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?
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The midpoint during this time interval would be 9 seconds. I get this by finding the median between 5 sec and 13 seconds. (13+5)/2= 9.
* What is the velocity at the midpoint of this interval?
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I found the velocity to be 28cm/second. I first found the average velocity of the object to be 3cm/sec / second. I found this average velocity by dividing the difference of the velocities and times between the 5 and 13 second time intervals. Knowing there's a 4 second difference between the starting point and the midpoint, I know that by multiplying 4 sec by 3cm/sec / second, there will be and increase in velocity of 12cm/second. So 16cm/sec + that 12cm/sec = a midpoint velocity of 28cm/sec.
* How far do you think the object travels during this interval?
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I think that the object would travel about 24cm during the 8 second spam. I get this by multiplying both 5sec and 13sec by the average velocity of 3cm/sec. I then get 15cm and 39cm respectively. Subtracting the two distances should give me the distance traveled between the time interval, 24 cm.
* By how much does the clock time change during this interval?
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THere's a 8 second clock time change between this interval. I get this by subtracting 13sec by 5sec to get 8 seconds.
* By how much does velocity change during this interval?
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The velocity appears to change +24cm/second. Using the information given for the graph, the velocity is 16cm/sec at the 5 second interval and 40cm/sec at the 13 second interval. So 40 - 16 = 24cm/second.
* What is the average rate of change of velocity with respect to clock time on this interval?
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I get an average rate of change of velocity of 3cm/second / second. I get this by finding the difference the velocities and times. 40cm/sec - 16cm/sec = 24cm/sec. 13seconds - 5seconds = 8 seconds. 24cm/sec / 8seconds = 3cm/sec / second.
* What is the rise of the graph between these points?
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It is a rise of 24 points, or 24cm/sec. I get this by simply subtracting the y-axis points of 40 and 16.
* What is the run of the graph between these points?
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The run is 8 points, or 8 seconds. I get this by subtracting the x-axis points of 13 and 5.
* What is the slope of the graph between these points?
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If the information I gave you above is correct, the slope should be 3, or 3/1. I get this by dividing the rise over run, which is 24 over 8. Reduced this comes out to be 3/1.
* What does the slope of the graph tell you about the motion of the object during this interval?
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The slope tells me, along with the information given that the graph yields a straight line, that this object is moving at a fairly consistent, positive rate.
* What is the average rate of change of the object's velocity with respect to clock time during this interval?
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This question confuses me because I feel as though I answered it just a few questions before. The only difference between the two questions is the first one asked for the average rate of change of the velocity, and this asks to the average rate of change of object's velocity. If these two are the same question, then the average rate of change of the object's velocity during the time interval should be 3cm/sec / second. I set this up and solved it just as I did in the question before.
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This assignment took me about 35 minutes to complete.
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Most of your answers are good. However you did get off track on one part.
&#Please compare your solutions with the expanded discussion at the link
Solution
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