### Day two of acceleration: the edge of integration

Building off of the distance and velocity versus time graphs from Monday, I worked from the velocity versus time graph backwards to what distance must have been covered in that time.

On the left board I began with a blank time versus distance graph and a linearly increasing velocity with a slope of two m/s on the time velocity graph. The other notes on the board above were added later, the board began with just the two graphs.

On the right board I diagrammed time versus velocity graphs for a speed of two meters per second. I made a hand waving argument that neither the start speed of 0 m/s nor the end speed would yield the correct distance from the d = vt formula. At 0 m/s the formula says I covered no distance. At the end of one second the 2 m/s suggests I would have covered two meters, but that would only be true if I had been at 2 m/s for the whole one second. And I was not. I started from 0 m/s. I suggested that halfway, the average speed, could be used in d = vt. Thus after one second I had covered 1 m/s times 1 second or one meter.

After 2 seconds I was moving at 4 m/s, thus my average speed was 2 m/s, or 2 m/s times two seconds, four meters covered. At three seconds I reached 6 m/s and I could use the average speed at 1.5 seconds, which was 3 meters per second. 3 times 3 is 9 meters covered. On the distance versus time graph I plotted (0,0), (1,1), (2,4), (3,9). I did explain that at time zero I was moving at zero speed and thus the distance covered was also zero, hence (0,0).

This plotted as a curve that looked like the result on the sidewalk on Monday

I argued that this suggested we were on the right track. Then I went back and showed that the formula I was using was actually the same at the formula for the area under a triangle: ½·base·height. Thus the distances were the areas under the line. Which is essentially what Newton realized and led him to develop integration - ahead of differentiation.

I do not expect that the students followed all of this. I then showed the students the physical science equation d=½at² and the full quadratic form. I could not help briefly touching on the slope and intercept for a quadratic, again noting to the students that this was not something they needed to calculate. I then circled back to the curve, the partial quadratic, and noted that this was really just half of the curve - as was suspected by a student on Monday. I told them I wanted them to see the other half of the curve.

I know this second run is always confusing for students. So I tried to prepare them by starting with a time versus velocity graph with a negative linear slope.

That half-hearted attempt was constrained by time and can be seen mid-center left on the board above. I noted that the graph meant my speed would drop to zero and then go negative. I would go backwards, so to speak. I ran quickly through the two term quadratic that would model the curve.

Time was tight, 33 minutes after the top of the hour in a 50 minute period, so I opted to time the run myself. I slapped down marks only at 3, 6, and 9, I did not mark the vertex nor did I time the vertex. Which was fortuitous as the vertex is usually the one point that does not fall on the curve - I arc around at just above 0 m/s, so the vertex is never correct.

The result was sufficiently quadratic as to be convincing when run into Desmos.

In the above the red points are time versus distance data points at 0, 3, 6, and 9 meters on the uphill into LRC. I then came back down the slope from 9 to 6, 3, and 0. In this instance I began with an initial velocity estimated by the quadratic regression to be about 2.5 m/s. I decelerated at an average -0.36 m/s throughout the run, which was an acceleration in the negative axis direction after the vertex.

The purple data points are estimated velocities, essentially an average velocity, over the duration of time to cover the three meters between the marks on the sidewalk. At the vertex the distance marks were 9 meters and 9 meters, hence a speed of zero at the vertex. The data is that which I foretold on the board. Note that the purple line is not a linear regression on the speed data. The purple line is v = at+v₀ where a was -0.36 m/s² and v₀ was 2.53 m/s, the values obtained from the quadratic regression. Those values produce a line that agrees reasonably well with the estimated average speed from point-to-point data.

The challenging part is to help lift the students up from bogging down in the mechanics and the specifics of the mathematics, as that is not my intention. The challenge is to have them see the patterns almost at a qualitative level. To see that a straight line does not well fit accelerating data, but that a quadratic causes the line to almost magically snap onto the points and predict the pattern. I will need to address this in class, to try to get them to lift their heads up out of the details of the mathematics and up to where they can see the patterns. To see that math is ridiculously good at modelling nature. For that is a core theme in the course.

