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Phy 201
Your 'cq_1_02.2' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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The problem:
A graph is constructed representing velocity vs. clock time for the interval between clock times t = 5 seconds and t = 13 seconds. The graph consists of a straight line from the point (5 sec, 16 cm/s) to the point (13 sec, 40 cm/s).
• What is the clock time at the midpoint of this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
The clock time at the midpoint of the interval would be 9. This justified in two ways you can say that 9 is the number in between 5-13 ….5 6 7 8 9 10 11 12 13, or by taking the average and saying that 5 + 13 = 18 / 2 = 9. This is the way I would think you would find the midpoint, but is this the right way and is there another way of finding is.
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• What is the velocity at the midpoint of this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
For this question would it be correct to use the change of velocity formula and say that (40cm/sec) - (16 cm/sec) / (13sec) - (5sec) = 24cm/sec / 8sec = 3cm/sec/sec? No know I see the ending answer I do not think that is right.
So could average the numbers again like for the first question. 5 + 13 = 18 / 2 = 9sec and 40 + 16 = 56 / 2 = 28cm/sec
@& Good.*@
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• How far do you think the object travels during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
Using the starting points and the ending points I would take 40cm/sec and subtract it by 16cm/sec to equal 24cm/sec. Then take 13sec and subtract it by 5sec to equal 8sec. Since the object travel for 8sec between point A and point B at a velocity of 24cm/sec, I would multiply 24 * 8 = 192 cm
@& The 8 second change in clock time is appropriate for this calculation.
However 24 m/s is the change in the object's velocity.
Multiplying the change in velocity by the change in clock time doesn't give you the change in position. For example an object which travels at a constant 10 m/s for 12 seconds would travel 120 meters; however the change in its velocity is zero, so multiplying change in velocity by change in clock time would give you 0.*@
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• By how much does the clock time change during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
The clock time changes by 8sec because 13-5=8
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• By how much does velocity change during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
Velocity changes by 24cm/sec because 40-16=24
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• What is the average rate of change of velocity with respect to clock time on this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
Average rate of change of velocity is :
(40cm/sec) - (16 cm/sec) / (13sec) - (5sec) = 24cm/sec / 8sec = 3cm/sec/sec
@& Good.*@
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• What is the rise of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
The graph has a positive Rise. Because the graph runs in a upward direction.
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• What is the run of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
The run of the graph is positive because it would go to the right.
@& You can give a quantitative answer for the run.*@
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• What is the slope of the graph between these points?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
The slope of the graph would be the distance between the two points in the y direction / the distance between the two points in the x direction.
@& You need to use the definition of slope; to do so you'll need to find a run to go with your rise, which is correct.*@
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• What does the slope of the graph tell you about the motion of the object during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
That as the time progresses the velocity of the object speeds up going faster and faster.
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• What is the average rate of change of the object's velocity with respect to clock time during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
I am uncertain as to answer this question. I understand that you are asking for the average change in rate of the velocity: (40cm/sec) - (16 cm/sec) / (13sec) - (5sec) = 24cm/sec / 8sec = 3cm/sec/sec. But I am unsure as to how to relate it with clock time. I am confused as to what the question is actually asking for. Not to say that the question is poorly worded but saying I do not fully comprehend.
@& This question repeats a question asked above, and you got it right both times.*@
@&
&#Please see my notes and, unless my notes indicate that revision is optional, submit a copy of this document with revisions and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes.
If my notes indicate that revision is optional, use your own judgement as to whether a revision will benefit you. &#
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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 line would start at the lower left hand side of the graph and as time increase the velocity would increase and the line would go up and to the right, with a straight and constant slope.
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*#&!
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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):
Average velocity = Length or distance of travel divided by the Time
= 30cm / 5sec = 6cm/sec.
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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):
Its final velocity would be double its average velocity: 6cm/sec * 2 = 12cm/sec
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• By how much did its velocity therefore change?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
6cm/sec / 2.5sec = 2.4cm/sec/sec
Or would you take the change in velocity and divide it by same clock time instead of dividing it by the half of the original clock time:
6cm/sec / 5sec = 1.2cm/sec/sec?????????????????????????????????
@& What is the initial velocity on the interval?
What is the final velocity on the interval?
What therefore is the change in the 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):
Again I am unsure as to how to answer this question. What do I look for that would give be the information I need to answer this question? Does the answer lie within the final answer of the average rate of change of velocity formula???????????????????????????????????
@& What is the definition of the average rate of change of velocity with respect to clock time?
How then can you use the information from previous questions to answer the present question?*@
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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 line would start at the lower left hand side of the graph and as time increase the velocity would increase and the line would go up and to the right, with a straight and constant slope.