Your 'cq_1_7.2' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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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:
The average rate of change of the automobile's acceleration is 9.75m/s.
I calculated the average velocity for each incline, then used the average velocities to calculate the acceleration for each incling. I graphed the acceleration vs. incline and solved for the slope of the graph. Below are my calculations:
vAve = v0 + vf/2 and vf = 2vAve - v0
vAve = `ds/`dt = 10m/8s = 1.25m/s
= = 10m/5s = 2 m/s
a = (vf - v0)/`dt = {[(2*1.25)-0] - 0}/8s = 0.3125m/s^2
= = {[(2*2) - 0] - 0}/5s = 0.8m/s^2
m = rise/run = (0.8m/s^2 - 0.3125m/s^2 / (0.10 - 0.05) = 9.75 m/s^2
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15min.
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Please respond relative to goal of C.
************* This note is added to each submission to ensure that the instructor has reviewed your work. The instructor must manually remove this note. If it hasn't been removed, and if the instructor has not made specific responses, then it is possible that your work hasn't been properly reviewed and you should notify the instructor to be sure.
Your responses have been reviewed and everything looks good.
Please let me know if you have any questions related to this orientation assignment.
Your 'cq_1_7.2' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** **
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:
The average rate of change of the automobile's acceleration is 9.75m/s.
I calculated the average velocity for each incline, then used the average velocities to calculate the acceleration for each incling. I graphed the acceleration vs. incline and solved for the slope of the graph. Below are my calculations:
vAve = v0 + vf/2 and vf = 2vAve - v0
vAve = `ds/`dt = 10m/8s = 1.25m/s
= = 10m/5s = 2 m/s
a = (vf - v0)/`dt = {[(2*1.25)-0] - 0}/8s = 0.3125m/s^2
= = {[(2*2) - 0] - 0}/5s = 0.8m/s^2
m = rise/run = (0.8m/s^2 - 0.3125m/s^2 / (0.10 - 0.05) = 9.75 m/s^2
** **
15min.
** **
Please respond relative to goal of C.
************* This note is added to each submission to ensure that the instructor has reviewed your work. The instructor must manually remove this note. If it hasn't been removed, and if the instructor has not made specific responses, then it is possible that your work hasn't been properly reviewed and you should notify the instructor to be sure.
Your responses have been reviewed and everything looks good.
Please let me know if you have any questions related to this orientation assignment.