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Phy 231
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: ->->->->->->->->->->->-> :
Trial 1:
'ds = 10m
'dt = 8s
m = .05
v0 = 0
vave = 10/8 - 1.25 m/s
1.25 = (vf + 0) /2
vf - 2.5 m/s
aave - 2.5 /8 = .3125m/s^2
Trial 2:
'ds = 10
'dt = 5
m = .1
v0 = 0 m/s
vave = 10/5 = 2m/s
2 m/s = (vf + 0) /2 = 4 m/s
aave = 4/5 = .8 m/s^2
Average rate of acceleration change:
(.8 - .3125)/(.1-.05) = 9.75 m/s^2
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Good.
You might think about why we would in fact expect the slope to be equal to the acceleration of gravity, assuming an ideal rolling object (no friction, tiny wheels so angular kinetic energy is minimized).
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