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phy121
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):
to find the midpoint, you would find the difference between the two clock times then divide it by 2 to find the middle.
13s-5s= 8s; 8s/2= 4s would be the midpoint.
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• What is the velocity at the midpoint of this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
40cm/s-16cm/s= 24cm/s; 24/2= 12cm/s
16cm/s+12= 28cm/s would be the velocity at the midpoint of the interval
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• How far do you think the object travels during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
‘ds=vAve*’dt
‘ds= (40cm/s-16cm/s)* (13s-5s)
‘ds= 24cm/s*8s
‘ds= 192cm
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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 changes by 8 seconds during this interval
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• By how much does velocity change during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
velocity changes by 24cm/s during this interval.
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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):
(24cm/s)/8s would be the average change of velocity to respect to clock time during this interval.
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• What is the rise of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
the rise of the graph is 40cm/s-16cm/s= 24cm/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 of the graph is 13s-5s= 8s
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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
slope= (24cm/s)/8s
slope= 3cm/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 of the graph tells us that the object is increasing in velocity in respect to clock time.
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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):
‘ds=vAve*’dt
‘ds= (40cm/s-16cm/s)* (13s-5s)
‘ds= 24cm/s*8s
‘ds= 192cm
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** **
15 minutes
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self-critique #$&*
#$&* self-critique
self-critique rating
`gr91 rating #$&*:
`c022
#$&*
phy121
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):
to find the midpoint, you would find the difference between the two clock times then divide it by 2 to find the middle.
13s-5s= 8s; 8s/2= 4s would be the midpoint.
#$&*
• What is the velocity at the midpoint of this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
40cm/s-16cm/s= 24cm/s; 24/2= 12cm/s
16cm/s+12= 28cm/s would be the velocity at the midpoint of the interval
#$&*
• How far do you think the object travels during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
‘ds=vAve*’dt
‘ds= (40cm/s-16cm/s)* (13s-5s)
‘ds= 24cm/s*8s
‘ds= 192cm
@&
You appear to have multiplied the change in velocity by the time interval.
This doesn't give a meaningless quantity.
For example a car moving for three hours at an unchanging velocity of 60 mph would, by this reasoning, end up going nowhere.
*@
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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 changes by 8 seconds during this interval
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• By how much does velocity change during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
velocity changes by 24cm/s during this interval.
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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):
(24cm/s)/8s would be the average change of velocity to respect to clock time during this interval.
@&
Good.
You should divide out the result, and be very careful about the units.
*@
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• What is the rise of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
the rise of the graph is 40cm/s-16cm/s= 24cm/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 of the graph is 13s-5s= 8s
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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
slope= (24cm/s)/8s
slope= 3cm/s
@&
Good, but you haven't divided the units correctly.
*@
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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 tells us that the object is increasing in velocity in respect to clock time.
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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):
‘ds=vAve*’dt
‘ds= (40cm/s-16cm/s)* (13s-5s)
‘ds= 24cm/s*8s
‘ds= 192cm
@&
40 cm/s - 16 cm/s isn't the average velocity, it's the change in velocity.
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You've fallen into a couple of the common errors, especially the one that fails to distinguish between average velocity and change in velocity. Very easy to do this and you have to be careful.
&#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. &#
*@