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course phy 121
2/10/13 1:51
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):
midpoint=(x1+x2/2, y1+y2/2)
=(5+13/2, 16 +40/2)
=(18/2, 56/2)
(9,28)
Midpoint time would be 9 seconds
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What is the velocity at the midpoint of this interval?
answer/question/discussion (start in the next line):
The velocity would be 28 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):(9,28)
D=sqareroot of (x2-x1)^2 + (y2-y1)^2
sqareroot of (9-5)^2+(28-16)^2
squareroot of (4^2) + 12^2
Square rood of 16 +144
squareroot of 160
12.65 cm
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Very good application of the Pythagorean Theorem, though this doesn't answer the question.
On the graph the x interval has units of seconds and the y interval has units of cm/s, so
(x2-x1)^2 + (y2-y1)^2
will not have consistent units. One number will have units of sec^2 and the other will have units of cm^2 / s^2. So the two numbers cannot be added.
They could be multiplied, in which case the units of the square root would come out in cm, but the Pythagorean Theorem doesn't involve multiplication of the squares.
So, bottom line, the Pythagorean Theorem can't be applied here because the squares you get have different units, making them unlike terms which cannot be added.
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By how much does the clock time change during this interval?
answer/question/discussion (start in the next line):
9-5= 4 seconds
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By how much does velocity change during this interval?
answer/question/discussion (start in the next line):
28-16= 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):
vAve= distance/ time
=12.65/ 4
=3.1625 cm/s
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Your rise is 12 cm/s, not 12.65.
Rise and run have units which must be included in the calculation.
The correct calculation would then be
12 cm/s / (4 s).
What is the result, and what are the units of the result?
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What is the rise of the graph between these points?
answer/question/discussion (start in the next line):
beginning to midpoint: 28-16=12
midpoint to end: 40-28=12
full line: 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):
beginning to midpoint:9-5=4
Midpoint to end:13-9=4
full line: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):
slope= 24/8 =3 12/4=3
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Very good, but note that these quantities have units, which must be included throughout the calculation.
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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):
For every 12 moves up, it goes 4 to the right. It is increasing at a constant rate.
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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):
(Change in velocity)/ (change in time)
40-16 / 13-5
24/8
Velocity =3
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24 indicates 24 cm/s, which is the change in velocity.
8 represents 8 seconds, which is the change in clock time.
So this calculation represents change in velocity / change in clock time.
That is not velocity, nor is it average velocity. The units of the calculation are not units of velocity, and the meaning isn't velocity.
What quantity do you get when you divide change in velocity by change in clock time?
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You're doing a lot of things right, and your use of the Pythagorean Theorem, though it doesn't apply to this situation, indicates good mathematical sophistication.
You have missed a number of details in these calculations, which is quite typical for students at this stage, but with your obviously good background and aptitude I don't think you'll have any trouble correcting them.
Still, some of these concepts are confusing at first, so don't hestitate to ask if you have difficulty.
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