phy 201
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.
** **
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 the ball travels at 8cm per second for 3 seconds then it has traveled 24cm if it started with a velocity of 12cm/s, we add the two and divide by two leaving us with a velocity of 18cm/s
The ball does not travel at 8 cm/second. It has an acceleration of 8 cm/s^2, and this is what dictates the change in its velocity.
You cannot add 24 cm to 12 cm/s, because these quantities have different units and are therefore unlike
terms.
What is the definition of acceleration, and how does it apply in this situation?
• Assuming that acceleration is constant, what will be its average velocity during this interval?
answer/question/discussion: First we take our overall distance traveled of 36cm/s and we subtract the acceleration of 24cm/s leaving us with a difference of 12cm/s if we divide this by the time of 3 seconds we have a average velocity during this time it would be 4cm/s per second
36 cm/s is not a distance. cm/s is not a unit of distance.
For what quantities could cm/s be the unit?
• How far will it travel during this interval?
answer/question/discussion: 36cm by taking the initial velocity of 12cm/s and the calculated velocity of 24cm/s and adding them gives us the total distance.
When you add two velocities you do not get a distance.
** **
30 minutes
** **
See if you can resolve the inconsistencies I point out in my notes within a reasonable time. You are of course welcome to ask questions. If you can't resolve these inconsistencies I'll send you a copy of the solution for self-critique, but you will learn more effectively if you can work it out yourself.
&#Please see my notes and submit a copy of this document with revisions and/or questions, and mark your insertions with &&&&. &#