Phy201
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?
9 sec because if you add them together to find average that is the midpoint.
What is the velocity at the midpoint of this interval?
28cm/s is the velocity at the midpoint of the line.
How far do you think the object travels during this interval?
224cm because the average of 28cm/s times the difference of 8 sec
By how much does the clock time change during this interval?
8sec is the difference in the clock time.
By how much does velocity change during this interval?
The velocity increases 24cm/s.
What is the average rate of change of velocity with respect to clock time on this interval?
3:1 is the average rate of velocity change with respect to clock time.
The rise of the graph represents the velocity change 24 cm/s and the run represents the 8 second change in clock time. So the slope represents the 3 cm/s^2 average rate of change of velocity with respect to clock time, i.e., the average acceleration.
What is the rise of the graph between these points?
The rise would be 3
What is the run of the graph between these points?
The run would be 1
What is the slope of the graph between these points?
The slope is equal to 3
What does the slope of the graph tell you about the motion of the object during this interval?
The graph shows that the velocity increases at a constant rate.
What is the average rate of change of the object's velocity with respect to clock time during this interval?
3:1 is the average rate of velocity change with respect to clock time.
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20min
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&#Good work. See my notes and let me know if you have questions. &#