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PHY 121
Your 'cq_1_06.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** CQ_1_06.1_labelMessages **
For each situation state which of the five quantities v0, vf, `ds, `dt and a are given, and give the value of each.
• A ball accelerates uniformly from 10 cm/s to 20 cm/s while traveling 45 cm.
answer/question/discussion: ->->->->->->->->->->->-> :
The average velocity is v0 (10cm/s) + vf (20cm/s) / 2 = 15cm/s
From the average velocity we can determine the time elapsed through `dt= `ds (45cm) / vAve (15cm/s) = 3 sec
The average acceleration is aAve = `dv OR (vf (20cm/s) - v0 (10cm/s) / `dt (3 sec) = 3.3 cm/s/s
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• A ball accelerates uniformly at 10 cm/s^2 for 3 seconds, and at the end of this interval is moving at 50 cm/s.
answer/question/discussion: ->->->->->->->->->->->-> :
First we find the change in velocity `dv = aAve (10cm/s^2) * `dt (3 sec) = 30cm/s
If the change in velocity is 30cm/s then the initial velocity must be 20cm/s
Now we can find the average velocity with vAve = v0 (20cm/s) + vf (50cm/s) / 2 = 35 cm/s
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• A ball travels 30 cm along an incline, starting from rest, while accelerating at 20 cm/s^2.
answer/question/discussion: ->->->->->->->->->->->-> :
Without an initial velocity or a change in time I do not know how to solve this, I believe there is a subsequent assignment that gives an equation on how to solve with only the change in position and the acceleration.
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Then for each situation answer the following:
• Is it possible from this information to directly determine vAve?
answer/question/discussion: ->->->->->->->->->->->-> :
To determine the average velocity we would need either the initial and final velocities or the change in time, which we do not currently have.
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• Is it possible to directly determine `dv?
answer/question/discussion: ->->->->->->->->->->->-> :
To find the change in velocity we would also need either the change in time or the initial and final velocities.
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&#This looks very good. Let me know if you have any questions. &#