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 down a ramp of length 30 cm, covering the distance in 5 seconds.
· What is its average velocity?
answer/question/discussion: 6 cm/sec (v = 30cm/5sec)
· If the acceleration of the ball is uniform then its average velocity is equal to the average of its initial and final velocities.
answer/question/discussion: I’m not sure about that. The initial velocity is 0 and it travels at an average of 6cm/sec, shouldn’t the velocity after 1 second be 6cm/sec, after 2 seconds be 12cm/sec, after 3 seconds be 18cm/sec, after 4 seconds be 24cm/sec and after 5 seconds be at 30cm/sec? The average of its initial and final velocities should be 15 cm/sec. Am I anywhere close on this?
If the ball's velocity changed at the rate of 6 cm/s^2, then the velocities would be as you say.
However this is not the rate at which the velocity changes.
The average rate at which the position changes is 6 cm/s; this is the average velocity, not average rate of change of velocity.
· You know its average velocity, and you know the initial velocity is zero. What therefore must be the final velocity?
answer/question/discussion: 30 cm/sec
Initial velocity is 0, average velocity is 6 cm/s, which is also equal to the average of the initial and final velocities.
What therefore is the final velocity?
· By how much did its velocity therefore change?
answer/question/discussion: The velocity changed by 6 cm/sec. After 5 seconds, it had changed from 0 to 30 cm/sec.
· At what average rate did its velocity change with respect to clock time?
answer/question/discussion: The velocity changes at a rate of 1.2cm/s/s. Average velocity/change in time
the average rate of change of the velocity would be change in velocity / change in clock time, not average velocity / change in clock time.
Your units, much to your credit, are correct.
· What would a graph of its velocity vs. clock time look like? Give the best description you can.
answer/question/discussion: This would graph as a very steep straight line, increasing at a constant rate. The points on the y-axis increase exponentially and the points on the x-axis increase a little at a time.
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15 min
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Hopefully you will be able to revise your answers according to my notes. However you used a couple of nearly-but-not-quite-correct ideas (which stumped Aristotle and a lot of other brilliant people for well over 1000 years), and it might be difficult to correct them based on just my notes. Give it a try then look at the following:
The average velocity is 6 cm/s, initial velocity is 0, and average of initial and final velocities is the average velocity, i.e. 6 cm/s.
What number would you average with 0 to get 6? The answer is 12.
So the final velocity is 12 cm/s.
The velocity changes from 0 to 12 cm/s, averaging 6 cm/s.
The change in velocity from 0 to 12 cm/s is 12 cm/s. This occurs in 5 sec so the average rate of change of velocity with respect to clock time is
ave rate = change in velocity / change in clock time = 12 cm/s / (5 s) = 2.4 cm/s/2 or 2.4 cm/s^2.