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Phy 121
Your 'cq_1_04.1' 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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The problem:
A ball is moving at 10 cm/s when clock time is 4 seconds, and at 40 cm/s when clock time is 9 seconds.
• Sketch a v vs. t graph and represent these two events by the points (4 sec, 10 cm/s) and (9 s, 40 cm/s).
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
Velocity goes on the y axis, time goes on the x axis. The x axis is numbered one to ten. The y axis, from the bottom, starts with 10 cm, 20cm, 30cm, and 40cm. Find 4 sec. on the x axis, then find 10 cm on the y axis. Mark that point, do the same with 9 sec (on x axis) and 40 cm (on y axis). Connect the dots, then shade the area.
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• Sketch a straight line segment between these points.
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
Draw a line from 4, 10 to 9,40
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• What are the rise, run and slope of this segment?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
Rise/run= 40-10/9-4= rise= 30 run=5
Slope=30/5= 6
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• What is the area of the graph beneath this segment?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
area= height* width
ave. height= 20, ave width=5, area= 20*5= 100
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Good reasoning, but be careful.
20 is not halfway between 10 and 40.
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You are generally doing well here, but you need to use units with all calculations.
&#See any notes I might have inserted into your document, and before looking at the link below see if you can modify your solutions. If there are no notes, this does not mean that your solution is completely correct.
Then please compare your old and new solutions with the expanded discussion at the link
Solution
Self-critique your solutions, if this is necessary, according to the usual criteria. Insert any revisions, questions, etc. into a copy of this posted document. Mark any insertions with &&&& so they can be easily identified.If your solution is completely consistent with the given solution, you need do nothing further with this problem. &#
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