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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: ->->->->->->->->->->->-> :
v_ave_1=10m/8s = 1.25 m/s so ‘dv = 2.50 m/s a_1=2.50 m/s / 8 s = 0.3125 m/s^2
v_ave_2 = 10m/5s = 2 m/s so ‘dv=4 m/s a_2= 4m/s / 5 s = 0.8 m/s^2
Average rate of changing acceleration with respect to time implies ‘da/change in slope(m) so ‘da/’dm = (a_2 - a_1)/(m_2 - m_1) = (0.8 m/s^2 - 0.3125 m/s^2)/(0.10 - 0.05) = 0.4875 m/s^2 / 0.05 = 9.75 m/s^2
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5 minutes
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