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course phy 121
10/5 around 5:00
cq_1_082#$&*
phy 121
Your 'cq_1_08.2' 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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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: ->->->->->->->->->->->-> :
vAve = `ds / `dt
= 20cm / 2s
= 10 cm/s
vAve = (vf + v0) / 2
2 * vAve = vf +v0
2 * vAve - v0 = vf
2 * 10 cm/s - 0 = vf
vf = 20 cm/s
a = (20 cm/s - 0) / 2s
= 20cm/s /2s
= 10 cm/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: ->->->->->->->->->->->-> :
2 s * 3% = 0.06 s
20 cm / 2.06 s = 9.71cm/s for vAve
vf = 2 * 9.71 cm/s - 0 = 19.42cm/s
a = (19.42 - 0 ) / 2.06s = 9.43 cm/s^2
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What is the percent error in each?
answer/question/discussion: ->->->->->->->->->->->-> :
the percent error is 3.
one has a 3% error, the other does not. Be sure you have actually calculated both.
O
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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: ->->->->->->->->->->->-> :
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If the percent errors are different explain why it must be so.
answer/question/discussion: ->->->->->->->->->->->-> :
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>????? i am not sure how to get percent error.
Divide the error by the original value and express as a percent. More detail is given at the link.
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20 min
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&#See any notes I might have inserted into your document, and before looking at the link below see if you can modify your solutions. If there are no notes, this does not mean that your solution is completely correct.
Then please compare your old and new 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.
If your solution is completely consistent with the given solution, you need do nothing further with this problem. &#
cq_1_082
#$&*
phy 121
Your 'cq_1_08.2' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** **
A ball is tossed upward at 15 meters / second from a height of 12 meters above the ground. Assume a uniform downward acceleration of 10 m/s^2 (an approximation within 2% of the 9.8 m/s^2 acceleration of gravity).
How high does it rise and how long does it take to get to its highest point?
answer/question/discussion: ->->->->->->->->->->->-> :
v0=0
a=10m/s^2
vf=15m/s
`dt = (vf - v0) / a
= (15m/s - 0) / 10m/s^2
= 15m/s / 10m/s^2
= 1.5s
`ds = (vf +v0) / 2 * `dt
= (15m/s + 0) /2 * 1.5s
= 7.5m/s * 1.5 s
= 11.25 m
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?????????
I don't think I am doing this right, I am confused on what is what in the question..
How fast is it then going when it hits the ground, and how long after the initial toss does it first strike the ground?
answer/question/discussion: ->->->->->->->->->->->-> :
&&&&&& v0 = +15 m/s
`ds = -12 m
a = -10 m/s^2
vf^2 = v0^2 + 2 a `ds
vf = +- sqrt( v0^2 + 2 a `ds)
= +- sqrt( (15 m/s)^2 + 2 * (-10 m/s^2) * (-12 m) )
= +- sqrt( 225 m^2/s^2 + 240 m^2/s^2)
= +- sqrt(465 m^2/s^2)
= +-21.6 m/s
vAve = (+15 m/s + (-21.6 m/s) ) / 2 = -3.3 m/s,
`dt = `ds / vAve = -12 m / (3.3 m/s) = 3.6 sec
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At what clock time(s) will the speed of the ball be 5 meters / second?
answer/question/discussion: ->->->->->->->->->->->-> :
&&&&&&
vf = v0 + a `dt
dt = (vf – v0) / a
= ( 5 m/s – 15m/s) / -10m/s
= 1 s
= (-5m/s – 15 m/s) / -10m/s
=2s
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At what clock time(s) will the ball be 20 meters above the ground?
How high will it be at the end of the sixth second?
answer/question/discussion: ->->->->->->->->->->->-> :
&&&&& `ds = 8 m,
a = -10 m/s^2 and
v0 = 15 m/s.
vf^2 = v0^2 + 2 a `ds, so
vf = +-sqrt(v0^2 + 2 a `ds) =
+-sqrt( (15 m/s)^2 + 2 * (-10 m/s^2) * 8 m) =
+- sqrt( 65 m^2 / s^2) = +-8.1 m/s
`dt = (vf - v0) / a = (8.1 m/s - 15 m/s) / (-10 m/s^2) = .69 s
15 cm/s + (-10 m/s^2) * 6 s = -45 m/s.
vAve = (15 m/s + (-45 m/s) ) / 2 = -15 m/s,
height = -15 m/s * 6 s = - 90 m.
78 m below the level of the ground.
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?????????
I don't think I am doing this right, I am confused on what is what in the question..
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You had a good start. Check the discussion at the link and submit a revision.
&#See any notes I might have inserted into your document, and before looking at the link below see if you can modify your solutions. If there are no notes, this does not mean that your solution is completely correct.
Then please compare your old and new 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.
If your solution is completely consistent with the given solution, you need do nothing further with this problem. &#
Very good revision.