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:
· Sketch a straight line segment between these points.
answer/question/discussion: This line is a rather steep straight line.
· What are the rise, run and slope of this segment?
answer/question/discussion:
Rise = 40cm/s - 10cm/s = 30cm/s
Run = 9s - 4s = 5s
Slope = rise/run = 30cm/s / 5s = 6cm/s/s
· What is the area of the graph beneath this segment?
answer/question/discussion: The area beneath the segment is found by multiplying the average heights by the width of the trapazoid = 15cm * 5s = 75cm/s.
The 'altitidues' of the trapezoids are 10 cm/s and 40 cm/s. The midpoint altitude, which since the graph is a straight line segment, is equal to the average of these 'altitudes' and is 25 cm/s. So the area would be 25 cm/s * 5 s = 125 cm/s.
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15 min
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&#Good work. See my notes and let me know if you have questions. &#