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phy 121
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** 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: ->->->->->->->->->->->-> :
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For .05 incline: Vave = 10m / 8 sec = 1.25m/s
a = 2.5m/s / 8sec = 0.3125 m/sec^2
Avg rate of change in acceleration = 0.3125m/sec^2 / .05 = 6.25
For .10 incline= Vave = 10 m / 5sec = 2m/s
a = 4m/s / 5s = .8m/s^2
Avg rate of change = .8m/s^2 / .10 = 8
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5 mins
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The question asked for the average rate of change of acceleration with respect to slope.
In both cases you seem to be dividing the acceleration by the slope, which does not give you an average rate of change.
This problem requires you so start from the definition of rate.
You have all the information you will need; you just need to correctly apply the definition of rate.
Wait until you've completed the required tests, then return to this problem and give it 10 more minutes. It shouldn't take any longer than that; if it does just submit what you have by that time and I'll tell you the rest.
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