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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.
** CQ_1_04.1_labelMessages **
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: ->->->->->->->->->->->-> scussion (start in the next line):
The graph has a y axis labeled velocity in cm/s. The x axis is labeled clock time in seconds. There are two points on the graph
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Sketch a straight line segment between these points.
answer/question/discussion: ->->->->->->->->->->->-> scussion (start in the next line):
The line segment between these two points is straight. It is increasing at a constant rate.
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What are the rise, run and slope of this segment?
answer/question/discussion: ->->->->->->->->->->->-> scussion (start in the next line):
The rise is 40cm/s - 10 cm/s = 30 cm/s
The run is 9s - 4 s = 5 sec
The slope is the rise/run or (30cm/s)/5 s = 6cm/2^2
This slope represents the average acceleration.
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What is the area of the graph beneath this segment?
answer/question/discussion: ->->->->->->->->->->->-> scussion (start in the next line):
The area of the graph is the average altitude * width. We can tell the width because it is the same as our run of 5 sec.
For average altitude we need to know the y coordinate of the midpoint of the line segment. For this we add the two y coordinates and divide by two, thus giving us the average altitude. (40 cm/s + 10 cm/s)/2 = (50cm/s)/2 = 25cm/s
Plug this in and the area = 25cm/s * 5 sec = 125 cm which represents distance traveled in the time interval.
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&#Your work looks very good. Let me know if you have any questions. &#