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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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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: ->->->->->->->->->->->-> :
25 m/s - 10 m/s2 = 15 m/s3
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• What will be its velocity at the end of two seconds?
answer/question/discussion: ->->->->->->->->->->->-> :
15 m/s3 - 10 m/s = 5 m/s3
@& 25 m/s - 10 m/s^2 cannot be calculated. The units are different, so these are unlike terms.
m/s - m/s^2 is not in any case equal to m/s^3. (suggest a review of addition and multiplication of fractions, which will help you avoid some of the frequently-recurring errors students typically make with units).
Your idea, however, is correct.
10 m/s^2 means that, every second, the velocity changes by 10 m/s in the downward direction.
So in the first second the velocity changes by -10 m/s, from 25 m/s to 25 m/s - 10 m/s = 15 m/s (note how 25 m/s and 10 m/s have the same units so they can be combined).
That reasoning will apply for the next several questions.
However m/s^3 will never occur in your work on these questions.
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• During the first two seconds, what therefore is its average velocity?
answer/question/discussion: ->->->->->->->->->->->-> :
(15 m/s3 - 5 m/s3) / 2 = 5 m/s3
@& That would be correct for the two-second interval between t = 1 second and t = 3 seconds.
However that interval does not span the first two seconds (though it does overlap).*@
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• How far does it therefore rise in the first two seconds?
answer/question/discussion: ->->->->->->->->->->->-> :
5 m/s3 x 2 = 10 m/s3
@& Good reasoning but see the preceding, and note that m/s^3 is not a unit of distance.
The unit of ave velocity will be m/s, and the unit of your 2-second time interval is s. The unit of your answer will be m/s * s, which you can then simplify.*@
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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: ->->->->->->->->->->->-> :
15 m/s and 20 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: ->->->->->->->->->->->-> :
Maximum height 2.5 seconds in and it rises to 70 meters
@& That is the right time, but you don't show how you got 70 meters, and that answer isn't correct.*@
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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: ->->->->->->->->->->->-> :
average velocity = (25 m/s -
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• How high will it be at the end of the sixth second?
answer/question/discussion: ->->->->->->->->->->->-> :
90 meters
@& That average velocity will be negative, as will the change in position.*@
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45 minutes
@& You're on the right track, but need to refine your units and some of your calculations.
&#Please see my notes and, unless my notes indicate that revision is optional, submit a copy of this document with revisions and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes.
If my notes indicate that revision is optional, use your own judgement as to whether a revision will benefit you. &#
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