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Phy 231
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.
** CQ_1_08.1_labelMessages **
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: ->->->->->->->->->->->-> :
vf1 = v0 + a*'dt = 25 m/s + (-10 m/s)* 1 s = 15 m/s
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Good, but you don't need a formula to figure this out.
In one second the ball's velocity will decrease by 10 m/s, from 25 m/s to 15 m/s.
I advocate using common sense and basic definitions when it is possible to do so, as this generally increases your understanding in ways formulas do not. Of course formulas are necessary in many cases, and if used correctly they can also expand understanding.
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What will be its velocity at the end of two seconds?
answer/question/discussion: ->->->->->->->->->->->-> :
vf2 = vo + a*'dt = 25 m/s + -10 m/s * 2s = 5 m/s
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During the first two seconds, what therefore is its average velocity?
answer/question/discussion: ->->->->->->->->->->->-> :
vAve 2seconds = (vf + vo) /2 = (5 m/s + 25 m/s) / 2 = 15 m/s
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How far does it therefore rise in the first two seconds?
answer/question/discussion: ->->->->->->->->->->->-> :
dy = vave * 'dt = 15 m/s * 2s = 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: ->->->->->->->->->->->-> :
vf3 = v0 + a*'dt = 25 m/s + (-10 m/s^2) * 3s = - 5m/s (its falling back to earth)
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Good.
Just to reinforce my previous note, you should also understand the alternative to the formulas. From the definition of acceleration you can reason out that the velocities at the ends of the first four seconds are 15, 5, -5 and -10 m/s. Average velocities can be easily calculated over any interval between these clock times, and displacements as well.
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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: ->->->->->->->->->->->-> :
vf = v0 + a*t, vf =0
0 = 25 m/s + (-10 m/s^2) *t
10 m/s^2 * t = 25 m/s
t = 25 m/s / 10 m/s^2 = 2.5 s
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Direct reasoning:
It takes 2.5 seconds to lose 25 m/s of velocity, during which average velocity is 12.5 m/s (half of the initial), leading to the conclusion that displacement is 2.5 s * 12.5 m/s.
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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: ->->->->->->->->->->->-> :
Vave4sec = (vf + vo) / 2 = ( -15 m/s + 25 m/s) / 2 = 5 m/s
'dy = vave * 'dt = 5 m/s * 4 s = 20 m
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How high will it be at the end of the sixth second?
answer/question/discussion: ->->->->->->->->->->->-> :
vf6 = v0 + a*'dt = 25 m/s + (-10 m/s^2) * 6 s = -35 m/s
vAve6 = (vf + vo) / 2 = (-35 m/s + 25 m/s) / 2 = -5m/s
dy = vAve6 * 'dt = -5 m/s * 6 s = -30 m, 30 meters below the release point.
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&#Good responses. See my notes and let me know if you have questions. &#