Phys 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:
The clock time is :(13-5/2) + 5 = 9secs
• What is the velocity at the midpoint of this interval?
answer/question/discussion:
The velocity is : (40-16/2) + 16 =28cm/s
good; alternatively (40 cm/s + 16 cm/s) / 2 = 28 cm/s.
• How far do you think the object travels during this interval?
answer/question/discussion:
Distance travelled = average velocity * time. 28*9= 252 cm
• By how much does the clock time change during this interval?
answer/question/discussion:
Change in clock time: 13-5= 8secs
• By how much does velocity change during this interval?
answer/question/discussion:
\Velocity changes by: 40-16= 24cm/s
• What is the average rate of change of velocity with respect to clock time on this interval?
answer/question/discussion:
Average rate of change: 24/8 = 3cm/s
• What is the rise of the graph between these points?
answer/question/discussion:
The rise is 24cm/s
• What is the run of the graph between these points?
answer/question/discussion:
The run is 8secs
• What is the slope of the graph between these points?
answer/question/discussion:
The slope is rise/run = 3cm/s
• What does the slope of the graph tell you about the motion of the object during this interval?
answer/question/discussion:
The slope informs me of the average velocity during this interval
slope of v vs. t graph is the ave rate of change of vel, not ave velocity
• What is the average rate of change of the object's velocity with respect to clock time during this interval?
answer/question/discussion:
The average rate of change can be determined by the slope. It is 3cm/s
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15 mins
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&#Your work looks good. See my notes. Let me know if you have any questions. &#