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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: ->->->->->->->->->->->-> (start in the next line):
Velocity is vertical axis, time represents horizontal
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Sketch a straight line segment between these points.
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
ok
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What are the rise, run and slope of this segment?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
rise= 40-10 cm/s = 30cm/s
run= 9sec-4sec = 5sec
slope= 30cm/s / 5sec = 6cm/s^2
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What is the area of the graph beneath this segment?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
A= (30cm/s^2 * 5s) / 2 = (150cm/s^2)/2 = 75cm/s^2
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30 cm/s^2 isn't a quantity associated with this situation.
Your graph consists of a trapezoid with 'graph altitudes' 10 cm/s and 40 cm/s, and width 5 s.
The area of this trapezoid is equal to that of a rectangle of the same width whose 'graph altitude' is the average of the two altitudes.
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Good answers on rise, run and slope.
You didn't get the area correctly, so I'm going to ask you to revise that
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