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PHY 201
Your 'cq_1_08.1' 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 Questions 1_08.1 submitted 11 Feb 11 around 11:04 PM.
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1 hr
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Seed Questions 1_08.1
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You should enter your answers using the text editor or word processor. You will then copy-and-paste it into the box below, and submit.
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A ball is tossed upward with an initial velocity of 25 meters / second. Assume that the acceleration of gravity is 10 m/s^2 downward.
• What will be the velocity of the ball after one second?
answer/question/discussion: ->->->->->->->->->->->-> :
Given: v0 = 25 m/s, g = -10 m/s^2, t = (1s & 2 s), vf =?
vf = v0 +gt
vf = 25 m/s + (-10 m/s^2) (1s)
vf = 25 m/s - 10 m/s
vf = 15 m/s
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@& You have a good solution. However there's an easier way to have answered this question.
You don't need equations to answer this question. If the ball's acceleration is 10 m/s^2, this means that the velocity changes by 10 m/s every second; since the acceleration is downward the change will be in that direction.
You're good with the equations, but you'll be better off if you also think about what the quantities and the results mean. I advocate reasoning before equations, then checking your reasoning with the equations.
Some questions on the Major Quiz spcifically ask you to reason out your results, some specifically ask you to use the equations. It's important to be able to do both.
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• What will be its velocity at the end of two seconds?
answer/question/discussion: ->->->->->->->->->->->-> :
Given: v0 = 25 m/s, g = -10 m/s^2, t = (1s & 2 s), vf =?
vf = v0 +gt
vf = 25 m/s + (-10 m/s^2) (2s)
vf = 25 m/s - 20 m/s
vf = 5 m/s
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• During the first two seconds, what therefore is its average velocity?
answer/question/discussion: ->->->->->->->->->->->-> :
vAve = (vf + v0) / 2s
vAve = (5 m/s + 25 m/s) / 2s
vAve = 30 m/s / 2 s
vAve = 15 m
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• How far does it therefore rise in the first two seconds?
answer/question/discussion: ->->->->->->->->->->->-> :
y - y0 = v0t - ½ g t^2
y - y0 = 25 m/s (2s) - ½ (10 m/s^2) (2s)^2
y - y0 = 50 m - ½ (40 m)
y - y0 = 50 m - 20 m
y - y0 = 30 m
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• What will be its velocity at the end of a additional second, and at the end of one more additional second?
answer/question/discussion: ->->->->->->->->->->->-> :
v = v0 - gt
v = 25 m/s - (10 m/s^2)(3s)
v = 25 m/s - 30 m/s
v = -5 m/s
v = v0 - gt
v = 25 m/s - (10 m/s^2)(4s)
v = 25 m/s - 40 m/s
v = -15 m/s
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• At what instant does the ball reach its maximum height, and how high has it risen by that instant?
answer/question/discussion: ->->->->->->->->->->->-> :
v = v0 - gt
0 = 25 m/s - (10 m/s^2) t
t = 25 m/s / 10 m/s^2
t = 2.5 s
y - y0 = v0t - ½ gt^2
y - y0 = (25 m/s) (2.5 s) - ½ (10 m/s^2) (2.5 s)^2
y - y0 = 62.5 m - ½ (62.5 m)
y - y0 = 62.5 m - 31.25 m
y - y0 = 31.25 m
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• What is its average velocity for the first four seconds, and how high is it at the end of the fourth second?
answer/question/discussion: ->->->->->->->->->->->-> :
vAve - (v0 + vf) / 2
vAve = (25 m/s -15 m/s) /2
vAve = 10 m/s / 2
vAve = 5 m/s
y - y0 = v0t - ½ gt^2
y - y0 = (25 m/s) (4 s) - ½ (10 m/s^2) (4 s)^2
y - y0 = 100 m - ½ (160 m)
y - y0 = 100 m - 80 m
y - y0 = 20 m
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• How high will it be at the end of the sixth second?
answer/question/discussion: ->->->->->->->->->->->-> :
y - y0 = v0t - ½ gt^2
y - y0 = (25 m/s) (6 s) - ½ (10 m/s^2) (6 s)^2
y - y0 = 150 m - ½ (360 m)
y - y0 = 150 m - 180 m
y - y0 = -30 m
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*#&!
@& Excellent use of equations, and all your answers are correct. Very good job.
However see my notes about the importance of direct reasoning.
See also the link below. There's no need for any revision, unless you have questions.
&#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. &#
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