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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Seed 7.2
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:
First I am going to find the 2 accelerations.
First hill:
'ds = 10m
slope = .05
'dt = 8 seconds
v0 = 0 m/s
vAve = 10/8
vAve = 1.25 m/s
vF = 2.5 m/s
a = 2.50/8
a = .31 m/s/s
Second hill:
'ds = 10 m
slope = .10
'dt = 5 seconds
v0 = 0 m/s
vAve = 10/5
vAve = 2 m/s
vF = 4 m/s
a = 4/5
a = .8 m/s/s
First hill aAve = .31 m/s/s
Second hill aAve = .8 m/s/s
I am not really sure how to find the average rate of change of acceleration with respect to the slope of the incline. I would assume this number would be the slope of a graph of accel vs. slope of hill.
So I am thinking that I need to divide the change in acceleration by the change in the slope of the hill (rise over run)
That would be:
'da = .23 m/s/s
'dslope = .05
'da/'dslope = 4.6
Good, but the difference in accelerations is about .49 m/s^2, which gives an average rate of change of about 9.8 m/s^2:
The average rate of change of acceleration with respect to ramp slope is
ave rate = change in acceleration / change in ramp slope
= ( .8 m/s^2 - .31 m/s^2) / ( .10 - .05 )
= (.49 m/s^2) / (.05)
= 9.8 m/s^2,
meaning 9.8 m/s^2 per unit of ramp slope.
I am not 100% sure this is right because I do not know what units to put with it. Slope carries no units and acccel has m/s/s. I know that the change cannot be associated with time as the slope has nothing to do with time. So I am going to say that it means the acceleration will change by 4.6 m/s/s per meter of change of slope.
slope isn't measured in meters; you can say 9.8 m/s^2 per unit of ramp slope.
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30 minutes
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Excellent work, but with one apparent arithmetic error. See my notes.