On the left board I began with a blank time versus distance graph and a linearly increasing velocity with a slope of two m/s on the time velocity graph. The other notes on the board above were added later, the board began with just the two graphs.

On the right board I diagrammed time versus velocity graphs for a speed of two meters per second. I made a hand waving argument that neither the start speed of 0 m/s nor the end speed would yield the correct distance from the d = vt formula. At 0 m/s the formula says I covered no distance. At the end of one second the 2 m/s suggests I would have covered two meters, but that would only be true if I had been at 2 m/s for the whole one second. And I was not. I started from 0 m/s. I suggested that halfway, the average speed, could be used in d = vt. Thus after one second I had covered 1 m/s times 1 second or one meter.

After 2 seconds I was moving at 4 m/s, thus my average speed was 2 m/s, or 2 m/s times two seconds, four meters covered. At three seconds I reached 6 m/s and I could use the average speed at 1.5 seconds, which was 3 meters per second. 3 times 3 is 9 meters covered. On the distance versus time graph I plotted (0,0), (1,1), (2,4), (3,9). I did explain that at time zero I was moving at zero speed and thus the distance covered was also zero, hence (0,0).

This plotted as a curve that looked like the result on the sidewalk on Monday

I argued that this suggested we were on the right track. Then I went back and showed that the formula I was using was actually the same at the formula for the area under a triangle: ½·base·height. Thus the distances were the areas under the line. Which is essentially what Newton realized and led him to develop integration - ahead of differentiation.

I do not expect that the students followed all of this. I then showed the students the physical science equation d=½at² and the full quadratic form. I could not help briefly touching on the slope and intercept for a quadratic, again noting to the students that this was not something they needed to calculate. I then circled back to the curve, the partial quadratic, and noted that this was really just half of the curve - as was suspected by a student on Monday. I told them I wanted them to see the other half of the curve.

I know this second run is always confusing for students. So I tried to prepare them by starting with a time versus velocity graph with a negative linear slope.

That half-hearted attempt was constrained by time and can be seen mid-center left on the board above. I noted that the graph meant my speed would drop to zero and then go negative. I would go backwards, so to speak. I ran quickly through the two term quadratic that would model the curve.

Time was tight, 33 minutes after the top of the hour in a 50 minute period, so I opted to time the run myself. I slapped down marks only at 3, 6, and 9, I did not mark the vertex nor did I time the vertex. Which was fortuitous as the vertex is usually the one point that does not fall on the curve - I arc around at just above 0 m/s, so the vertex is never correct.

The result was sufficiently quadratic as to be convincing when run into Desmos.

In the above the red points are time versus distance data points at 0, 3, 6, and 9 meters on the uphill into LRC. I then came back down the slope from 9 to 6, 3, and 0. In this instance I began with an initial velocity estimated by the quadratic regression to be about 2.5 m/s. I decelerated at an average -0.36 m/s throughout the run, which was an acceleration in the negative axis direction after the vertex.

The purple data points are estimated velocities, essentially an average velocity, over the duration of time to cover the three meters between the marks on the sidewalk. At the vertex the distance marks were 9 meters and 9 meters, hence a speed of zero at the vertex. The data is that which I foretold on the board. Note that the purple line is not a linear regression on the speed data. The purple line is v = at+v₀ where a was -0.36 m/s² and v₀ was 2.53 m/s, the values obtained from the quadratic regression. Those values produce a line that agrees reasonably well with the estimated average speed from point-to-point data.

The challenging part is to help lift the students up from bogging down in the mechanics and the specifics of the mathematics, as that is not my intention. The challenge is to have them see the patterns almost at a qualitative level. To see that a straight line does not well fit accelerating data, but that a quadratic causes the line to almost magically snap onto the points and predict the pattern. I will need to address this in class, to try to get them to lift their heads up out of the details of the mathematics and up to where they can see the patterns. To see that math is ridiculously good at modelling nature. For that is a core theme in the course.

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