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.
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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: In this problem I think you use the equation vf = v0 + at then just fill in what you know and solve. Initial velocity is 12cm/s time is 3sec and acceleration is 8cm/s^2 so vf= 12cm/s + (8cm/s^2)(3sec) so vf= 36 cm/s
• Assuming that acceleration is constant, what will be its average velocity during this interval?
answer/question/discussion: vAve = vf + v0/2 so again just fill in the equation vAve = 36cm/s + 12cm/s / 2 so vAve is 24 cm/s
• How far will it travel during this interval?
answer/question/discussion: I think to get this answer you just multiply the acceleration by the time period. So 8cm/s^2 * 3sec = 24cm/s
Acceleration is rate of change of velocity with respect to clock time (aAve = `dv / `dt) so this calculation would give you change in velocity (`dv = aAve * `dt), not change in position.
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20 min
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Very well reasoned. Your last answer is a valid calculation, but not the one asked for, so see the link below.
&#At least part of your solution does not agree with the solution and comments given at the link below. You should view the solution at that link and self-critique as indicated there.
Solution
This link also expands on these topics and alerts you to many of the common errors made by students in the first part of this course. &#