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Phy 241
Your 'cq_1_02.2' 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_02.2_labelMessages **
The problem:
A graph is constructed representing velocity vs. clock time for the interval between clock times t = 5 seconds and t = 13 seconds.
The graph consists of a straight line from the point (5 sec, 16 cm/s) to the point (13 sec, 40 cm/s).
What is the clock time at the midpoint of this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
(t2 + t1) / 2 = 9 seconds
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What is the velocity at the midpoint of this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
(v2 + v1) / 2 = (40 cm/s + 16cm/s) / 2
= (56cm/s) /2
= 28 cm/s
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How far do you think the object travels during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
28 cm/s * 9 s = 252 cm
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By how much does the clock time change during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
'dt = 13 s - 5 s = 8 s
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By how much does velocity change during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
'dv = 40 cm/s - 16 cm/s = 24 cm/s
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What is the average rate of change of velocity with respect to clock time on this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
average rate of change of velocity with respect to clock time
= 'dv / 'dt
= (24 cm/s) / 8s
= 3 cm/s^2
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What is the rise of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
y_2 - y_1 = 40 - 16 = +24
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What is the run of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
x_2 - x_1 = 13 -5 = +8
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What is the slope of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
m = slope
= rise/run
= 24/8
= 3
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What does the slope of the graph tell you about the motion of the object during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
the slope of the graph is equal to acceleration and is positive
now, if the graph was distance vs time, then the slope would be velocity
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What is the average rate of change of the object's velocity with respect to clock time during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
Its the same as the question before
a = 'dv / 'dt
= (24 cm/s) / 8 s
= 3 cm/s^2
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Note that the axes of the graph have units, so the coordinates of the graph points have units. It follows that rise, run, slope and area all have units, which should be included in any calculations of these quantities.
As you'll understand, those units are consistent with the units you've obtained for the various quantities.
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