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:
Answer: ave vel1 = 10m/8sec = 1.25m/sec, vf = 2.5m/sec, dv = 2.5m/s, acceleration1 = 2.5m/s / 8s = .31m/s/s. ave vel2 = 10m/5s = 2m/s, vf = 4m/s, dv = 4m/s, acceleration2 = 4m/s / 5 s = .8m/s/s.
Your accelerations are calculated correctly.
ave acceleration of both runs = (.31m/s/s + .8m/s/s)/(2) = .6m/s/s.
Average slope = (.10 + .05)/2 = .08. Average rate of the automobile's acceleration changing with respect to the slope of the incline is .6m/s/s / .08 = 7.5.m/s/s.
Dividing the average of your accelerations by the average of your ramp slopes does not give the average rate of change of acceleration with respect to ramp slope.
What is the definition of the average rate of change of A with respect to B?
How does this definition applied to the question asked here?
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15 minutes
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You did not calculate the final result correctly, though your accelerations are done correctly.
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