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course PHY 201
cq_1_022#$&*
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: ->->->->->->->->->->->-> (start in the next line):
The clock time at the midpoint of the interval 5 sec and 13 sec is 9 sec. This can be found by finding the middle number between 5 and 13.
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• What is the velocity at the midpoint of this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
The velocity at the midpoint of the two intervals is about 28 cm/s. This can be found by constructing a graph and drawing a line between the two points. Then on the horizontal axis find the number 9 then go up until you reach the line drawn and see what point on the vertical axis it corresponds to.
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• How far do you think the object travels during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
I would guess the object traveled about 30 cm during this time interval. I guessed this number because it is in between how fast the object was going when it started and when it had finished.
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To figure out how far something goes you have to use information about how fast it's moving, and for how long.
How long did this time interval last?
&&&& this time interval lasted 8 s. 13 s - 5 s = 8 s
So would we multiply how fast the object was moving and for how long??? then it would be 28 cm/s * 8 s = 224 cm/s^2 &&&&
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Right. Very good.
And no ambiguous equations to confuse the issue.
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• By how much does the clock time change during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
The clock time changes by 8 seconds from the first point to the second. The clock time increases at a constant rate, a second will always be a second.
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• By how much does velocity change during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
The velocity changes in total 24 cm/s during the interval between the two points.
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• What is the average rate of change of velocity with respect to clock time on this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
The average rate of change of velocity with respect to clock time would be 3 cm/s every second. vAve = 'ds / 'dt = 24 cm/s / 8 cm/s
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Good reasoning, but be careful to group the denominator:
24 cm/s / (8 s).
What is the result of that calculation?
24 cm/s / (8s) = 3cm/s/s
I must have not paid very close attention when I was typing out my response.
Note that this result is not the average velocity. Average velocity is the average rate of change of position with respect to clock time.
What do you call the quantity you are calculating here?
&&& the quantity I am calculating is acceleration I think. &&&&
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• What is the rise of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
The rise of the graph is 24. 40 - 16 = 24
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• What is the run of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
The run of the graph is 8. 13 - 5 = 8
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• What is the slope of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
The slope of the graph is as shown
slope = rise/run = 24/8 = 3/1 =3
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Good, but rise and run have units, which need to be included when you state those quantities, as well as in your calculation of slope.
&&&& Note to self: Use Units! Got it. Thank you. &&&&
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• What does the slope of the graph tell you about the motion of the object during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
The slope of the graph tells about the average acceleration from the two time intervals.
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There is only one time interval involved here. It starts at t = 5 seconds and ends at t = 13 seconds.
The slope is the acceleration, as you correctly say.
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• What is the average rate of change of the object's velocity with respect to clock time during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
The average rate of change of the object's velocity with respect to clock time between these two points is going to also equal the same value as the slope. I think. So the average rate of change would also be 3 cm/s
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Good, but the units of the slope are not cm/s.
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*#&!
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You've done a lot of good things here. You've also made a few errors, and it will be well worth your effort to correct them.
&#Please see my notes and submit a copy of this document with revisions, comments and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes.
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Good revision. Check my note.
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