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Phy 231
Your 'cq_1_03.1' 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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The problem:
A ball starts with velocity 0 and accelerates uniformly down a ramp of length 30 cm, covering the distance in 5 seconds.
• What is its average velocity?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
average velocity = (v0 + vf) / 2
vAve = ?
v0 = 0 cm/s
vf = final position / final time = 30 cm / 5 s = 6 cm/s
vAve = (0 + 6 cm/s) / 2 = 3 cm/s
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• If the acceleration of the ball is uniform then its average velocity is equal to the average of its initial and final velocities.
You know its average velocity, and you know the initial velocity is zero.
What therefore must be the final velocity?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
vAve = (v0 + vf) / 2
v0 = 0
vAve = 3 cm/s
(vAve)*2 = vf
Vf = (3 cm/s)*(2) = 6 cm/s
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• By how much did its velocity therefore change?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
Therefore its velocity changed 6 cm/s
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• At what average rate did its velocity change with respect to clock time?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
average rate velocity changes = average acceleration = a
a = [(2*vAve) – v0] / dt
a = ?
vAve = 3 cm/s
v0 = 0
dt = 5 s
a = [(2)*(3 cm/s)] / 5 s = (6 cm/s) / 5 s = 1.2 cm/s^2
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• What would a graph of its velocity vs. clock time look like? Give the best description you can.
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
The graph of its velocity vs. clock time would be an increasing straight line starting at the point (0,0) and ending at the point (5,30). The line would have aslope of 1.2 cm/s^2 because the slope of the line is equal to the acceleration of the object. The velocity coordinates or vertical coordinates will be found along the y-axis. The clock time coordinates or the horizontal coordinates will be found along the x-axis.
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15 min
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&#Very good responses. Let me know if you have questions. &#