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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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:
Answer: The clock time is at 9 seconds because it is halfway between 5 and 13 seconds.
What is the velocity at the midpoint of this interval?
answer/question/discussion:
Answer: The velocity at 9 seconds is: 40 cm/sec + 16 cm/sec = 56 cm/sec, 56 cm/sec/2 = 28 cm/sec.
How far do you think the object travels during this interval?
answer/question/discussion:
Answer: The object travels 28 cm/sec * 8 sec = 224 cm.
By how much does the clock time change during this interval?
answer/question/discussion:
Answer: 13 sec 5 sec = 8 seconds.
By how much does velocity change during this interval?
answer/question/discussion:
Answer: 40 cm/sec 16 cm/sec = 24 cm/sec.
What is the average rate of change of velocity with respect to clock time on this interval?
answer/question/discussion:
Answer: (24 cm/sec) / (8 sec) = 3 cm/sec.
What is the rise of the graph between these points?
answer/question/discussion:
Answer: 40 cm/sec 16 cm/sec = 24 cm/sec.
What is the run of the graph between these points?
answer/question/discussion:
Answer: 13 sec 5 sec = 8 sec.
What is the slope of the graph between these points?
answer/question/discussion:
slope = rise/run = (24 cm/sec) / (8 sec) = 3 cm/sec. I am not sure these correct, why isnt it cm/sec^2? Ill post this question later in the form.
What does the slope of the graph tell you about the motion of the object during this interval?
answer/question/discussion:
Answer: For each second, on average the object is moving 3 cm.
What is the average rate of change of the object's velocity with respect to clock time during this interval?
answer/question/discussion:
Answer: This question was answered in a previous question but the slope of the line and the average rate of change of the objects velocity with respect to clock time are equal. They are both 3 cm/sec.
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20 minutes
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For the calculation for the slope and average rate of change of velocity with respect to time, the calculation is: 24 cm/sec) / (8 sec) = 3 cm/sec. I don't understand why the units wouldn't be cm/sec^2. If the units cm/sec are correct, why are they correct?
I believe the units you gave on your original solution were cm/s, not cm/s^2. The correct units are cm/s^2.
Be sure to mark your insertions as requested, so I can tell what has been newly inserted and what was part of the original document.