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:
For the 0.05 slope:
We know the v0 = 0 m/s. We can also determine the vAve from the problem statement, by calculating 'ds / 'dt, which = 10 m / 8 s = 1.2 m/s. Since the auto started from rest (0 m/s), we know that 2 * vAve = vf, therefore, vf = 2 * 1.2 m/s = 2.4 m/s. Now that we know the vf and v0, we can calculate acceleration, which is change in velocity with respect to time. [vf - v0] / 'dt = 2.4 m/s / 8 s = 0.3 m/s/s.
you can't calculate an average rate of change of A with respect to B unless you have a change in A and a change in B. You're only using one slope and one acceleration here, so you don't have the change in either quantity.
For the 0.10 slope:
Again, the v0 = 0 m/s. vAve = 10 m / 5 s = 2 m/s. 2 m/s * 2 = 4 m/s = vf. Now that we know vf and v0, we calculate acceleration: ( 4 m/s - 0 m/s ) / 5 s = 0.8 m/s/s
The automobile on the 0.10 slope is accelerating at 0.8 m/s/s, while the auto on the 0.05 slope is only accelerating at 0.3 m/s/s.
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20 mins
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&#At least part of your solution does not agree with the solution and comments given at the link below. You should view the solution at that link and self-critique as indicated there.
Solution
This link also expands on these topics and alerts you to many of the common errors made by students in the first part of this course. &#