Phy 201
Your 'cq_1_7.2' 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.
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At what average rate is the automobile's acceleration changing with respect to the slope of the incline?
answer/question/discussion: change in A / change in B; the average velocity of both runs are 1.25 m/s (run 1) and 2.0 m/s (run 2); the difference between the two runs equals to .075 m/s^2 if divided out over the length of the run (10 cm). .075 / .05 = 1.5 m /s^2 acceleration change with respect to the change in slopes
The difference in the average velocities would be in m/s, not in m/s^2 (subtracting two quantities with the same units give you a quantity with the same units). The slope is unit less, so there is no way your calculation could represent change in acceleration/change in slope.
However you have the right idea. Calculate the average acceleration on each ramp, then use the same reasoning you used here and you'll get a good result.
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10 minutes
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Almost right but not quite. I don't think you have any trouble.
&#Let me know if you have questions. &#