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phy 231
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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: ->->->->->->->->->->->-> :
Use the equation: `ds = v0*`dt + .5a*`dt
Rearrange to (`ds - v0*`dt) / (.5*dt^2) = a
Plug in the values from each problem:
#1 (10m - 0*8s) / (.5 * 8^2) = .3125 m/s^2
#2 (10m - 0*5s) / (.5 * 5^2) = .8 m/s^2
Change in a / change in slope
(.8 m/s^2 - .3125 m/s^2) / (.10 - .05) = .4875/.05 = 9.75 m/s^2
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20 min
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&#Very good data and responses. Let me know if you have questions. &#