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Phy 201
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 **
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):
average velocity=30 cm/5 sec=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): final velocity=0 + 6 cm/s=6 cm/s*2=12 cm/s
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By how much did its velocity therefore change?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line): change in velocity=12 cm/s - 0=12 cm/s
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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):
average rate=change in velocity/time
=(12 cm/s)/(5 s)=2.4 cm/sec^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):
graph shows a straight line rising from 0 to 30 cm/s and time changes by 5 second.
rise of graph=30 cm/s
run of graph=5 sec
slope=rise/run=(30 cm/s)/(5 s)=6 cm/sec^2
acceleration=average rate of change in velocity with respect to clock time, which is equal to slope
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20 min
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