Your 'cq_1_8.1' 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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cq_1_81
Your 'cq_1_8.1' 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: 25m/s – 10m/s/s after 1 sec = 15m/s ???
· What will be its velocity at the end of two seconds?
answer/question/discussion: 25m/s – 10m/s/s – 10m/s/s (after 2 seconds) = 5m/s/ ???
· During the first two seconds, what therefore is its average velocity?
answer/question/discussion: 25m/s + 5m/s / 2 = 15m/s
· How far does it therefore rise in the first two seconds?
answer/question/discussion: ‘dv / ‘dt = 20m/s / 2sec = 10m
20m/s / 2sec = 10 (m/s) / s = 10 m/s * 1/s = 10 m/s^2, not 10 m.
This would be the acceleration of an object whose velocity changes by 20 m/s in 2 s. If the upward direction is positive, this object has a velocity change of -20 m/s so its acceleration is -10 m/s.
However the question asked for displacement. You can't get displacement from change in velocity and change in clock time.
You can get displacement from average velocity and change in clock time.
What therefore is the displacement of this object?
&&&& The average velocity should be (25m/s + 5m/s) / 2 = 15m/s
&&&& The displacement should be 15m/s / 2s =7.5m
15 m/s / (2 s) would be 7.5 m/s^2, not 7.5 m. Done correctly the unit calculation would alert you to an error.
vAve = `ds /` dt so `ds = vAve * `dt, not vAve / `dt.
· What will be its velocity at the end of a additional second, and at the end of one more additional second?
answer/question/discussion: after 3 seconds = -5m/s, after 4 seconds = -15m/s
· At what instant does the ball reach its maximum height, and how high has it risen by that instant?
answer/question/discussion: the ball reaches its maximum height at v = 0. v^2 – vo^2 / 2a = (25m/s)^2 / 2(-10m/s) = 31.25m
· What is its average velocity for the first four seconds, and how high is it at the end of the fourth second? (25+-5) / 2 =10m/s
answer/question/discussion: (25+-5) / 2 =10m/s
displacement = average velocity * change in clock time
You have just found the average velocities for these intervals.
What is the change in clock time for each of the intervals?
&&&& The change in clock time is 4 sec.
What therefore is the displacement for each of the intervals?
&&&& Displacement = 10m/s / 4s = 2.5m
Your error here was similar to your error on the preceding.
· How high will it be at the end of the sixth second?
answer/question/discussion: ???
&&&& after 6 seconds, -35m/s. vAve = (-35m/s + 25m/s) / 2 = -10m/s / 2 = -5m/s
&&&& displacement should be –5m/s / 6sec = -.83m
same error here
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1 hour, maybe a little less.
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I spent way too much time on this. I just don't get it. I tried to use the book and it helped a little with the height of the ball at v=0 but i'm kinda lost.
You got everything except the answers related to displacement. In each of those situations you know average velocity and time interval, so you can use the definition of average velocity to find the displacement.
You also know initial velocity, final velocity and change in clock time for each of those intervals, so you could use the equatios of uniformly accelerated motion. However it's better that you reason out the answers conceptually, using the definitions.
Please see my notes and submit a copy of this document with revisions and/or questions, and mark your insertions with &&&&.
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Good, but you made one error, which however you repeated in three places. It should take you just a couple of minutes to correct it, but to avoid this error in subsequent work be very sure you understand what my notes are telling you.