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.
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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):
The average velocity is determined by how far the object moved and in what time. The object moved 30 cm over a course of 5 seconds. Therefore the average velocity is 30cm/5 seconds = 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):
If the average velocity = initial velocity + final velocity /2, then our equation would look like the following:
6cm/s = 0+ final velocity /2
To get the final velocity on its own, we need to multiply each side by 2.
This gives us 12 cm/s = final velocity
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• By how much did its velocity therefore change?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
If our initial velocity is zero and our final velocity is 12 cm/s, then we know there is a change of 12 cm/s. (12cm/s – 0 = 12cm/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):
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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):
Is this question asking for the change in velocity ( 30 cm)
30 cm is a change in position, not a change in velocity
and the clock time change of 5 seconds???If so then the rise of the graph would be 30 cm and the run of the graph would be 5 seconds. So to determine the slope 30
cm/ 5 seconds = 6 cm/second.
A rise of 30 cm would occur on a position vs. clock time graph,
which would indeed have a slope of 6 cm/s. However the question here
concerns a velocity vs. clock time graph, and a rate of change of
velocity with respect to clock time.
If the question is asking us for the average rate of change of velocity with respect to clock time then our slope would be
the same as the average rate of change of velocity in regard to clock time. Our equation would be set up as follows: 12cm/s
/ 5 sec = 2.4 cm/s^2
Very good. This is what the question, which referred to a velocity vs. clock time graph, asked.
The rise of the graph represents the change in velocity, the run
represents the change in clock time.
Be sure you understand the error you made in your other solution (mistaking a change in position for a change in velocity), and how the units of your quantity did not match the units of the quantity to which you attributed it (i.e., cm is not a unit of velocity or change in velocity).
How do you know exactly which equation the question is asking you to use????
The question asked for the slope of the v vs. t graph. The
rise of the v vs. t graph would be change in velocity, the run would
be change in clock time.
If the question had asked you for the average rate of change of velocity with respect to clock time,
then your interpretation must therefore be guided by the definition of average rate of change of A with respect to B.
The A quantity would be velocity and the B quantity would be clock time, so the requested average rate will be
(change in A) / (change in B) = (change in velocity) / (change in clock time).
I believe you will understand this, but if not be sure to let me know.
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30-40 minutes
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Your thinking is very good. You have a couple of points of confusion. See my notes above, and also check out the link below (your question and my responses have been edited into that document). No need for a revision unless you have additional questions.
&#Please compare your solutions with the expanded discussion at the link
Solution
Self-critique your solutions, if this is necessary, according to the usual criteria. Insert any revisions, questions, etc. into a copy of this posted document. Mark any insertions with &&&& so they can be easily identified. &#