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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):
I have sketched a graph of v vs. t and plotted the two given points (4 sec, 10 cm/s) and (9 sec, 40 cm/s).
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• Sketch a straight line segment between these points.
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
I have draw a straight line between the two plotted points.
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• What are the rise, run and slope of this segment?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
The rise of this graph is the vertical difference in the graph between the two points. The distance between the two points vertically is 40 cm/s - 10 cm/s = 30 cm/s. So the rise is 30 cm/s. This amount also represent the change in velocity for the graph.
The run of this graph is the horizontal difference between the two points on the graph. The distance between the two points horizontally is 9 sec - 4 sec. = 5 seconds. So the run is 5 sec. This number represents the change in clock time for the graph.
The slope of this line segment can be found by doing rise/run. In this problem we would have 30 cm/s / 5 seconds = 6 cm/s/s. So the slope of this line segment in 6 cm/s/s. This number represents the average rate of change of velocity in relation to time.
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• What is the area of the graph beneath this segment?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
To find the area of the graph beneath the segment we treat it like a trapezoid. To find the area of a trapezoid we must find the average of the heights and multiply that number by the width. In our problem, the heights are 10 cm/s and 40 cm/s. So to get the average we do (10cm/s + 40 cm/s) / 2 = 25 cm/s. Now that we have our average height we multiply it by our width, 5 seconds, and get 25 cm/s * 5 seconds = 125 cm. So our area underneath the graph is 125 cm, which happens to represent the position change of the object.
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&#Very good responses. Let me know if you have questions. &#