Phy 201
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):
I completed the sketch
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• Sketch a straight line segment between these points.
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
I sketched the straight line
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• What are the rise, run and slope of this segment?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
My rise is 40cm/s – 10cm/s = 30cm/s
My run is 9s – 4s = 5 s
My slope is rise/run (30cm/s)/5s = 6cm/s^2
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• What is the area of the graph beneath this segment?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
I first must calculate the area of the trapezoid from 10cm/s to 40cm/s. I get that my average height of this is 30cm/s/2 = 15cm/s. This multiplied by our 5 seconds of change gives us an area of 75cm/s^2 directly under our trapezoid, however we didn’t start at 0 so we have to calculate the area from 0cm to 10cm. Since our height is 10cm/s and our time is still going to be 5 seconds we get 50cm/s^2 of area. This added to our first measurement gives us a total of 125 meters/second^2 of area.
When you multiply cm/s by s you get cm, not cm/s^2.
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20 Minutes
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Very good work, but be sure you undestand the algebra of the units.
&#Please compare your 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. &#