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.
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Seed Question: Asst 2 Question 2
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: ->->->->->->->->->->->-> :
The clock time at the midpoint is 9 sec.
• What is the velocity at the midpoint of this interval?
answer/question/discussion: ->->->->->->->->->->->-> :
The velocity at the midpoint is 28 cm/sec
• How far do you think the object travels during this interval?
answer/question/discussion: ->->->->->->->->->->->-> :
The object travels 252 cm/sec/sec during this interval.
Proof: `dv = Vave * `dT = 28cm/sec * 9 sec = 252 cm/sec/sec
• By how much does the clock time change during this interval?
answer/question/discussion: ->->->->->->->->->->->-> :
The clock time changes 6 secs during this interval.
19 secs – 13 secs = 6 secs
right definition, but 19 sec is not a clock time related to this interval
• By how much does velocity change during this interval?
answer/question/discussion: ->->->->->->->->->->->-> :
The velocity changes 12 cm/sec
28 cm/sec – 16 cm/sec = 12 cm/sec
initial velocity is 16 cm/s but final velocity is not 28 cm/s
• What is the average rate of change of velocity with respect to clock time on this interval?
answer/question/discussion: ->->->->->->->->->->->-> :
The averate rate of change of velocity with respect to clock time on this interval is 2 cm.
Vave = `dv/`dt = 12 cm/sec / 6 sec = 2 cm
the units of your result aren't correct; see my previous note related to `dv
• What is the rise of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> :
The rise of the graph between these points is 12 cm/sec
the 'vertical' coordinate of the graph changes by more than 12 cm/s
• What is the run of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> :
The run of the graph between these points is 6 sec
• What is the slope of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> :
The slope of the graph between these points is 2 cm
• What does the slope of the graph tell you about the motion of the object during this interval?
answer/question/discussion: ->->->->->->->->->->->-> :
The slope of the graph tells that the motion of the object during this interval is increasing.
• What is the average rate of change of the object's velocity with respect to clock time during this interval?
answer/question/discussion: ->->->->->->->->->->->-> :
The average rate of change of the objects velocity with respect to clock time during this interval is 188 cm/sec/sec.
Proof: area= average (height) * width) = (12+9)cm/sec/2 *6sec = 63 cm/sec/sec
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1.5
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Would you check to see if I'm doing the Average rate of change with respect to clock time correctly?
You're using the right definitions and getting many of the correct quantities; any errors should be easy to correct.
&#Please compare your solutions with the expanded discussion at the link
Solution
Self-critique your solutions, if this is necessary, according to the usual criteria. Insert any revisions, questions, etc. into a copy of this posted document. Mark any insertions with &&&& so they can be easily identified. &#