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Phy 231
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=(vInitial+vFinal)/2
2(vAve)=vInitial+vFinal
2(vAve)-vInitial=vFinal
2(6cm/s)-0=vFinal
12cm/s=vFinal
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By how much did its velocity therefore change?
answer/question/discussion: ->->->->->->->->->->->-> scussion (start in the next line):
The velocity went from 0cm/s to 12cm/s, so its velocity changed by 12cm/s during the interval.
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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):
ave. rate of change in velocity= (change in velocity)/(change in clock time)
=(12cm/s)/5sec
=2.4cm/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: ->->->->->->->->->->->-> scussion (start in the next line):
If the acceleration of the ball is uniform, that means its velocity is changing by a set amount during each second of this interval. So the graph of acceleration will be a straight line, increasing at a steady rate.
Here it is increasing steadily from 0cm/sec to 12cm/sec over the course of 5 sec. It increases by 2.4cm/s each second.
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*#&!
&#Very good responses. Let me know if you have questions. &#