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Phy 241
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):
I sketched a graph with the x axis from 0 to 9 with 1 s increments and the y axis from 0 to 40 with 10 cm/s increments. I plotted to points on this graph (4,10) and (9,40)
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Sketch a straight line segment between these points.
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
I sketched a straight line segment from (4,10) to (9,40)
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What are the rise, run and slope of this segment?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
rise = 30 cm/s
run = 5 s
slope = rise/run = 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):
y - y_1 = m(x - x_1)
v(t) = 6t - 14
integral from 4 to 9 of 6t -14 = 125 cm/s^2
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Also average 'graph altitude' * width, where 'graph altitudes' are 10 cm/s and 40 cm / s so average 'graph altitude' is 25 cm/s, and width is 5 s.
25 cm/s represents average velocity, 5 s represents time interval, so area is 25 cm/s * 5 s = 125 cm, which in agreement with your integral represents the displacement.
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