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Phy 231
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):
t=9sec
midpoint clock time calculated = (t2 + t1) / 2 = (13 + 5) sec / 2 = 9 sec
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What is the velocity at the midpoint of this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
28 cm/s
midpoint velocity calculated by = (vf + vi) /2 = (40 + 16)cm/s / 2 = 28 cm/sec
discussion. I tried to use the point slope formula y = mx + b, but could not figure out the y intercept or b.
m = rise / run = (40 - 16) cm /s / ( 13 -5) s = 3 cm /s^2.
From here I plugged in the 9 for x, y = 3cm /s^2* 9 sec + b = 27 cm /sec + b.
could this mean my yintercept would be at 1 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):
192 cm
distance calculated by:
'dv = 'ds /'dt
'ds = 'dv * 'dt = (40 - 16) cm/s * (13 -5)s = 192 cm
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This is not correct.
The change in its velocity is not related to how far an object travels.
*@
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By how much does the clock time change during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
'dt = t2 - t1 = 13s - 5s = 8 sec
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By how much does velocity change during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
'dv = v2 - v1 = 40 cm/s - 16 cm/s = 24 cm/sec
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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 = 'dv / 'dt = (24 cm/s) / 8 s = (3 cm/s)/s
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What is the rise of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
rise = 'daltitude = 40 cm/s - 16 cm/s = 24 cm /s
rise = +24
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What is the run of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
run = 'dtime = t2 - t1 = 13s - 5s = 8s
run = + 8 seconds
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What is the slope of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
slope = rise / run = (24 cm/s) / 8 s = (3 cm /s) /s
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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 tells me that the velocity is increasing at an increasing rate. Its speeding up.
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The velocity is increasing, but it might well be increasing at a constant or even a decreasing rate. All you know from the given information is the rate at which the velocity is increasing.
The position of the object is in fact increasing at an increasing rate.
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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):
average rate of change of velocity to clock time = 'dv / 'dt = (24 cm/s) / (8 s) = (3 cm /s) /s
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*#&!
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Very good overall, but you've repeated your error from the preceding document when finding the displacement of the object. This needs to be revised.
See also my note on the nature of the rate at which the velocity changes; however no revision is required there.
&#Please see my notes and submit a copy of this document with revisions, comments and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes. &#
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