#$&*
PHY 201
Your 'cq_1_02.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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SEED QUESTION 2.1 (1_02.1)
The problem:
A ball starts with velocity 4 cm/sec and ends with a velocity of 10 cm/sec.
• What is your best guess about the ball's average velocity?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
vAve = ds / dt, where `ds represents change in position and `dt represents change in clock time
vAve = (10 cm - 4 cm) / 1 sec
vAve = 6 cm/sec
#$&*
@& If an object's velocity changes from 99 cm/s to 100 cm/s, you wouldn't guess that the average velocity would be the 1 cm/s difference between the two.
However you can answer the remaining question based on your guess.*@
• Without further information, why is this just a guess?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
Because the only variables given to solve the question are velocities, which are the change in position. I used the formula anyway knowing that it would yield the same answer.
#$&*
• If it takes 3 seconds to get from the first velocity to the second, then what is your best guess about how far it traveled during that time?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
vAve = ds / dt,
vAve = (10 cm - 4 cm) / 3 sec
vAve = 6 cm/ 3 sec
vAve = 2 cm/s
#$&*
@& The calculation you have done is an important calculation, but it doesn't answer the present question.
2 cm/s isn't an answer to 'how far'.
The question 'how far' would be answered by a distance, and 2 cm/s is not a distance.*@
• At what average rate did its velocity change with respect to clock time during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
Ave. rate of change of position with respect to clock time = (change in position) / (change in clock time), by the definition of average rate
Ave. rate = (change in position) / (change in clock time)
Ave. rate = 6 cm / 3 sec
Ave. rate = 2 cm/s
#$&*
@& That is the ave rate of change of position with respect to clock time, for an object whose position changes by 6 cm in an interval lasting 3 seconds, and it is very well reasoned out. Good start.
However your reasoning doesn't apply to the situation of this problem, because nothing changes position by 6 cm in 3 seconds.
Your result is an average rate of change of position with respect to clock time, not an average rate of change of velocity with respect to clock time.
Can you apply your reasoning to the situation given here? The steps, and even the numbers, will be very similar, though there will be some differences (especially in meaning).*@
*#&!
@& You did some good thinking with definitions. Your results didn't quite come out right, so see my notes and please submit a revision.
&#Please see my notes and, unless my notes indicate that revision is optional, submit a copy of this document with revisions and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes.
If my notes indicate that revision is optional, use your own judgement as to whether a revision will benefit you. &#
*@