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course Phy 121
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
vAve=(20cm/s+10cm/s)/2=15cm/s
vAve=ds/dt
15cm/s=45cm/dt
Dt=3s
A=dv/dt
A=(20cm/s-10cm/s)/(3s)
A=3.33cm/s^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/s^2
dt=3s
vf=50cm/s
a=dv/dt
10cm/s^2=dv/3s
Dv=30cm/s
Dv=vf-v0
30cm/s=50cm/s-v0
20m/s=v0
vAve=(50cm/s+20cm/s)/2=35m/s
vAve=ds/dt
35m/s=ds/3s
105m=ds
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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
v0=0cm/s
a=20cm/s^2
vf^2=v0^2+2a*ds
vf^2=0m/s^2+2(20cm/s^2)*30cm
vf=34.6cm/s
vf=v0+a*dt
34.6cm/s=0cm/s+(20cm/s^2)*dt
(34.6cm/s)/(20cm/s^2)=dt
dt=1.73s
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Then for each situation answer the following:
Is it possible from this information to directly determine vAve?
answer/question/discussion:
vAve can be determined in the first situation by the equation vAVe=(vf+v0)/2. The second situation requires going through several equations to get to vAve, such as finding vf and then using (vf+v0)/2. The third situation requires that dt be found before vAve can be found.
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Is it possible to directly determine `dv?
answer/question/discussion:
The first situation only requires the simple calculation (vf-v0) to find dt. The second situation can also be directly found by the equation a=dv/dt/ The third scenario requires either vf or dt to be found before dv can be solved for.
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Very good. In addition to answering the slightly more limited questions asked, you provided well-constructed solutions for each case.
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This puts you ahead of the curve. Keep up the great work.
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