#$&*
PHY 201
Your 'cq_1_02.2' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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SEED QUESTION 2.2(1_02.2)
The problem:
A graph is constructed representing velocity vs. clock time for the interval between clock times t = 5 seconds and t = 13 seconds. The graph consists of a straight line from the point (5 sec, 16 cm/s) to the point (13 sec, 40 cm/s).
• What is the clock time at the midpoint of this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
The midpoint is (x2+x1) /2 , (y2+y1) /2
Midpoint = (13 sec + 5 sec) / 2, (40 cm/s + 16 cm/s) / 2
Midpoint = (18 sec / 2), (56 cm/s / 2)
Midpoint of this interval = 9 sec, 28 cm/s
#$&*
@& Goodl*@
• What is the velocity at the midpoint of this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
Midpoint = 9 sec, 28 cm/s
Average velocity is represented as slope between two points on a graph of position vs. clock time.
Velocity is rate of change of position with respect to clock time.
vAve = ds / dt
vAve = 28 cm/s / 9 sec
vAve = 3.11 cm/s^2
#$&*
@& 28 cm/s is the velocity at the midpoint, not a change in position.
You've already given the velocity at the midpoint as 28 cm/s.*@
• How far do you think the object travels during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
vAve = ds / dt
ds = vAve * dt
ds = 3.11 cm/s^2 * 9 sec
ds = 28 cm/s
#$&*
@& If the velocity changes at a constant rate, its midpoint value is the average velocity on the interval.
You already gave this as 28 cm/s.
How long does the interval last (i.e., what is `dt)?
What do you conclude is the displacement `ds?*@
• By how much does the clock time change during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
vAve = ds / dt
ds = vAve * dt
dt = ds / vAve
dt = 28 cm/s / 3.11 cm/s^2
dt = 9 sec
#$&*
@& What is the initial clock time?
What is the final clock time?
What therefore is the change in clock time?*@
• By how much does velocity change during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
vAve = ds / dt
vAve = 28 cm/s / 9 sec
vAve = 3.11 cm/s^2
#$&*
@& What is the initial velocity?
What is the final velocity?
What therefore is the change in velocity?*@
• 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):
Ave. rate = (change in position) / (change in clock time)
Ave. rate = 28 cm/s / 9 sec
Ave. rate = 3.11 cm/s^2
#$&*
@& Your reasoning is correct. Your quantities aren't quite right.
If you use the correct change in velocity and change in clock time you'll get a slightly different answer.*@
• What is the rise of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
The rise of the graph between these points is 28 cm/s, the change in y.
#$&*
• What is the run of the graph between these points?
The run of the graph between these points is 9 seconds, the change in x.
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
#$&*
@& I don't think the clock time changes by 9 seconds, as you will probably have discovered by answering some of my preceding questions.*@
• What is the slope of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
The slope = rise/run
Slope = 28 cm/s / 9 sec
Slope = 3.11 cm/s^2
#$&*
@& Again your numbers aren't quite right, though both are pretty close.*@
• 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):
slope with rise over run.
First the slope = rise over run.
The rise mean how many units you will move up or down from point to point, the changes in the Y values. The run mean how far to the left or right you move from point to point, the changes in X values.
#$&*
@& Right, but you haven't said what this tells you about the object.*@
• What is the average rate of change of the object's velocity with respect to clock time during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
Ave. rate = (change in position) / (change in clock time)
Ave. rate = 28 cm/s / 9 sec
Ave. rate = 3.11 cm/s^2
#$&*
@& This answer will change somewhat, but your reasoning and your units are good.*@
*#&!
@& You've got an excellent approach and you're doing a lot of good reasoning. You still need to be a little more careful with some of the quantities, and you need to sort these ideas out just a little more, but you're very much on the right track.
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Be sure to include the entire document, including my notes.
If my notes indicate that revision is optional, use your own judgement as to whether a revision will benefit you. &#
*@