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Phy 231
Your 'cq_1_07.2' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** CQ_1_07.2_labelMessages **
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: ->->->->->->->->->->->-> sion:
First incline:
vAve=10m/8s=1.25m/s
1.25m/s=(v0+vf)/2
vf=2.5m/s
a=`dv/`dt
a=2.5m/s/8s
a=.3125m/s^2
slope=.05
Second incline:
vAve=10m/5s=2m/s
2m/s=(v0+vf)/2
vf=4m/s
a=`dv/`dt
a=4m/s/5s
a=.8m/s^2
slope=.10
Change in acceleration with respect to slope = (.8m/s^2-.3125m/s^2)/(.10-.05)
=(.4875m/s^2)/(.05)
Acceleration is changing by about .49m/s^2 for every +.05 change in slope.
I can divide that out for a result of 9.75m/s^2 for every +1 change in slope. But I'm not sure what kind of unit I should have here. Technically slope would be in units of meters rise per meters run, or just meters per meter, it seems. So do they cancel each other out, leaving us with no unit at all?
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5 minutes
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Very good.
Slope is unitless, so the units of the result are just m/s^2.
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A little more information:
For an object on a frictionless incline, moving at speeds where air resistance is negligible, the slope of such a graph would be equal to the acceleration of gravity.
For an automobile there would be about a 2% friction loss, so the slope would be about 9.6 m/s^2.
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