Your 'cq_1_05.1' 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 accelerates at 8 cm/s^2 for 3 seconds, starting with velocity 12 cm/s.
• What will be its velocity after the 3 seconds has elapsed?
answer/question/discussion: If we multiply the 8cm/s/s by 3s, we get 24 cm/s. This is the velocity after 3 seconds.
• Assuming that acceleration is constant, what will be its average velocity during this interval?
answer/question/discussion: If acceleration is constant, then the average velocity during this interval will be ˝ of the final velocity. So, this will be equal to ˝ * 24 cm/s = 12 cm/s.
That is the ball's starting velocity. It speeds up, so this cannot be its average velocity.
For uniform acceleration, average velocity is understood at the velocity represented by the midpoint of a v vs. t trapezoid. The 'altitudes' of that trapezoid are the initial and final velocities.
Note that you average two number by adding them and dividing by 2, not subtracting and dividing by 2. How does this change your answer?
• How far will it travel during this interval?
answer/question/discussion: During this interval, the object will travel ‘ds=’dt * ‘dv = 3s * (24cm/s-8cm/s) = 48 cm.
`ds is not equal to `dv * `dt.
You need to connect your reasoning to the definition of average velocity as the average rate of change of position with respect to clock time.
Thus `vAve = `ds / `dt, and `ds = vAve * `dt. `ds is not equal to `dv * `dt; and object can move a long way without 0 change in velocity.
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This took me about 15 minutes.
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Some of your answers are inconsistent with the definition of velocity. Please see my notes and submit a copy of this document, inserting revisions and/or questions, and mark you insertions with &&&&.