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Phy 231
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An automobile rolls 10 meters down a constant incline with slope .05 in 8 seconds, starting from rest. The same automobile requires 5 seconds to roll the same distance down an incline with slope .10, again starting from rest.
• At what average rate is the automobile's acceleration changing with respect to the slope of the incline?
answer/question/discussion: ->->->->->->->->->->->-> :
(8s + 5s)/2= 6.5s
(.10m/s + .05m/s)/2= .075m/s
10m/6.5s= 1.54m/s= vAve
2* 1.54m/s= 3.08m/s= vf
aAve= (3.08m/s – 0m/s)/6.5s= .474m/s^2
the average aAve of both gives .474m/s^2? I’m not sure about the way that I came to this conclusion. But if its right then the acceleration is changing at a rate of (.474m/s^2)/(.075m/s)= 6.32m^2/s^3? Not sure about this one.
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@& Acceleration is inversely proportional to the time interval, which makes it a nonlinear function of the time interval. So while is would be valid to average time intervals for identical trials in order to compensate for measurement uncertaintites, it wouldn't be useful to average time intervals for two different trials under two different conditions.
Each trial results in an acceleration, and for each there is an incline slope. You get two accelerations and two incline slopes, and need to use these to answer the given question.*@
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