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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.
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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: ->->->->->->->->->->->-> :
#Slope problem 1
`ds = 10 m
slope = .05
`dt = 8 s
v0 = 0
For the first situation, where the car rolls down a constant incline with slope .05, its average acceleration is :
vAve = `ds / `dt
vAve = 10 m / 8 s
vAve = 1.25 m/s
vAve = (v0 + vf) / 2
1.25 m/s = (0 + vf ) / 2
2.5 m/s = vf
# aAve = `dv / `dt
aAve = 2.5 m/s / 8 s
aAve = .3125 m/s/s
For the second situation, where the car rolls down a constant incline of slope .10 in 5 seconds, its average acceleration is:
vAve = `ds / `dt
vAve = 10 m / 5 s
vAve = 2 m/s
vAve = (v0 + vf) / 2
2 m/s = (0 + vf) / 2
4 m/s = vf
# aAve = `dv / `dt
aAve = 4 m/s / 5 s
aAve = .8 m/s/s
Rate = change in acceleration / change in slope
Rate = .4875 m/s/s / .05
Rate = 9.75
?? Not sure of the units to use here or if I performed the correct calculation.
the slope .05 is unitless
your calculation is absolutely the right one, and since the denominator is unitless (you can therefore think of the unit as being just 1), the resulting unit is the unit of the denominator, m/s/s
if this experiment is actually done then the expected result is that, to the extent the kinetic energy of the rotating tires is negligible compared to that of the moving car, the slope of that graph should be very nearly equal to the acceleration of gravity.
#$&*
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20 minutes
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Very well done. See my note about the units, and about the meaning of the result.