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phy121
Your 'question form' 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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question on qa06
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below is my revised work on the question I had difficulty on. Please let me know if my reasoning is correct or if I still have errors.
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Question: `q012. An object starts at position x = +10 cm with a velocity of +5 cm/s, and accelerates uniformly, ending up at position x = -30 cm after a time interval of 8 seconds. What is its velocity at this point, and what was its acceleration during this interval?
Principles of Physics students should not spend over 5 minutes on this problem, General College Physics students should not spend over 10 minutes.
University Physics students are expected to be able to solve this problem, but if it hasn't been solved within 15 minutes, should submit their best thinking and await the instructor's notes.
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Your solution:
`ds = -30 - 10 = -40 cm
V0 = +5cm/sec
`dt = 8 sec
Vel = `ds/`dt = -40/8 = 5 cm/sec
Vf^2 = v0^2 + 2a*`ds
Vf^2 = 25 + 2a * -40cm
I am confused
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Self-critique (if necessary):
I am not sure how to proceed with this problem????
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This would be vAve.
vAve = 5 cm/s would imply that velocity is unchanging, in which case it would not be possible to end up at the negative position.
In any case -40 cm / (5 s) = -5 cm/s, not 5 cm/s.
Knowing vAve and v0 you can find vf. There is no need to resort to the fourth equation of uniformly accelerated motion, though that is an option is you have sufficient information. At this point, though, you don't. Both vf and a are at this point unknown.
If you find vf, then you can apply either the second or fourth equation of motion, or you can easily reason out a from the definitions.
I recommend that you submit a copy of this question, your original solution, my notes and your modified solution (and/or questions) using the question form. However if you're sure you have a good solution and understand it completely, as you well may, there is no need to submit anything additional.
#### revised answer
The x0 = +10 cm which is the initial starting position, xf = -30 cm which is the ending position, so .30 cm - 10 cm = -40 cm = `ds
So we have
`ds = -40 cm
V0 = +5cm/sec
`dt = 8 secondsd
‘ds = [(v0 + vf)/2] * ‘dt
‘ds/’dt = (v0 + vf)/2
vo + vf = 2 (`ds/`dt)
vf = 2 (‘ds/’dt) - v0
vf = 2 (-40 cm / 8sec) - 5cm/sec
vf = -15 cm/sec
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You could also have concluded this from the previous result that vAve = -5 cm/s. If v0 = 5 cm/s and vAve = -5 cm/s, then vf has to be -15 cm/s; since vAve is halfway between v0 and vf, the change from v0 to vAve is equal to the change from vAve to vf.
Of course you can also use the equations to obtain this result.
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‘dv = vf - v0
`dv = -15 cm/sec - d cm/sec = -20 cm/sec
a = `dv/`dt
a = -20 cm/sec / 8 sec
a = -2.5 cm/sec^2
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Very good. Check my notes.
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