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.
At what average rate is the automobile's acceleration changing with respect to the slope of
the incline?
answer/question/discussion:
10m/8s=1.25m/s^2
1.25m/s^2/.05=25
10m/8s=1.25m/s, not 1.25 m/s^2, and this quantity represents the change in clock time for the first trial. It does not represent an acceleration, nor does it represent a rate of change of acceleration with respect to the slope of the incline.
You have two trials. For each one you can calculate an acceleration.
Then you should apply the definition of rate of change to the question. If you do everything step by step you should get the correct answer and the meaning should become apparent. If not, of course I'll help, but it's best if you figure it out independently.
10m/5s=2m/s^2
2m/s^2/.10=20
Your calculation does not give you an acceleration, as noted for a similar calculation on the first ramp.
You do not get the average rate of change of acceleration with respect to incline slope by dividing an acceleration by an incline slope. The definition of rate of change does not apply to a single value of quantity A and a single value of quantity B.
I'm not really for sure what Im trying to get here?
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25 minutes
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Don't spend too much time on a revision, and don't hesitate to ask additional questions, but see if you can work this out based on my notes. You have two tasks here, one is two calculate your accelerations correctly from the given information, and the other is to calculate the requested rate of change.
&#Please see my notes and submit a copy of this document with revisions and/or questions, and mark your insertions with &&&&. &#