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phy 121
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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: ->->->->->->->->->->->-> :
a = 2`ds / `dt^2
= 2 (10 m) / 8 s^2
= 20 m / 8 s^2
= 2.5 m/s^2
a = 2 (10 m) / (8 s)^2. You have to square `dt, not just the units of `dt.
You get about .3 m/s^2.
aAve = 0.05 m / 10 m
= 0.01 m
.05 m / (10 m) = .005.
m is divided by m so the result is unitless.
However this is not an acceleration.
a = 2 (10 m) / 5s^2
= 20 m / 5 s^2
= 4 m/s^2
aAve = .10m / 10 m
= .10 m
#$&*
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15 min
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You used the right relationship for the accelerations, except that you didn't square the time interval; you only squared the unit of the time interval. This is easily corrected (see my note above for the first calculation of acceleration, and apply the same idea to the second).
You didn't apply the definition of rate of change to answer the overall question.
Do submit a revision on this one.
&#See any notes I might have inserted into your document. If there are no notes, this does not mean that your solution is completely correct.
Then please compare your solutions with the expanded discussion at the link
Solution
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