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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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):
13+5=18/2=9 sec
9 sec. is at the midpoint of this interval.
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What is the velocity at the midpoint of this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next
On the graph that I drew which is far from perfect it looks like about 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):
The object is going for 8 sec. and goes from 16 cm/s to 40 cm/s so that is a change by 24 cm/s so basically the speed could be increasing at the rate of about 3 cm/s so the
equation I will use is 19+22+25+28+31+34+37+40=236 cm. Am I really off base on this???
You have good insight into what's happending here, though you're off by 12 centimeters. For example for the first second the velocity increases from 16 cm/s to 19 cm/s, so the average velocity during the first second would be 17.5 cm/s, halfway between the initial and final velocities. So the object would move 17.5 cm during the first second. Distances in subsequent seconds would be 20.5 cm, 23.5 cm, 26.5cm, ..., 38.5 cm and the sum would be 224 cm.
See the link at the end of the document for a more efficient way of reasoning this out. It's not always practical to do a second-by-second breakdown.
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By how much does the clock time change during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
There is an 8 sec. time change.
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By how much does velocity change during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
Velocity changes by 24 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):
Since it changes by a rate of 24 cm/s in the time of 8 sec. this would be approximately 3cm/s.
That would be 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 is 40-16 = 24cm/s
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What is the run of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
Run is 13 sec - 5 sec = 8 sec
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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=24/8=3
the slope has units
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 the same answer as the rate of increase in velocity.
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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/sec.
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about 30 min.
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2-3-10 at 7:18 p.m.
I'll leave it to you whether to submit a revision or not. If you're sure you understand everything at the given link, there's no need for a revision. If not you should submit a revision indicating your thinking on the parts you're not sure you understand.
&#Please compare your solutions with the expanded discussion at the link
Solution
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