#$&*
course Phys201
6/14@12:34pm
q_1_022#$&*
Phys201
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.
** CQ_1_02.2_labelMessages **
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):
I think it would be 24 cm/sec. I got this because we know theres a point at 40 cm/s and another at 16 cm/sec. I subtracted 40-16=24 which would be half way between the two points.
@&
You need to include units with all calculations.
24 is in any case not halfway between 16 and 40.
&&&&13sec+5sec/2=9 sec is at midpoint of this interval&&&&
*@
#$&*
• What is the velocity at the midpoint of this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
vAve=`ds/`dt
vAve=24/ (13-5)
vAver=24/8
vAve=6 cm/s
@&
24, even with correct units, does not represent a change in position.
Had the position changed by 24 cm, then your calculation would represent the average velocity.
One reason to use units throughout is to help focus on the meanings of the quantities.
*@
#$&*
• How far do you think the object travels during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
vAve=`ds/`dt
I would manipulate the equation to find `ds.
vAve/`dt=`ds
6/(13-5)=`ds
6/8=.75 cm
@&
If you solve vAve = `ds / `dt for `ds, you don't get vAve / `dt = `ds. Be sure to do the algebra correctly.
And use units with every quantity at every step.
*@
&&&&vAve=`ds/`dt=vAve*`dt=`ds=9cm*(13sec-5sec)=72cm/sec
@&
You have, among other things,
vAve=`ds/`dt=vAve*`dt=`ds
The conclusion would be that vAve = `ds.
This is not the case.
You are misusing the = sign, which needs to be used solely and strictly to indicate equality between two quantities.
vAve = `ds / `dt
is however algebraically equivalent to
`ds = vAve * `dt.
However
`ds=9cm*(13sec-5sec)
does not represent an application of this equation. 9 cm does not represent vAve. 9 seconds is the midpoint of the interval, but the equation `ds = vAve * `dt does not involve the midpoint clock time.
(13 s - 5 s) = 8 s is, however, the correct quantity to use for `dt.
So if you can correctly find vAve from the given information, you should be able to use this equation to find `ds.
*@
@&
Note also that
9cm*(13sec-5sec)= 72 cm * sec, not 72 cm/sec.
A correct unit calculation here would have revealed that the result does not have units of `ds.
*@
#$&*
• By how much does the clock time change during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
vAve*`ds=`dt
6*.75=4.5 cm/sec
@&
You can find the change in clock time directly from the given information.
*@
&&&& Change in clock times t=13 sec, t=5sec, 13sec-5sec=8 sec
@&
Right.
*@
#$&*
• By how much does velocity change during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
vAve/`dt=`ds
6/4.5=1.3 cm/sec
??? Not sure if I am using correct units.
@&
This can also be found directly from the given information, without calculating any intermediate quantities.
*@&&&&40 cm/sec-16cm/sec=24cm/sec
@&
Correct.
*@
#$&*
• What is the average rate of change of velocity with respect to clock time on this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
vAve=`ds/`dt
@&
If you find the correct change in position and the correct change in clock time, this will give you the correct result.
By including the units throughout your calculation you will also end up with the correct units.
&&&&vAve=`ds/`dt (24cm/sec)/(8sec)=3 cm/sec
@&
What you have calculated with
(24cm/sec)/(8sec)
is change in velocity / change in clock time.
This gives you an average rate of change, but not an average velocity.
What average rate of change does this calculate?
How does this fit with the question you are answering here?
What are the correct units of this calculation?
*@
@&
24 cm/sec is not `ds.
`ds would be a change in position, and would have units of distance.
24 cm/s has units of velocity.
The units of your calculation are not cm/sec.
vAve = `ds/ `dt does not calculated the average rate of change of velocity with respect to position. It calculates average velocity, which is not at all the same thing.
*@
*@
vAve=.75/1.3
vAve=.58 cm/sec
#$&*
• What is the rise of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
Would you use `dt as rise? So in this case, would it be 4.5.
#$&*
@&
The graph is of v vs. t.
Which quantity therefore goes on the vertical axis, and which on the horizontal?
4.5 has no units, so it can't be the rise or the run of this graph. In any case, 4.5 is not the number that would be associated with either the rise or the run.
The rise and the run can be identified directly from the given information, provided you sketch and label the graph.
*@&&&& velocity would go on vertical axis, time on horizontal axis. Rise would be 40cm/s-16cm/s=24cm/s. Run would be 13 sec-5 sec=8 sec&&&&
• What is the run of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
I think here you would use position change for run so your answer would be 1.3
#$&*
@&
The graph is of v vs. t.
The position change is not 1.3, nor is it 1.3 cm, and it is not the rise or the run of the graph.
*@
• What is the slope of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
(4.5,1.3) are your points
so to get slope you would divide 4.5 by 1.3 and get a slope of 3.5
@&
The graph is of v vs. t.
The position change is not 1.3, nor is it 1.3 cm, and it is neither the rise or the run of the graph.
*@
#$&*
&&&&Slope=rise/run=24cm/sec/8sec=3cm/sec
@&
(cm / s) / s is not cm/s
What is it?
*@
• What does the slope of the graph tell you about the motion of the object during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
The motion of the object I believe would be constant. I am not sure how I would determine that. I think it might be because I got a positive number that is half way between the rise and the run.
#$&*
&&&& The slope tells us that the object is increasing at an increasing rate.&&&&
@&
Something is increasing at an increasing rate, but it is not the object.
It could be the acceleration, the velocity, or the position.
One of these quantities is constant, one is increasing at a constant rate and one is increasing at an increasing rate.
Which is which?
*@
• 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):
vAve=`ds/`dt would be the equation I assume.
vAve=3.5 (which is your slope)/4.5
vAve=.8?
Not quite sure if I did this correct. The answer seems to be a bit off.
#$&*
@&
vAve is the average rate of change of position, not velocity. &&&& I think the answer to this question would be as the clock time increases, the average rate of change of the object’s velocity also increases.&&&&
@&
The average rate of change of the velocity is the acceleration.
The condition of the problem is that the graph of v vs. t is a straight line.
If a quantity has an increasing straight line graph, is that quantity increasing at a constant, an increasing or a decreasing rate?
How is this related to the answer to the given question?
*@
*@
*#&!
@&
You are in the process of sorting out a lot of definitions and concepts.
You need to use units to keep track of the meanings of the various quantities.
You will see these same ideas frequently in this and other upcoming assignments, and by the time you're through them you will have sorted these things out.
In fact, having revised your work on this problem you will have gone a long ways toward doing so.
&#Please see my notes and submit a copy of this document with revisions, comments and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes.
@&
You are making progress, but you are still making a few errors in procedure:
You aren't being careful enough to distinguish which average rate you are using.
You aren't doing the units calculations. You need to do those calculations and explain how you are doing them.
You sometimes misidentify quantities, for example identifying a displacement as a velocity or a clock time, and when you do you seem to sometimes neglect to check whether the units of the quantity you have chosen are the units of the quantity you are looking for.
You sometimes make mistakes in the algebra of manipulating your equations.
With a little more practice, and attention to these details, you should be able to correct errors of this sort and correctly apply the definitions.
Along with the notes I've provided here, the notes I provided on your preceding submission should be helpful.
I will ask for another revision, and ask you to use #### for your new insertions.
*@