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

In this Friday exploration I sought to demonstrate that the acceleration calculated from d=½at² produces the velocity predicted by v=at. From the first equation a = 2d÷t². That acceleration a can then be used to produce the predicted velocity from v=at. The demonstration was physically based on laboratory four, thus students were already familiar with using speed traps at the bottom of a slope to obtain a velocity.


Class began in A101 where I explained the layout, theory, and times that would be needed. I asked whether the velocity predicted from the acceleration would be the same as the speed trap velocity. One student ventured the guess that they would not be the same, and I agreed. I felt that the two values would be far apart. The class then moved out onto the sidewalk. 


The 17.17 meter slope. Two runs from the post just below the sidewalk junction produced times of 12.03 and 12.12 seconds. 


The three meter speed trap is just after the turn. This had times of 1.13, 1.16, and 1.18 seconds.  Up for debate is whether a shorter speed trap would produce a more accurate velocity as the RipStik is slowing down after the turn. The trade-off would be less accuracy for a faster split time.


I have no selfie or TikTok experts in the course, so this was as good an image as I got of this demonstration. The class then returned to the classroom.

acceleration a = 2 × 17.15 ÷ 12.07² = 0.24 m/s²

That same acceleration predicts a velocity v = 0.24 × 12.07 = 2.85 m/s.  

The speed trap measured a velocity v = 3 ÷ 1.16 = 2.57 m/s. The difference in those is just under 10%, a difference fairly readily explained by frictional slowing of the RipStik via a demonstration in the classroom. 

This exercise now presages the Monday "change in momentum is force" demonstration. Might gut sense is to reduce the force as much as possible to avoid fast back pedaling which is always problematic. 



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