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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.
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cq_1_022
#$&*
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):
The clock time is 9 seconds. I drew the graph and found the midpoint. You can also find
the mid point by adding 5 and 13 and dividing the sum by 2, which gets you 18/2 or 9.
&&&&It looks like I may have actually gotten this one correct.&&&&
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What is the velocity at the midpoint of this interval?
answer/question/discussion: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> (start in the next line):
The velocity is 3 cm/sec. I took the difference between this point and the starting
point to determine this (28-16)/(9-5)=14/4 = 3 cm/sec.
&&&&If I'm looking at the graph correctly, it's just a matter of seeing that at 9
seconds, the velocity is 28 cm/sec. I'm not clear on the explantion below, although I
do see that my original calculation does not yield cm/sec. I do understand the cm/s/s
now. It took a while for that to sink in as a concept.&&&&
@&
Your calculation does not lead to units of cm / sec.
(28-16)/(9-5) is therefore not a good calculation, though it's on the right track with
the right numbers in the right places.
(28 cm/s -16 cm/s)/(9 s - 5 s) would be a good calculation, since the change in velocity
28 cm/s - 16 cm/s would take place in 4 seconds.
However the more appropriate calculation would be
(40 cm/s - 16 cm/s) / (13 s - 5 s)
since this applies to the entire interval. You're simply less likely to make an error if
you use the whole interval, not to mention that this calculation involves only given
numbers as opposed to numbers you calculated.
The answer would in any event be the same as for your calculation.
*@
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How far do you think the object travels during this interval?
answer/question/discussion: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> (start in the next line):
It traveled from 16 to 28, which is a difference of 14 cm.
&&&& This time I made the trapezoid on my graph. I ended up with a height of 28 cm/s
and base of 8 s. I multiplied base times height. Because the height is in cm/s and the
base is in seconds, when I multiply them, the seconds cancel and I ended up with cm.
The distance traveled is 224 cm.&&&&
@&
The units of 16 and 28 aren't cm, and those numbers don't represent distances.
*@
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By how much does the clock time change during this interval?
answer/question/discussion: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> (start in the next line):
The clock time changed from 5 to 9 during this interval, or 4 sec.
&&&&The clock time changes from 5 to 13, which is 8. I misinterpreted the question to
go with the on that asked about mid-point.&&&&
@&
That's for only half the given interval.
*@
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By how much does velocity change during this interval?
answer/question/discussion: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> (start in the next line):
The velocity stays at 3 cm/sec during this interval.
&&&&The velocity changes from 16 cm/s to 40 cm/sec or 24 cm/s during the 8 second
interval.&&&&
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@&
The velocity clearly changes during the interval.
The rate of change of the velocity with respect to clock time stays constant, but that
rate does not have the units you give here.
*@
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):
(28-16)/(9-5) = 12/4 = 3 cm/s
&&&&(40 cm/s-16 cm/s)/(13 s-5 sec)=24 cm/sec/8 sec or 3 cm/s/s or 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):
The rise is 3. It goes up 3 cm for each second.
&&&&The rise is 3. It goes up 3 cm/sec for each second.&&&&
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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 1. For each 3 cm gained it takes 1 second.
&&&&The run is 1. For each 3 cm/sec gained it takes 1 second.&&&&
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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 3; rise/run is 3/1 or 3.
&&&&The slope is 3; rise/run is 3/1 or 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 object is moving at a rate of 3 cm/sec during this interval.
&&&& For each 3 cm/sec gained in velocity it took 1 second.&&&&
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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 velocity remains 3 cm/s throughout the interval.
&&&&The rate of change of the object's velocity with respect to clock time is 3
cm/s/s.&&&&
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*#&!
@&
Very good.
It's also clear from the recent work you've submitted that you now understand this.
*@
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