#$&*
course Phy 121
3.21.11 at 7:51pm
Resubmit 71#$&*
course PHY 121
3.12.11 at 6:10pm
cq_1_071#$&*
PHY121
Your 'cq_1_07.1' report has been received. Scroll down through the document to see any
comments I might have inserted, and my final comment at the end.
** **
A ball falls freely from rest at a height of 2 meters. Observations indicate that the ball
reaches the ground in .64 seconds.
Based on this information what is its acceleration?
answer/question/discussion: ->->->->->->->->->->->-> :
2m/0.64sec = 3.125m/s
a=3.125m/sec/0.64sec = 4.882m/sec^2
##### I used 3.125m/sec because v0=0 in this case. THe change in velocity would be the same
as vAve, right?#####
@& If the initial velocity is 0 and the final velocity is, say, 10 cm/s, then what is the average velocity?
Is the average velocity equal to the final velocity?
It should be clear that it is not.
How are the final and average velocities related when the initial velocity is zero?*@
@& Acceleration is based on change in velocity, not aveage velocity.*@
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Is this consistent with an observation which concludes that a ball dropped from a height of
5 meters reaches the ground in 1.05 seconds?
answer/question/discussion: ->->->->->->->->->->->-> :
5m/1.05se=4.762m/sec
a=(4.762m/sec)/1.05sec = 4.535m/sec^2
#$&*
Are these observations consistent with the accepted value of the acceleration of gravity,
which is 9.8 m / s^2?
answer/question/discussion: ->->->->->->->->->->->-> :
No, these figures are significantly lower than the accepted value, assuming they are
correct.
#$&*
** **
10 minutes
** **
Are my calculation correct? If not, what did I do wrong?
@& See my note. You used average velocity where you should have used the change in velocity.
&#Please see my notes and, unless my notes indicate that revision is optional, submit a copy
of this document with revisions and/or questions, and mark your insertions with &&&& (please
mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes.
If my notes indicate that revision is optional, use your own judgement as to whether a
revision will benefit you.
&#
*@""
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
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Self-critique rating:
________________________________________
#$&*
@& Average and final velocities will not be equal in this situation.
See my note and see if you can answer the questions I pose, and use your reasoning to revise your solution to the given problem.
Please indicate new revisions using &&&&.*@"
Self-critique (if necessary):
------------------------------------------------
Self-critique rating:
________________________________________
#$&*
@&
&#See any notes I might have inserted into your document, and before looking at the link below see if you can modify your solutions. If there are no notes, this does not mean that your solution is completely correct.
Then please compare your old and new 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.If your solution is completely consistent with the given solution, you need do nothing further with this problem. &#
*@
@& Average velocity and change in velocity are two different things.
Change in velocity is final velocity - initial velocity.
Average velocity is change in position / change in clock time.
Consider this:
If the initial velocity is 0 and the final velocity is 10 m/s, then what is the average velocity, and what is the change in velocity?
Clearly the average velocity is 5 m/s.
It should also be clear that the change in velocity is 10 m/s
See also the link above..*@