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PHY 201
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: ->->->->->->->->->->->-> :
'ds_1 = 10 m, 'dt_1 = 8 s
'ds_2 = 10 m, 'dt_2 = 5 s
`ds = v0 `dt + .5 a `dt^2
a_1 = ('ds_1/ (0.5 * 'dt_1^2) - v0'dt_1 = (10 m / (0.5 * (8 s)^2)) - 0 m/s*8 s = 0.31 m/s^2
a_2 = ('ds_2/ (0.5 * 'dt_2^2) - v0'dt_2 = (10 m / (0.5 * (5 s)^2)) - 0 m/s*8 s = 0.80 m/s^2
The average rate of change of the automobile's acceleration is equal to change in acceleration divided by the change in slope.
Average rate of change = (0.80 m/s^2 - 0.31 m/s^2) / (0.10 - 0.05) = 9.8 m/s^2
&#Very good responses. Let me know if you have questions. &#