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course phy 121
seed 22#$&*
course Phy 121
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
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What is the velocity at the midpoint of this interval?
answer/question/discussion (start in the next line):
28 cm/sec
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How far do you think the object travels during this interval?
answer/question/discussion (start in the next line):
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This isn't a reasonable answer to the given question. 24 cm/s is the change in the velocity, and you might have been thinking along those lines, but the result wouldn't be 24 cm.
The interval lasts 8 seconds, and the minimum velocity is 16 cm/s. So the object travels at least 16 cm/s * 8 s = 128 cm.
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vAve = 40cm/s +16cm/s / 2
vAve = 28cm/s
28cm/s(8s) = ~224cm
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By how much does the clock time change during this interval?
answer/question/discussion (start in the next line):
8 seconds
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By how much does velocity change during this interval?
answer/question/discussion (start in the next line):
24cm/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):
The balls velocity changed 3cm/sec.
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The velocity changed by 24 cm/s.
The average rate at which the velocity changed is (24 cm/s) / (8 s) = 3 cm/s^2.
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Average rate of change of velocity with respect to clock time = vf-v0 / t-t0
=40cm/s-16cm/s / 13s-5s
=24cm/s / 8s
=3cm/s/s
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What is the rise of the graph between these points?
answer/question/discussion (start in the next line):
8
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The rise has units.
The run is 8 seconds, but the rise doesn't have numerical value 8.
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24cm/s is the rise between these points.
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What is the run of the graph between these points?
answer/question/discussion (start in the next line):
24
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What is the slope of the graph between these points?
answer/question/discussion (start in the next line):
1/3
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You're on the right track, but you've reversed the rise and the run.
Rise and run also have units.
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The run of these points is 8 seconds.
The slope of this graph 24cm/s / 8s = 3cm/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):
The slope of the graph tells me that velocity is rapidly getting higher, much more rapidly than the time interval increases.
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The slope does tell you how rapidly the velocity is rising.
The slope is the change in velocity / change in clock time, which is the average rate of change of velocity with respect to clock time, or acceleration.
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Along with my previous answer, the slope shows the acceleration during the interval. If it is a straight line, the acceleration is constant.
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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 balls velocity changed 3cm/sec.
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See my previous notes for correct units and terminology.
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Solution
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