phy201
Your 'cq_1_02.1' 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 21
course phy201
Seed 2.1A ball starts with velocity 4 cm/sec and ends with a velocity of 10 cm/sec.
What is your best guess about the ball's average velocity?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
my best guess at the balls average velocity is 4 cm/s - 10 cm/s = - 6cm/s
average velocity is not found by subtracting velocities
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Without further information, why is this just a guess?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
Without further information this would be just a guess because we are not given the total time it took for the ball to change velocities. Since average velocity is equal to the change in position / the chnage in time we are not given enough information to make it more that a guess.
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If it takes 3 seconds to get from the first velocity to the second, then what is your best guess about how far it traveled during that time?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
The distance traveled = change in velocity * time.
On way to remember that this doesn't work: If this was the case then traveling for 3 hours at a steady 60 mph would result in a displacement of 0, since the change in velocity would be 0.
Then how far it traveled depends on the rate at which the ball was traveling. So if it went from a velocity of 3cm/s to 4 cm/ s we would subract 3cm/s -4 cm/s = -1cm/s and then multiply that by the time, which was 3 seconds to get -3cm
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Self critique
Ok I should have use the average of velocities here to find the solution....... 'ds= ave vel * change in clocktime.........'ds= 7cm/s * 3s 'ds=21cm
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At what average rate did its velocity change with respect to clock time during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
average rate is equal to the change in velocity / change in time
so the average rate = change from the first velocity to the second velocity/ 3 seconds
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your response &&&&&&&&&&&&&&&&&&
to my solution
You appear to be confusing average velocity with change in velocity. Check out the link given below.
&#Please compare your solutions with the expanded discussion at the link
Solution
Self-critique your solutions, if this is necessary, according to the usual criteria. Insert any revisions, questions, etc. into a copy of this posted document. Mark any insertions with &&&& so they can be easily identified. &#
Note that seed question is in its own form. To ensure a response that matches the question, you should use the form for the question to submit the question.
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ok, I think I understand. We needed the average velocity here because we have 2 different velocityies we simply must average them together. I think I am confusing average velocity with average rate
Average velocity is (change is position) / (change in clocktime)
Average rate is (change in velocity)/ (change in clocktime)
Unless the average rate is clearly defined, you need to be specific about just what average rate is involved.
It isn't strictly correct to say that
'Average rate is (change in velocity)/ (change in clocktime)'
because you aren't specifying the quantities A and B in the definition of average rate. (The definition is 'average rate of change of A with respect to B is (change in A) / (change in B). To apply this definition you have to be explicit about what the A quantity is, and what the B quantity is.)
A correct statement would be
'Average rate of change of velocity with respect to clock time is (change in velocity)/ (change in clocktime)'.
The average velocity you found is the average rate of change of position with respect to clock time, which is (change in position) / (change in clock time).
Your statement 'Average rate is (change in velocity)/ (change in clocktime)' would not be correct if the average rate to which you are referring is the average rage of change of position with respect to clock time. Nor would the statement 'Average rate is (change in position)/ (change in clocktime)' give you the right result for the present situation.
For the time being, you should never use the phrase 'average rate' with out the phrase 'average rate of change of __ with respect to __", with the blanks filled in by the appropriate quantities.
So i was finding the average rate, not the average velocity. since we had velocitis that went from initial at 4cm/s to final 10cm/s to find the average we must do the average adding them and then dividing by 2. So, the velocities had started from 0 to find the average velocity we would multiply by 2 to get the final velocity.
I actually was getting ready to send a question asking how do we know when to average the velocities together and how do we know when to subtract, but I think I see now that we subtrace them when we are finding the average rate, we average them to find the average velocity.
That's a pretty good way to put it.
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ok, I made some revisions to this document. I put it at the bottom of the page in between ##### marks. I think i understand what I am doing wrong here. Thanks
I sent this once before on accident I was not finished filling out the identity information
Your last statement is good.
However you really need to be careful with the 'average rate' terminology. Make sure you are clear on just what average rate you are trying to find, the average rate of change of what quantity with respect to what other quantity. See my note.