#$&*
PHY 121
Your 'cq_1_03.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** CQ_1_03.1_labelMessages.txt **
cq_1_031
#$&*
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):
vAve = 'ds/'dt; vAve = (30-0)/(5-0); vAve=30/5;
&&&&The average velocity is 30 cm/5 seconds or 6 cm/s&&&&
@&
You do need to include the units in your calculation. This is important for a number of
reasons, one is that \it reminds you of the meanings of the various quantities.
*@
vAve = 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):
Because it started at zero, its final velocity would be 6 cm/sec.
&&&&Since the average velocity is 6 cm/s, then the final velocity would be twice that,
or 12 cm/s.&&&&
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By how much did its velocity therefore change?
answer/question/discussion: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> (start in the next line):
It changed by a rate of 6 cm/sec from the start.
&&&&The velocity changed from 0 cm/s to 12 cm/s.&&&&
@&
6 cm/s is its average velocity.
The average velocity is in between the initial and final velocities.
Unless the object is staying still, its initial and final velocities will be different
and the final velocity will have to differ from the average velocity.
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#$&*
At what average rate did its velocity change with respect to clock time?
answer/question/discussion: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> (start in the next line):
For every second, the ball moved 6 cm.
&&&&The average velocity was 6 cm/s and that changed over a 5 second time span, so the
change in velocity with respect to clock time is 6 cm/s / 5 s, or 1.2 cm/s/s.&&&&
@&
6 cm/s is the average velocity, not the change in velocity.
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Perhaps the most common error I see in these problems is confusion between average velocity and change in velocity.
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On the average, that is so.
But that doesn't answer the question of the average rate at which the velocity changed.
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What would a graph of its velocity vs. clock time look like? Give the best description
you can.
answer/question/discussion: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> (start in the next line):
The graph starts at (0,0). The next point is (1,6), which is a steep slope--a slope of
6. The next point would be (2,12). The line starts at (0,0) and stays close to the y
axis as it climbs slowly away from it.
&&&&It starts at (0,0). The end point is at (5, 12). The slope is 1.2, which means
that for every 1.2 cm/s increase, it takes 1 sec.&&&&
#$&*
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Between (0, 0) and (5, 12):
What is the rise?
What is the run?
What therefore is the slope?
How does this relate to a previous error in your reasoning?
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*#&!
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There is an important distinction you've missed here.
I'm going to ask for another revision.
This time use #### to indicate your insertions.
*@