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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.
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SEED Question 2.2
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):
(t1 + t2)/ 2 = midpoint
(5 +13) / 2 = 9sec
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What is the velocity at the midpoint of this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
(v1 + v2) / 2 = midpoint velocity
right, but there are no 0 and 3 cm/s velocities involved, so your next line appears to be incorrect.
(0cm/s + 3cm/s) / 2 = 1.5cm/s
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How far do you think the object travels during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
Im guessing that this question refers to the previous question about the midpoint.
(d2 d1) / 2 = midpoint distance
(40cm -16cm) / 2 = 12cm
the velocity is never less than 16 cm/s so it's going to move a lot further than that
40 cm and 16 cm are not quantities associated with this motion. 40 cm/s and 16 cm/s are.
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By how much does the clock time change during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
(t2 t1) / 2 = midpoint time interval
(13sec -5sec) / 2 = 4sec
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By how much does velocity change during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
The ball is traveling at a constant rate so the rate of change stays the same (0)
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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):
(dist 2 dist 1) / total time = Ave velocity
(40cm -16cm) / 8sec = 3cm/sec
40 cm and 16 cm are not given quantities
40 cm/s and 16 cm/s are
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What is the rise of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
(d2 d1) = rise
40cm -16cm = 24cm
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What is the run of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
(t2 t2) = run
13s 5s = 8 sec
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What is the slope of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
(d2 d1) / (t2 t1) = slope
(40 16) / (13 5) = 3
Right calculation, but all quantities have units.
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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):
that the average velocity of the ball is 3cm/sec
if 40 and 16 were positions the since 13 and 5 were clock times then this would be an average velocity
but the aren't so it isnt
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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):
There is no change the ball is moving at a constant rate.
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1 hr
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