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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.
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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: ->->->->->->->->->->->-> :
V1_avg = 10m / 8 sec = 1.25m/s
Vf = 1.25m/s * 2 = 2.5m/s
A = `dv / `dt
A = 2.5m/s / 8 sec
A = .3125m/s^2
V2_avg = 10m / 5 sec = 2m/s
Vf = 4m/s
A = `dv / `dt
A = 4m/s / 2m/s
A = 2m/s^2
@& 4 m/s / (2 m/s) = 2, not 2 m/s^2.
You don't get acceleration by dividing one velocity by another.
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The average rate the car’s acceleration changes with respect to slope is (2m/s^2 - .3125m/s^2) / (.05) = 33.75 m/s^ per unit slope change
@& Had your second acceleration been correct, this would be correct. You have interpreted the rate of change correctly; most students at this stage do not. So you're doing well here, despite the glitch on the preceding calculation.*@
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5 minutes
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&#Good responses. See my notes and let me know if you have questions. &#