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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= 30cm / 5 s = 6cm/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):
The final velocity must be double the average velocity.
vFinal = 6cm/s *2 = 12cm/s
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• By how much did its velocity therefore change?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
12cm/s - 6cm/s = 6cm/s. Therefore, it was double 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):
(12cm/s - 0cm/s) / 5 s = 2.5 cm/s^2
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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):
Clock time would be on the x axis and velocity would be on the y axis. A graph of its velocity vs. clock time would be increasing at a constant rate since the object accelerated uniformly. As clock time increases so does velocity.
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&#Your work looks good. Let me know if you have any questions. &#
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However, note that the change in velocity is 12 cm/s, not 6 cm/s; you were a little ambiguous on this but used the 12 cm/s in your calculation of acceleration, so it appears that you understand.
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