Acceleration from d=½at² produces velocity in v=at

An initial trial was done before class using the nine meter downhill by the LRC and a three meter speed trap. This failed rather miserably in showing that v = 2d÷t, which is the result of solving d=½at² for the acceleration and then substituting that into the equation for the velocity v = at. The failure appeared to in part due to my acceleration not being constant.


I started the class in A101 with the integration on the left side of the board above leading to the equation that the velocity at the bottom of a slope should be 2d₁÷t₁.

The slope on which the acceleration would occur was termed d₁ and the time to traverse d₁ was referred to as t₁. Thus d₁ = ½at₁². That acceleration should produce a velocity v at the bottom of d₁ as a result of v=at₁. Call this predicted velocity v₁. Today d₁ was 17.2 meters with the assistance of Ashmiranda sitting on the tape measure. This would turn out to help immensely.


At the bottom of slope d₁ is a speed trap to determine my actual speed on the RipStik at the bottom of the slope. The speed trap had a length d₂ of three meters. A time t₂ would be used to obtain the speed trap velocity v₂ = d₂÷t₂.


Two timers put my descent time t₁ at 11.31 and 11.41 seconds respectively. Here I am about to enter the speed trap, Wayne providing a time of 11.31 seconds. Note that the speed trap line is angled so as to not lose speed as I corner onto the flat. I would later run my calculation on the 11.41 second time.


I crossed the three meter d₂ speed trap in 1.00 or 1.06 seconds according to two different timers. I brought stopwatches enough for many timers. 


I had a Desmos spreadsheet prepared so I could get a quick glimpse at the numbers. 

Desmos

When I ran the predicted velocity from 2d₁÷t₁ I obtained 3.01 m/s. When I heard that Ian had a time of 1.00 seconds for t₂ I opted to use that time for the actual velocity calculation of d₂÷t₂. That was just too tempting: to have obtained accidental 


Back in the classroom I plugged in the numbers and then tried to explain that my acceleration on the slope of 0.26 m/s² explained my actual velocity at the bottom of the slope, that 2d÷t yielded the exact velocity I actually had at the bottom of the slope, and this was possible because the acceleration equations held true. 


Another view of the angles lines for the three meter speed trap. They were as parallel and I could make them.  

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