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: In the first run the average speed is 10 m / 8 s = 1.25 m/s which is then divided by the time elapsed (1.25 m/s)/ 8 s = 0.15625 m/s^2. This gives us the acceleration with respect to slope as 3.125 m/s^2/ degrees = a / slope = (.15625 m/s^2)/ .05 degrees. In the second run the average speed is 10 m / 5 s = 2 m/s which is then divided by the time elapsed (2 m/s)/ 5 s = .4 m/s^2. This gives us the acceleration with respect to slope as 4 m/^2/ degrees = a / slope = (.4 m/s^2)/.1 degrees.
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10 minutes
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1.25 m/s is not `dv, so when divided by `dt is doesn't give you a.
Slope is .05, not .05 degrees. Slope is rise / run; both quantities are measured in the same units so the result is unitless. The angle is the arctangent of the slope (in this case about 3 degrees), but the question didn't involve angles.
Similar notes apply to the second.
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