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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):
the average velocity or average rate of change of position is vAve = `ds(30 cm) / `dt(5 sec) = 6cm / sec
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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):
the final velocity must be the average of the initial (0) and final velocities (6cm/sec) which because the acceleration is uniform, there is no average rate of change of velocity acceptbetween the 0 and 6cm / sec which = 6cm / sec. the final velocity is 6cm/sec
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If the initial velocity was 0 cm and the final was 6 cm/s then the average velocity would be 3 cm/s.
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• By how much did its velocity therefore change?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
the average rate of change of velocity or acceleration is 6cm/sec
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6 cm/s is the average rate of change of position with respect to clock time.
What was the change in velocity, what was the change in clock time, and what therefore was the average rate of change of velocity with respect to clock time?
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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):
6cm / sec
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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 would at first curve as the ball got up to speed and from there it would increase at an increasing rate for as long as the ball continues to travel since the acceleration is uniform.
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Acceleration is the slope of the v vs. t graph.
So uniform acceleration would imply a straight-line graph.
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*#&!
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This is where things get confusing for about 99% of students.
It's really important here to be very careful about identifying quantities, and about keeping the terminology straight.
I don't think you'll have any trouble making corrections, but if you do don't spend over about 15 minutes before you submit your best revision accompanied by your questions.
&#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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