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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.
** 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).
• Sketch a straight line segment between these points.
• What are the rise, run and slope of this segment?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
rise = ‘dv = (40 cm/s)-(10 cm/s) = 30 cm/s
run = ‘dt = (9s - 4s) = 5 s
slope = rise/run = (30 cm/s)/(5 s) = 6 cm/s/s
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• What is the area of the graph beneath this segment?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
The area will represent the position change of this object for this time interval. To calculate we have to find the average velocity as this represents the average height or altitude of the trapezoid on the graph. Assuming constant acceleration, the average velocity will be 15 cm/s which is half of the ‘dv. The we multiply this by the time for the interval or ‘dt which is 5 s. 15(cm/s) * 5s = 75 cm (the seconds cancel) So the position change from 4 s to 9 s is 75 cm
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&#Very good responses. Let me know if you have questions. &#