Phy 231
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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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 equation for finding a midpoint is ( (x1+x2)/2, (y1+y2)/2 ) Therefore, the midpoint for the given values would be (5 sec + 13 sec)/2 = 9 sec for the x value, and (16 cm/s + 40 cm/s)/2 = 28 cm/s for the y value. The clock time at the midpoint of the interval is 9 seconds.
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• What is the velocity at the midpoint of this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
The equation for finding a midpoint is ( (x1+x2)/2, (y1+y2)/2 ) Therefore, the midpoint for the given values would be (5 sec + 13 sec)/2 = 9 sec for the x value, and (16 cm/s + 40 cm/s)/2 = 28 cm/s for the y value. The velocity at the midpoint comes out to be 28 cm/s.
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• How far do you think the object travels during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
To find how far the object travels during this interval, I see that it travels 9 sec – 5 sec = 4 sec total. The velocity changes from 28 cm/s – 16 cm/s = 12 cm/s also. We can find the rate at which it travels by doing 12 cm/s / 4 sec which amounts to 3 cm/sec/sec. After finding this, the total distance can be shown by taking the first velocity which is 16 cm/s and add 3 cm/sec/sec to that value until we get 28 cm/s then sum the numbers up. 16 cm + 19 cm + 22 cm + 25 cm + 28 cm = 110 cm total.
Your thinking isn't bad. However among other things the object travels for 8 seconds and ends up moving at 40 cm/s. The time interval runs from the initial instant to the final instant, not to the midpoint instant.
See the link at the end of this document for more.
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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 change is just found by finding the difference. 9 sec – 5 sec = 4 sec. The clock time changes by 4 sec.
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• By how much does velocity change during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
The velocity ends at 28 cm/s and starts at 16 cm/s. To find the change I just need to take the difference. 28 cm/s - 16 cm/s = 12 cm/s. The velocity changes 12 cm/s during the time 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):
To find how far the object travels during this interval, I see that it travels 9 sec – 5 sec = 4 sec total. The velocity changes from 28 cm/s – 16 cm/s = 12 cm/s also. We can find the rate at which it travels by doing 12 cm/s / 4 sec which amounts to 3 cm/sec/sec.
Once more you need to consider these values at the initial and final instants, not the midpoint instant. However you made the same error in calculating both the numerator and denominator quantities, and as a result your final answer is correct.
You will understand when you check the previously mentioned link.
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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 for the midpoint is 28 cm/s – 16 cm/s which amounts to 12 cm/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 for the midpoint is 9 sec – 5 sec which amounts to 4 sec.
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• What is the slope of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
The equation of the slope is rise/run. In this case, the rise is 12 cm/s and the run is 4 sec which makes the slope 12cm/s / 4 sec = 3 cm/sec/sec
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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 me that the slope is positive and increasing at a constant rate. The slope also happens to be the same as the rate of change in velocity.
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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 average rate of change of the objects velocity with respect to clock time is 3 cm/sec/sec.
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38 mins
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Good work overall, but see my notes and be sure to compare your work with that in the discussion at the link. Revision is not necessary unless you have questions.
&#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. &#