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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):
The average velocity is the change in distance/change in time = (30 cm - 0 cm)/5 seconds = 6 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):
v_Ave = (vi + vf)/2, therefore vf = 2*v_Ave - vi = 2*(6 cm/s) - 0 cm/s = 12 cm/s or twice the average velocity.
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By how much did its velocity therefore change?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
Its velocity changed from 0 cm/s to 12 cm/s over a time interval of 5 seconds.
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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):
Its average rate of change is therefore (12 cm/s - 0 cm/s)/5 seconds = 2.4 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 velocity vs. clock time would have clock time on the x-axis and velocity on the y-axis. The plot would be a straight line starting at time 0 and velocity 0. Its slope would be 2.4 cm/s^2 because the rise would be the change in velocity and the run would be the change in time, which is the average rate of change of the velocity.
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