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PHY 121
Your 'cq_1_03.1' 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_03.1_labelMessages.txt **
The problem:
A ball starts with velocity 0 and accelerates uniformly down a ramp of length 30 cm,
covering the distance in 5 seconds.
What is its average velocity?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
vAve = 'ds/'dt; vAve = (30-0)/(5-0); vAve=30/5;
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You do need to include the units in your calculation. This is important for a number of reasons, one is that \it reminds you of the meanings of the various quantities.
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vAve = 6 cm/s
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If the acceleration of the ball is uniform then its average velocity is equal to the
average of its initial and final velocities.
You know its average velocity, and you know the initial velocity is zero.
What therefore must be the final velocity?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
Because it started at zero, its final velocity would be 6 cm/sec.
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By how much did its velocity therefore change?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
It changed by a rate of 6 cm/sec from the start.
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6 cm/s is its average velocity.
The average velocity is in between the initial and final velocities.
Unless the object is staying still, its initial and final velocities will be different and the final velocity will have to differ from the average velocity.
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At what average rate did its velocity change with respect to clock time?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
For every second, the ball moved 6 cm.
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On the average, that is so.
But that doesn't answer the question of the average rate at which the velocity changed.
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What would a graph of its velocity vs. clock time look like? Give the best description
you can.
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
The graph starts at (0,0). The next point is (1,6), which is a steep slope--a slope of 6. The next point would be (2,12). The line starts at (0,0) and stays close to the y axis as it climbs slowly away from it.
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The interval lasts for 5 seconds. The average velocity doesn't occur after 1 second.
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*#&!
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You'll want to revise this. Once you nail down these ideas the analysis goes quickly.
&#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. &#
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