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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.
** CQ_1_02.2_labelMessages **
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):
'dt / 2 = (13 - 5) / 2 = 8 s / 2 = 4 sec
The clock time at the midpoint of this interval would be 9 sec.
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What is the velocity at the midpoint of this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
I would venture to guess it would be somewhere around 28 cm/s.
'dv = 40 cm/s - 16 cm/s = 24 cm/s
There are 8 seconds between these two data points, so divide 24 cm/s by 8 data points = 3 cm/s / data point.
The midpoint between these two coordinates is 4 seconds (data points). so 4 data points * 3 cm/s / data point = 12 cm /s.
Add 12 cm/s to the 16 cm/s velocity at t=5 to get 28 cm/s at t = 9 sec to get the coordinates (9 sec, 24 cm/s)
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How far do you think the object travels during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
a = 'dv / 'dt = (28 cm/s - 16 cm/s) / 4 s = 12 cm/s / 4 s = 3 cm/s^2
'ds = v0 'dt + 0.5 a 'dt^2
'ds = 16 cm/s(4 s) + 0.5(3 cm/s^2)(4 s)^2
'ds = 64 cm + 24 cm = 88 cm
This is obviously wrong because my vAve is only 22 cm/s
vAve = 88 cm / 4 s = 22 cm/s
if its the interval between t1 and t2, then
a = 'dv / 'dt = (40 cm/s - 16 cm/s) / 8 s = 3 cm/s^2
'ds = 16 cm/s(8 s) + 0.5(3 cm/s^2)(8 s)^2 = 128 cm + 96 cm = 224 cm
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The interval is from the first clock time to the second. The average velocity is 28 cm/s and the time interval is 8 seconds. This gives you 224 cm.
You don't need to formula `ds = v0 `dt + 1/2 a `dt^2, and you should be able to reason this out without the formula, as I believe you are able to do.
Once you know how to reason this out, then it's beneficial to also know the formula, so you're probably a step ahead at this point.
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By how much does the clock time change during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
I am confused, which interval are we talking about? The interval between t1 and midpoint, or the interval between t1 and t2, which is 8 seconds?
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The only interval to which the questions refer is the interval from t = 5 seconds to t = 13 seconds.
You do include a correct solution for that interval.
You also have a solution for the interval from the initial point to the midpoint. That isn't the interval requested in the question, but your solution for that interval is correct.
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By how much does velocity change during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
Sorry, I am confused by this question. Between t1 and t2, the velocity changes by 24 cm/s. Between t1 and midpoint it is estimated to change by 12 cm/s
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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):
a = 'dv / 'dt = (40 cm/s -16 cm/s) / 8 s = 3 cm/s^2
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What is the rise of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
rise = 40 cm/s - 16 cm/s = 24 cm/s
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What is the run of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
Run = 13 s - 5 s = 8 s
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What is the slope of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
slope = 24 cm/s / 8 s = 3 cm/s^2
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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):
It is increasing at a rate of 3 cm/s^2 in the positive direction from left to right.
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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):
3 cm/s^2
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You've actually covered every possibility in your solutions, and have done so correctly.
However only some of your solutions address the question asked in this problem.
Check my notes for clarification.
In any case you're doing very well here.
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