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phy121
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: ->->->->->->->->->->->-> :
for trial 1:
slope of the incline = .05 m
`ds = 10m
V0 = 0
V = `ds/`dt = 10m/8sec = 1.25m/sec
Vf = 2*v = 2.5 m/sec
`dv = 2.5 m/sec
a =`dv/`dt = 2.5m/sec / 8sec = .3125 m/sec^2
for trial 2:
`ds = 10m
Slope of incline = .10 m
`dt = 5sec
v0 = 0
v= `ds/`dt
= 10m/5sec = 2m/sec
vf = 2*2m/sec = 4m/sec
`dv = 4 m/sec
a = 4m/sesc / 5sec
= .8m/sec^2
The average rate of acceleration changing with respect to the slope of the incline is simply
Rate of change = (change in acceleration) / (change in slope)
(.8-.3125)/(.10-.05) = .4875/.05 = 9.75 m/sec^2
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15 minutes
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4:30 12/10
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Very good.
Note that the result is close to the acceleration of gravity. For small inclines, under certain conditions on the object descending the incline, this rate of change provides a measure of the acceleration of gravity.
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