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Phy 121
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** 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: ->->->->->->->->->->->-> :
v0 = 10 cm/s
vf = 20 cm/s
‘ds = 45 cm
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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: ->->->->->->->->->->->-> :
aAve = 10 cm/s^2
‘dt = 3 seconds
Vf = 50 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: ->->->->->->->->->->->-> :
‘ds = 30 cm
v0 = 0 cm/s (at rest)
aAve = 20 cm/s^2
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Then for each situation answer the following:
• Is it possible from this information to directly determine vAve?
answer/question/discussion: ->->->->->->->->->->->-> :
A) For situation one:
• v0 = 10 cm/s
• vf = 20 cm/s
• ‘ds = 45 cm
Yes it is possible to find vAve because acceleration is uniform so vAve = (vf - v0) / 2 = (20 cm/s - 10 cm/s) / 2 = 5 cm/s
B) Situation two:
• aAve = 10 cm/s^2
• ‘dt = 3 seconds
• Vf = 50 cm/s
Yes it is possible to find vAve after we find v0.
aAve = (vf- v0) / ‘dt
10 cm/s^2 = (50 cm/s - v0) / 3s
Multiple both sides by 3s so
30 cm/s = 50 cm/s - v0
Subtract 50 cm/s from both sides to get,
-20 cm/s = -v0
We are not finished because this gives us -v0 so, divide both sides by -1 to find v0
20 cm/s = v0
Now that we have both v0 and vf and acceleration is uniform, we can find vAve. vAve = (vf - v0) / 2 = (50 cm/s - 20 cm/s) / 2 = 15 cm/s
C) Situation 3:
• ‘ds = 30 cm
• v0 = 0 cm/s (at rest)
• aAve = 20 cm/s^2
It is not possible to find vAve here. If we had vf, it would be easy to find vAve by (v0 + vf) / 2. If we had ‘dt we could have either found vf by (aAve * ‘dt) + v0 = vf or vf = ‘ds / ‘dt. Once we had vf we could find vAve by (v0 + vf) / 2.
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• Is it possible to directly determine `dv?
answer/question/discussion: ->->->->->->->->->->->-> :
A) For situation one:
• v0 = 10 cm/s
• vf = 20 cm/s
• ‘ds = 45 cm
Yes it is possible to find ‘dv because both v0 and vf are given. ‘dv = (vf - v0) = 20 cm/s - 10cm/s = 10 cm/s.
B) Situation two:
• aAve = 10 cm/s^2
• ‘dt = 3 seconds
• Vf = 50 cm/s
Yes it is possible to find ‘dv after we find v0.
aAve = (vf- v0) / ‘dt
10 cm/s^2 = (50 cm/s - v0) / 3s
Multiple both sides by 3s so
30 cm/s = 50 cm/s - v0
Subtract 50 cm/s from both sides to get,
-20 cm/s = -v0
We are not finished because this gives us -v0 so, divide both sides by -1 to find v0
20 cm/s = v0
Now that we have both v0 and vf and acceleration is uniform, we can find ‘dv. ‘dv = (vf - v0) = 50 cm/s - 20 cm/s = 30 cm/s.
C) Situation 3:
• ‘ds = 30 cm
• v0 = 0 cm/s (at rest)
• aAve = 20 cm/s^2
It is not possible to find ‘dv here. If we had vf, it would be easy to find ‘dv by (vf-v0). If we had ‘dt we could have either found vf by (aAve * ‘dt) + v0 = vf, or vf = ‘ds / ‘dt. Once we had vf we could find ‘dv by (vf - v0).
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15 minutes
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&#Very good responses. Let me know if you have questions. &#