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Phy 121
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.
** CQ_1_02.1_labelMessages.txt **
The problem:
A 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):
I would say that the ball's average velocity is the sum of the two given velocities divided by 2.
(4 cm/s + 10 cm/s) / 2 = 14 cm/s / 2 = 7 cm/s But I also read in a previous assignment that the average velocity is half of the final velocity. Which in this case would be 5 cm/s.
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For uniform acceleration the average velocity is half the final velocity only in the special case that the initial velocity is zero.
So the first result is the one you would want to use.
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Without further information, why is this just a guess?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
Because I am not sure that I am doing what it asks correctly.
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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):
I would say that if it took 3 seconds to get from the first velocity to the second, then that would have traveled at the velocity of 6 cm/sec.
So estimating this would allow me to find the estimated distance it traveled. `ds = `dt * vAve = 3 sec * 6 cm/sec = 18 cm
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You would multiply the time interval by the average velocity, according to the definition of average velocity.
There is no reason to expect the difference of the velocities, which is 6 cm/sec, to be the average velocity.
Your 7 cm/s estimate is a valid estimate of the average velocity.
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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 = 6 cm/sec / 3 sec = 2 cm/sec^2
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Good.
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&#Good responses. See my notes and let me know if you have questions. &#