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Phy 231
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: ->->->->->->->->->->->-> :
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: ->->->->->->->->->->->-> :
a = 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, a = 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: ->->->->->->->->->->->-> :
vAve1 = (vf + v0) / 2 = (20 cm/2 + 10 cm/s) / 2 = 15 cm/s
Vave2,
1st 'dv = a*dt = vf -v0
v0 = vf -dv
vAve = (vf + v0) / 2
'dv = 10 cm/s^2*3s = 30 cm/s
v0 = 50 cm/s - 30 cm/s = 20 cm/s
vAve2 =(50 cm/s + 20 cm/s) / 2 = 35 cm /s
vAve3
vf^2 = v0^2 + 2*a*'ds
vf = (v0^2 +2*a*'ds)^.5 = (0^2 + 2*20cm/s^2*30 cm)^.5 = ((1200 cm/s)^2)^.5
vAve = (v0 + vf) / 2 = ( 34.641 cm/s /2) = 17.32 cm/s
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Is it possible to directly determine `dv?
answer/question/discussion: ->->->->->->->->->->->-> :
Situation 1 yes it posiible to directly determine 'dv = vf - v0
Situation 2 yes it is possible by 'dv = a*'dt given acceleration times the change in time
Situation 3 yes it is possible by vf^2 = v0^2 + 2*a*'ds
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1/2 hour
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Flow diagrams are helping alot.
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Good. They are very helpful to many students, not as helpful to some. I'm glad they're helpful to you.
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