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Phy 121
Your 'cq_1_05.1' 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_05.1_labelMessages **
The problem:
A ball accelerates at 8 cm/s^2 for 3 seconds, starting with velocity 12 cm/s.
• What will be its velocity after the 3 seconds has elapsed?
answer/question/discussion: ->->->->->->->->->->->-> :
aAve = (vf - v0) / ‘dt.
Here we are trying to find vf after 3 seconds.
8 cm/s^2 = (vf - 12 cm/s) / 3s.
Multiply both sides by 3s so,
24 cm/s = vf - 12cm/s
Add 12 cm/s to both sides and,
36 cm/s = vf
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• Assuming that acceleration is constant, what will be its average velocity during this interval?
answer/question/discussion: ->->->->->->->->->->->-> :
vAve = (v0 + vf) / 2
vAve = (12cm/s + 36 cm/s) / 2 = 24 cm/s
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• How far will it travel during this interval?
answer/question/discussion: ->->->->->->->->->->->-> :
To find how long it will travel during this interval we must first find ‘ds for both velocities.
‘ds = v0 * ‘dt
‘ds = 12 cm/s * 3s
‘ds = 36 cm
‘ds = vf * ‘dt
‘ds = 36 cm/s * 3 s
‘ds = 108 cm
Now we know that it travelled 108 cm - 36 cm = 72 cm during this interval.
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Good.
Note that you would get the same result by averaging initial and final velocities, which when acceleration is uniform gives you the average velocity.
You would get average velocity of 24 m/s, which multiplied by 3 s gives you the 72 m displacement.
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&#This looks good. See my notes. Let me know if you have any questions. &#