#$&*
course phy121
12/8/13 1:30 pm
For each situation state which of the five quantities v0, vf, `ds, `dt and a are given, and give the value of each. •A ball accelerates uniformly from 10 cm/s to 20 cm/s while traveling 45 cm.
answer/question/discussion:
`ds = 45 cm
v0 = 10 cm/sec
vf = 20 cm/sec
`dv = 20cm/sec - 10cm/sec = 10 cm/sec
`dt = `ds/vAve
`dt = `ds/vAve
= 45cm/15 cm/sec
`dt = 3 sec
a = `dv/`dt = 10cm/sec/(3sec) = 3.33 cm/sec^2
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• A ball accelerates uniformly at 10 cm/s^2 for 3 seconds, and at the end of this interval is moving at 50 cm/s.
answer/question/discussion:
a = 10cm/sec^2
`dt = 3 sec
`vf = 50 cm/sec
`vf = `v0 +a*`dt
50cm/sec = v0 + 10cm/sec^2 * 3sec
50cm/sec = v0 + 30cm/sec
v0 = 20 cm/sec
‘dv = 50cm/sec - 20cm/sec = 30cm/sec
vAve = 50cm/sec + 20cm/sec = 70cm/sec /2 = 35cm/sec
`ds = vAve * `dt
= 35cm/sec * 3 sec
`ds = 105 cm
v = `ds/`dt
= 105cm/3sec
v = 3.33 cm/sec
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• A ball travels 30 cm along an incline, starting from rest, while accelerating at 20 cm/s^2.
answer/question/discussion:
`ds = 30cm
a = 20cm/sec^2
v0 = 0 cm/sec
vf^2 = v0^2 + 2*a*`ds
= 0^2 + 2*20*30
=sqrt 1200
`vf = 34.647 cm/sec
vAve = (34.64+0)/2
vAve = 17.32 cm/sec
‘dv = 34.64-0 = 34.64 cm/sec
a = `dv/`dt
20cm/sec^2 = 34.64 cm/sec / `dt
20`dt = 34.64 cm/sec
`dt = 1.732 sec
v = `ds/`dt
= 30cm/1.732 sec
v= 17.32 cm/sec
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Then for each situation answer the following:
• Is it possible from this information to directly determine vAve?
answer/question/discussion:
yes (vf+v0)/2
#$&*
• Is it possible to directly determine `dv?
answer/question/discussion:
yes, vf-v0
#$&*
"
Self-critique (if necessary):
------------------------------------------------
Self-critique rating:
#$&*
course phy121
12/8/13 1:30 pm
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:
`ds = 45 cm
v0 = 10 cm/sec
vf = 20 cm/sec
`dv = 20cm/sec - 10cm/sec = 10 cm/sec
`dt = `ds/vAve
`dt = `ds/vAve
= 45cm/15 cm/sec
`dt = 3 sec
a = `dv/`dt = 10cm/sec/(3sec) = 3.33 cm/sec^2
#$&*
• A ball accelerates uniformly at 10 cm/s^2 for 3 seconds, and at the end of this interval is moving at 50 cm/s.
answer/question/discussion:
a = 10cm/sec^2
`dt = 3 sec
`vf = 50 cm/sec
`vf = `v0 +a*`dt
50cm/sec = v0 + 10cm/sec^2 * 3sec
50cm/sec = v0 + 30cm/sec
v0 = 20 cm/sec
‘dv = 50cm/sec - 20cm/sec = 30cm/sec
vAve = 50cm/sec + 20cm/sec = 70cm/sec /2 = 35cm/sec
`ds = vAve * `dt
= 35cm/sec * 3 sec
`ds = 105 cm
v = `ds/`dt
= 105cm/3sec
v = 3.33 cm/sec
#$&*
• A ball travels 30 cm along an incline, starting from rest, while accelerating at 20 cm/s^2.
answer/question/discussion:
`ds = 30cm
a = 20cm/sec^2
v0 = 0 cm/sec
vf^2 = v0^2 + 2*a*`ds
= 0^2 + 2*20*30
=sqrt 1200
`vf = 34.647 cm/sec
vAve = (34.64+0)/2
vAve = 17.32 cm/sec
‘dv = 34.64-0 = 34.64 cm/sec
a = `dv/`dt
20cm/sec^2 = 34.64 cm/sec / `dt
20`dt = 34.64 cm/sec
`dt = 1.732 sec
v = `ds/`dt
= 30cm/1.732 sec
v= 17.32 cm/sec
#$&*
Then for each situation answer the following:
• Is it possible from this information to directly determine vAve?
answer/question/discussion:
yes (vf+v0)/2
#$&*
• Is it possible to directly determine `dv?
answer/question/discussion:
yes, vf-v0
#$&*
@&
Very good. You actually went beyond the stated question, which is fine.
See also the discussion at
&#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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