cq_1_061

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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.

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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: ->->->->->->->->->->->-> :

v0 = 10 cm/s

vf = 20 cm/s

`dt = 45 cm

vAve = (10 cm/s + 20 cm/s) / 2 = 15 cm/s

`dv = 20 cm/s - 10 cm/s = 10 cm/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: ->->->->->->->->->->->-> :

`dt = 3 s

vf = 50 cm/s

a = 10 cm/s^2

10 = ( 50 - 20) / 3

v0 = 50 cm/s - 10 cm/s^2 * 3 s = 20 cm/s

vAve = (20 cm/s + 50 cm/s) / 2 = 35 cm/s

`dv = 50 cm/s - 20 cm/s = 30 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: ->->->->->->->->->->->-> :

`dt = 30 cm

v0 = 0 cm/s

a = 20 cm/s^2

vf = 20 cm/s^2 * 30 cm + 0 cm/s = 600 cm/s

vAve = (0 cm/s + 600 cm/s) / 30 cm/s = 20 cm/s^2

`dv = 600 cm/s - 0 cm/s = 600 cm/s

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@& All you're really being asked to do is identify the quantities, which you have done correctly.

Note, however, that 20 cm/s^2 * 30 cm is 600 cm^2 / s^2, not 600 cm/s.

The equation you are using is for vf^2, not vf.*@

Then for each situation answer the following:

• Is it possible from this information to directly determine vAve?

answer/question/discussion: ->->->->->->->->->->->-> :

Yes. I will determine the vAve in each of the above problems.

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• Is it possible to directly determine `dv?

answer/question/discussion: ->->->->->->->->->->->-> :

Yes. I will determine `dv above.

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@& No need for revision, but check my notes as well as the discussion at the link.*@

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&#See any notes I might have inserted into your document, and before looking at the link below see if you can modify your solutions. If there are no notes, this does not mean that your solution is completely correct.

Then please compare your old and new solutions with the expanded discussion at the link

Solution

Self-critique your solutions, if this is necessary, according to the usual criteria. Insert any revisions, questions, etc. into a copy of this posted document. Mark any insertions with &&&& so they can be easily identified.

If your solution is completely consistent with the given solution, you need do nothing further with this problem. &#

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