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: I did this.
• Sketch a straight line segment between these points.
answer/question/discussion: I did this.
• What are the rise, run and slope of this segment?
answer/question/discussion: The rise of this segment is 40-10= 30 cm/s and the run is 9-4= 5 s and the slope is the rise divided by the run, which is 30 cm/s / 5 s. The slope is then 6 sm/s. I obtained these numbers by viewing the coordinates on the graph.
Good, but 30 cm/s / (5 s) = 6 (cm/s) / s = 6 cm/s * (1/s) = 6 cm/s^2, not 6 cm/s.
• What is the area of the graph beneath this segment?
answer/question/discussion: The area of the graph beneath this segment is representative of the position of the object. To get the number of the area, we simply multiply ½ * time * velocity to get ½ * 10sec * 40cm/sec = 200cm/sec/sec.
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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Approximately 10 minutes was spent on this problem.
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&#Your work looks good. Let me know if you have any questions. &#