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: quantity A=the change of the position of the ball (going from 10 cm to 20 cm has a 10cm change). Quantity B=the change in time (from 4 seconds to 9 seconds is a 5 second change. Therefore 10cm/5sec = 2 cm/sec change. For every second that elapses, the ball will roll 2 cm
You understand the definition and did obtain the correct result. However to avoid possible future confusion you need to be careful about terminology.
quantity A is the position, not the change in position. Quantity B is the clock time, not the change in clock time.
The definition of rate of change refers to 'change in quantity A' and 'change in quantity B'. If quantities A and B were themselves change in position and change in clock time, then change in A would be change in change in A, the change in B would be change in change in B, and the average rate of change of A with respect to B would be (change in change in A) / (change in change in B).
&#Quantity A is velocity, quantity B is clock time. The change in quantity A is (40 cm/s - 10 cm/s) = 30 cm/s. The change in quantity B is 3 s. So
Ave rate of change of velocity with respect to clock time = (change in velocity / change in clock time) = (30 cm/s) / (3 s) = 10 cm/s^2.
The unit calculation is (cm/s) / (s). Division is the same as multiplication by the reciprocal, so (cm/s) / (s) = (cm/s) * (1/s).
To multiply fractions we multiply numerators and multiply denominators, so (cm / s) * (1 / s) = (cm * 1) / (s * s) = cm / s^2. &#
· 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: quantity A=the change in the velocity (40 cm/sec – 10 cm/sec = 30 cm/sec). Quantity B= the change in time which is 3 seconds. Therefore, for every second that elapses, the velocity of the ball changes by 10 cm/second.
· 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: The average rate is 5cm/sec. We don’t know quantity A. Quantity B is the time change (10 seconds). Therefore, 5cm/sec = A/10 sec, multiply both sides by 10 sec and solve for A. A (the position change) = 50 cm.
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15 minutes
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Very good work, but see my note to clarify a couple of important points.
&#Let me know if you have questions. &#