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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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: ->->->->->->->->->->->-> :
9sec would be the time at the midpoint of this above interval.
In order to find the midpoint we use the average of the data by adding the number and then divide them with how many no. are there.
• What is the velocity at the midpoint of this interval?
answer/question/discussion: ->->->->->->->->->->->-> :
28cm/sec would be the velocity at the midpoint of this interval.
In order to find the midpoint we use the average of the data by adding the number and then divide them with how many no. are there.
• How far do you think the object travels during this interval?
answer/question/discussion: ->->->->->->->->->->->-> :
‘dv = vAVR * tv= 28cm/sec* 9sec =252cm
• By how much does the clock time change during this interval?
answer/question/discussion: ->->->->->->->->->->->-> :
tv =’dv /vAVR = 252cm/28cm/sec = 9
the clock time changes from 5 sec to 13 sec, a change of 8 sec, not 9 sec
Can you identify the inconsistency in your solution?
• By how much does velocity change during this interval?
answer/question/discussion: ->->->->->->->->->->->-> :
change in velocity = final velocity – initial velocity
= 40cm/sec- 16cm/sec = 24
• What is the average rate of change of velocity with respect to clock time on this interval?
answer/question/discussion: ->->->->->->->->->->->-> :
average rate of velocity = change in velocity / change in time
=24/8=3cm/sec^2
• What is the rise of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> :
Rise is 40cm/sec
• What is the run of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> :
Run is 13sec
13 seconds is a clock time, which is not a run
to get a 'run' you need two clock times
• What is the slope of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> :
slop = rise/run = 3.0
• What does the slope of the graph tell you about the motion of the object during this interval?
answer/question/discussion: ->->->->->->->->->->->-> :
The slop of the graph tells that the object contently moving.
• What is the average rate of change of the object's velocity with respect to clock time during this interval?
answer/question/discussion: ->->->->->->->->->->->-> :
average rate of velocity = change in velocity / change in time
=24/8=3cm/sec^2
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25 minute.
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Good reasoning overall, but you have a couple of errors.
&#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. &#