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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: ->->->->->->->->->->->-> scussion (start in the next line):
vAve = `ds/`dt
vAve = 30cm/5s
vAve = 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: ->->->->->->->->->->->-> n/discussion (start in the next line):
vAve =( v_i + v_f ) /2
6 cm/s = ( 0 + v_f)/2 multiply both sides by 2
12cm/s = 0 + v_f
v_f = 12 cm/s
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By how much did its velocity therefore change?
answer/question/discussion: ->->->->->->->->->->->-> scussion (start in the next line):
Change in velocity = v_f - v_i = 1c2m/s - 0 cm/s = 12cm/s
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At what average rate did its velocity change with respect to clock time?
answer/question/discussion: ->->->->->->->->->->->-> scussion (start in the next line):
It changed 12cm/s in 5 seconds
(12cm/s)/5 s = 2.4cm/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: ->->->->->->->->->->->-> scussion (start in the next line):
The graph would be linear. It would begin at the origin and increase at a constant rate as it moved to the left. It would increase in rise by 2.4 cm for every 1 second increase in run. The formula for the line would be y=2.4x.
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&#Very good responses. Let me know if you have questions. &#