number of students recorded the RipStik acceleration data, the above times were from Kamaloni. The data is clearly non-linear. The curvature is well matched by a quadratic equation. This doesn't prove that the data is parabolic, but the data provides support for this model. The acceleration was 0.45 meters per second per second, or 0.45 meters/second². The ½ in the equation for distance versus time can be conceived of having originated in the formula for the area of a triangle. This is not strictly true, but sorting this out properly is usually done with calculus.

An example of this demonstration can be seen in the following video.

Did my speed increase at a constant rate? Was there a linear rise in velocity?

Graph

The orange data points show that the rate of increase in speed of the RipStik was unsteady. The data, however, is roughly linear and does not appear to curve in a particular direction. Accelerating a RipStik in a steady fashion is challenging.