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phy 201
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):
(13+5)/2=9
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• What is the velocity at the midpoint of this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
(40+16)/2=28
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• How far do you think the object travels during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
16=distance/5=80cm
40=distance/13=520cm
440cm
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You've calculated the difference in the distances the object would travel at 16 cm/s and at 40 cm/s.
By this reasoning an object which increased its velocity on the same interval from, say, 27 m/s to 29 m/s would travel only 26 meters, even though it has the same average velocity as the first.
Can you correct your reasoning and your result? Don't spend an inordinate amount of time before you simply resubmit and ask for more help, but I think you can resolve this in a few minutes and will profit more by doing so than you would from a more extensive hint. If you go over 10 or 15 minutes, that's enough. At that point just ask for an additional hint or explanation.
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• By how much does the clock time change during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
13sec-5sec= 8sec
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• By how much does velocity change during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
40cm/s-16cm/s=24cm/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):
28/9=3cm
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You don't have units on either your numerator or denominator, and a rate of change of velocity with respect to clock time won't have units of cm.
You need to use the units at every step of the calculation.
You also need to be sure you are correctly applying the definition of rate of change. It appears that you are dividing the average velocity by the midpoint clock time. Other than the question of units, your reasoning is in the right ballpark but doesn't quite fit the definition of average rate.
Again, give this up to 10 minutes and try to resolve it yourself. Then show my your best reasoning and if necessary ask for additional hints/explanations.
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• What is the rise of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
24cm/s
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• What is the run of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
8sec
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• What is the slope of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
24/8=3
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This calculation is correct, except that the units are missing and need to be included. This should take you only a few seconds to correct.
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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):
moving in a positive direction
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The graph shows velocity vs. clock time. If the graph is above the t axis the object is moving in a positive direction. The direction of motion is not related to the slope of this graph.
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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):
3cm
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cm is not a unit of rate of change of velocity with respect to clock time. The units can be reasoned out from the definition of average rate.
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*#&!
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You have a number of fundamenatl errors in this document. If you correct them now it will save you effort in the long run and improve your performance. I know you are pressed for time, but it's worth a reasoable amount of time to clarify these issues.
You are on the right track. You just need to see what details require your further attention.
&#Please see my notes and, unless my notes indicate that revision is optional, submit a copy of this document with revisions 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.
If my notes indicate that revision is optional, use your own judgement as to whether a revision will benefit you. &#
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