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Phy 231
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 = change in V = (40 - 10)cm/s = 30cm/s
run = change in T = (9 - 4)s = 5 s
slope = rise / run = 'dv / 'dt = change in velocity / change in time = 30 cm/s / 5 s = 6 cm/s/s = 6 cm/s^2
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What is the area of the graph beneath this segment?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
Area = (Average Height) * Length = ((30 cm/s + 0 cm/s) / 2) * 5 s = 75 cm^2
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When you multiply cm/s by s you don't get cm^2.
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10 mins
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For the Area I used the parallelogram formula. I found the average rise being in cm/s, but the run was in s. So my result was just in cm units.
Did i miss a step in my evaluation?
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Your work was fine. You didn't miss any steps, but your area units as originally given were not correct. You are correct on your last statement above, so I assume the former was just a typo.
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&#This looks very good. Let me know if you have any questions. &#