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Phy 121
Your 'cq_1_01.1' 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:
Here is the definition of rate of change of one quantity with respect to another:
The average rate of change of A with respect to B on an interval is
average rate of change of A with respect to B = (change in A) / (change in B)
Apply the above definition of average rate of change of A with respect to B to each of the following. Be sure to identify the quantity A, the quantity B and the requested average rate.
If the position of a ball rolling along a track changes from 10 cm to 20 cm while the clock time changes from 4 seconds to 9 seconds, what is the average rate of change of its position with respect to clock time during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
Quantity A deals with the distance traveled while Quantity B deals with the clock time. The change in Quantity A can be found by to following: 20cm - 10cm = 10 cm. So the change in A is 10 cm.
The change in Quantity B can be found by doing 9 sec - 4 sec = 5 seconds. So the change in Quantity B is 5 seconds.
We can now find the average rate by dividing the change in A by the change in B. We get 10 cm / 5 seconds = 2 cm/sec. So our average rate is 2 cm/sec.
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If the velocity of a ball rolling along a track changes from 10 cm / second to 40 cm / second during an interval during which the clock time changes by 3 seconds, then what is the average rate of change of its velocity with respect to clock time during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
Quantity A refers to the velocity of the ball rolling along a track and Quantity B is the clock time. The change in Quantity A is 40 cm/sec - 10 cm/sec = 30 cm/sec.
The change in Quantity B is 3 seconds - 0 seconds = 3 seconds.
To find the averate rate we divide the change in A by the change in B. 30 cm/sec / 3 sec = 10 cm/sec/sec. So our average rate is 10 cm/sec/sec.
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If the average rate at which position changes with respect to clock time is 5 cm / second, and if the clock time changes by 10 seconds, by how much does the position change?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
Quantity A in the distance and Quantity B is the clock time.
If we know the average rate and the change in clock time, we can multiply the two to find the distance.
5 cm/sec * 10 seconds = 50 centimeters. So the object changes position of 50 centimeters over a 10 second time span.
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You will be expected hereafter to know and apply, in a variety of contexts, the definition given in this question. You need to know this definition word for word. If you try to apply the definition without using all the words it is going to cost you time and it will very likely diminish your performance. Briefly explain how you will ensure that you remember this definition.
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
I will ensure that I remember the definition of rate of change for one quantity with respect to another not only by looking at it time and again, but by also using it. Practicing with the equation will help me remember it.
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You are asked in this exercise to apply the definition, and given a general procedure for doing so. Briefly outline the procedure for applying this definition, and briefly explain how you will remember to apply this procedure.
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
I will apply this definition anytime that I need to find an average rate for two quantities that have a change. To use the definition, I just find the amount of change for each quantity and then divide the change of the quantities by each other in the appropriate fashion to get the correct answer.
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&#Your work looks very good. Let me know if you have any questions. &#