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phy 121
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): First find the difference in the time and the velocity.
13 sec- 5 sec= 8 sec
40 cm/sec-16 cm/sec= 24 sec
Then divide both differences by the number of data points (2):
8/2=4 24/2=12
Finally take the 2 numbers that you get and add them to the clock time data points (5+4=9) and (16+12=28)
Therefore the midpoint is (9,28).
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• What is the velocity at the midpoint of this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
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• How far do you think the object travels during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line): 28cm/sec from the previous question the data points are set up as (time,velocity).
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• By how much does the clock time change during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line): From the first data point (5,16) to (13,40) the clock time changed 8 seconds (13-5=8).
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• By how much does velocity change during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line): From the first data point (5,16) to (13,40) the velocity changed 24 cm/sec (40-16=24).
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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): vAve=(24 cm/sec)/( 8 sec)= 3 cm/sec
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• What is the rise of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line): Rise is the change in the y, so it would be 24 cm/sec (40cm/sec-16 cm/sec=24 cm/sec).
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24 cm/s / (8 s) = 3 cm/s^2, or 3 cm/s/s, not 3 cm/s.
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• What is the run of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line): The run is the change in x, so it would be 8 sec (13 sec-5 sec=8 sec).
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• What is the slope of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line): The slope is represented by (rise/run)
change in y/change in x
24/8=3
Slope=3
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Rise has units, run has units, and slope therefore has units.
Units always need to be included when they are present.
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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 is positive, therefore the object's velocity is increasing.
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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): The object changes a total 24 cm/sec over 8 seconds.
vAve = `ds/`dt.
(24cm/sec)/ (8 seconds) = 3cm/sec
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24 cm/s is not `ds.
24 cm/s is `dv, the change in velocity.
You are not asked here to calculate average velocity, but the average rate of change of velocity with respect to clock time.
Just to be sure, I'm going to ask you to do revise this calculation, showing details and being sure to apply the definition of rate of change.
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Good, but see my notes, and please revise that last question per my note.
&#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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