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phy 121
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):
9 seconds. Halfway between t= 13 and t =5 is t =9 this can be easily seen by plotting the graph.
Or you can reason that 13 - 5 = 8 seconds. So there is a total of 8 seconds between the two points. half-way would be 4 seconds. t_1 + 4 seconds = 9 seconds. Likewise, t_2 - 4 seconds = 9 sec
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Good. Also (13 s + 5 s) / 2 = 9 s. It's worth understanding why all three ways of calculating the midpoint clock time are equivalent.
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What is the velocity at the midpoint of this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
the acceleration = (40 cm/sec - 16cm/sec_ / (13 s - 5 s) = 3 cm/s^2
So every second, the velocity is increasing 3 cm/sec. If we start at 16 cm/sec and increase 3 cm/sec for 4 seconds, which is the difference in time to the midpoint, we get 28cm/sec at time = 9 seconds.
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How far do you think the object travels during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
distance is the area under the curve = 'dt * 1/2 (V0 + Vf) which gives (13 sec - 5 sec) * 1/2(40 cm/sec - 16 cm/sec) = 96 cm
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'dt * 1/2 (V0 + Vf) is correct. However your calculation (13 sec - 5 sec) * 1/2(40 cm/sec - 16 cm/sec) is equivalent to `dt * 1/2 (vf - v0), which is not correct.
Average velocity is 28 cm/s, which is what you would get from (vf + v0) / 2. Multiplying this by the 8 second time interval gives you 224 cm.
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By how much does the clock time change during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
'dt = 13-5 = 8 seconds
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By how much does velocity change during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
change in velocity = 40cm/sec - 16cm/sec = 24cm/sec
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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):
avg rate of change = change in velocity / change in time
24 cm/sec / 8 seconds = 3 cm/sec/sec = 3 cm/sec^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/sec - 16 cm/sec = 24 cm/sec
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What is the run of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
run = 13 seconds - 5 seconds = 8 seconds
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What is the slope of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
slope = rise /run = 24 cm/sec / 8 sec = 3 cm/sec^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):
the slope tells us that the object is increasing its velocity at a constant rate. its graph is linear
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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):
24 cm/sec / 8 sec / 3cm/sec^2
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It seems clear that you mean
24 cm/sec / 8 sec = 3cm/sec^2
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Good thinking throughout. See my notes about a couple of simple errors you'll want to avoid.
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