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: To find midpoint of clock time you simply find the mean of the clock times. To do this you would add 13seconds to 5 seconds and divide by 2. This gives a mid-point of 9 seconds.
• What is the velocity at the midpoint of this interval?
answer/question/discussion: The velocity at the mid-point would be the mean values of the velocity. In this case, it would be (16cm/s +40cm/sec)/2 which is equal to 28 cm/sec
• How far do you think the object travels during this interval?
answer/question/discussion: You can find estimated distance by multiplying the average speed (or the speed at the midpoint) by the total number of seconds. The object was moving for a total of 8 seconds with an average velocity of 28 cm/sec. Multiplying 28cm/sec by 8 sec gives 224cm.
• By how much does the clock time change during this interval?
answer/question/discussion: The change in clock time is found by subtracting 5 seconds from 13 seconds. This would give 8 seconds.
• By how much does velocity change during this interval?
answer/question/discussion: The initial velocity in this system was 16cm/sec and the final velocity was 40cm/sec. You can find the change in velocity by subtracting 16cm/sec from 40 cm/sec which gives a change in velocity of 24cm/sec.
• What is the average rate of change of velocity with respect to clock time on this interval?
answer/question/discussion: This would be found by taking the change in velocity over the change in time. The expression is set up as (24 cm/sec)/8 sec which gives 3cm/sec/sec
• What is the rise of the graph between these points?
answer/question/discussion: rise in this graph would correspond to change in velocity, so the rise in the graph would be 24 cm/sec
• What is the run of the graph between these points?
answer/question/discussion: run would correspond to the change in time. This would make the run 8 seconds.
• What is the slope of the graph between these points?
answer/question/discussion: The slope of the graph is the same as the average rate of change of velocity with respect to time. In this case it is 3cm/sec/sec.
• What does the slope of the graph tell you about the motion of the object during this interval?
answer/question/discussion: The slope of 3cm/sec/sec tells you that the object is speeding up during this interval by 3 cm/sec each second. This means that the object is accelerating.
• What is the average rate of change of the object's velocity with respect to clock time during this interval?
answer/question/discussion: This would be found by taking the change in velocity over the change in time. The expression is set up as (24 cm/sec)/8 sec which gives 3cm/sec/sec
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15 mins
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&#Good responses. Let me know if you have questions. &#