Phy 121
Your 'cq_1_09.1' 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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Seed Question 9.1
A ball accelerates uniformly as it rolls 20 cm down a ramp, starting from rest, in 2 seconds.
• What are its average velocity, final velocity and acceleration?
answer/question/discussion: ->->->->->->->->->->->-> :
We are given vo=0, ‘ds=20cm, and ‘dt=2 sec.
I found my average velocity by taking the ‘ds/’dt. So 20cm/2s= 10cm/s
To find my vf, I used the formula ‘ds=(vf+v0)/2+2’dt
I rearranged it to read as vf=’ds*2’dt-vo.
When I plugged in the given information I had, I got a value of vf=80cm/s.
If initial velocity is 0 and average velocity is 10 cm/s, then final velocity is 20 cm/s. You don't need formulas to reason this out, but it's very desirable to also be able to use the formulas.
You're making a good attempt with the formulas. However there is no formula like `ds = (vf + v0) / 2 + 2 `dt.
Among other things, the units of `ds are meters (units of length) while the unit of (vf + v0) / 2 is m/s (length units / time units), and the unit of 2 `dt is sec (time units). So you couldn't add (vf + v0) / 2 to 2 `dt; since the units are different these would be unlike terms.
If you multiplied (vf + v0) / 2 by 2 `dt you would get units m/s * s = m (length units) and your two sides would have the same units, but the formula would still be incorrect, since `ds = (vf + v0) / 2 * `dt (first equation of motion).
If you are going to use equations, use the four equations of uniformly accelerated motion as your basis.
An equation that applies to this situation, in which you know v0, `ds and `dt, is the first equation `ds = (vf + v0) / 2 * `dt. I think this is the equation you intended.
To solve this equation for vf:
Starting with
`ds = (vf + v0) / 2 * `dt
we multiply both sides by 2 / `dt to get
`ds * 2 / `dt = (vf + v0) / 2 * `dt * (2 / `dt), which simplifies to
(vf + v0) = 2 `ds / `dt. Then subtract v0 from both sides to get
vf = 2 `ds / `dt - v0.
Substitute and you will get vf = 20 cm/s.
To find my average acceleration, I took 80cm/s and divided it by my time (2sec). I got an acceleration of 40cm/s^2
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• If the time interval is in error so that it is 3% longer than the actual time interval, then what are the actual values of the final velocity and acceleration?
answer/question/discussion: ->->->->->->->->->->->-> :
I found what the actual time interval would be if it was 3% longer than the actual time interval. I got a value of 2.6 sec for this. I then recalculated my vf and got a value of 82.4 cm/s and an average acceleration of 41.2cm/s^2
You probably used 2.06 sec to get your final velocity.
You should then have divided by the same interval 2.06 sec, not by the previous 2 sec, to get your acceleration.
#$&*
• What is the percent error in each?
answer/question/discussion: ->->->->->->->->->->->-> :
To find the percent error, I used percent error= obtained result-accepted value/accepted value *100
My percent error for vf= -2.9%
My percent error for average acceleration = -2.9%
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• If the percent error is the same for both velocity and acceleration, explain why this must be so.
answer/question/discussion: ->->->->->->->->->->->-> :
I think that percent error is the same for velocity and acceleration because the acceleration is uniform.
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• If the percent errors are different explain why it must be so.
answer/question/discussion: ->->->->->->->->->->->-> :
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30-45 minutes
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Overall not bad, but your equation and your algebra weren't quite right (and weren't really necessary) in the first part. See my notes.
&#Please compare your solutions with the expanded discussion at the link
Solution
Self-critique your solutions, if this is necessary, according to the usual criteria. Insert any revisions, questions, etc. into a copy of this posted document. Mark any insertions with &&&& so they can be easily identified. &#