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.
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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=10cm.s
vf=20cm/s
`ds=45cm
‘dt=1.6s
A=6.25cm/s^2
vAve=20+10/2= 15cm/s
We know that `ds/vAve=`dt so 45cm/15cm/s= 3s
`dv= 20-10= 10cm/s
aAve=`dv/`dt so 10cm/s/3s= 3.33cm/s^2
Yes I could directly determine vAve because I was given both v0 and vf values
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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: ->->->->->->->->->->->-> :
v0= 20cm/s
vf=50cm/s
a=10cm/s^2
`ds=105cm
`dt= 3s
If we know acceleration and time and one velocity we are able to solve for the other velocity. So 10cm/s^2= (50cm/s-v0)/3s when we solve for v0 we get 20cm/s. If we find vAve= 20+50cm/s/2= 35cm/s Then we can find `ds by multiplying vAve by `dt= 35cm/s*3s= 105cm
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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: ->->->->->->->->->->->-> :
v0= 0cm/s
vf=34.6cm/s
a=20cm/s^2
`dt=1.73s
`ds=30cm
We can use equation to find vf
Vf^2= 0^2+2(20)(30cm)= 34.6cm/s
Now we can find vAve (34.6cm/s)/2=17.3cm/s then I can solve for `dt by dividing `ds by vAve= 30cm/ 17.3cm/s= 1.73s
I could not directly determine vAve from these values. I had to do some solving for toher values first.
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Then for each situation answer the following:
• Is it possible from this information to directly determine vAve?
answer/question/discussion: ->->->->->->->->->->->-> :
The first one yes, second no.
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• Is it possible to directly determine `dv?
answer/question/discussion: ->->->->->->->->->->->-> :
The first yes second no.
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15 min
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&#Good work. Let me know if you have questions. &#