course Phy 201
Below is my corrections for cq_1_03.1
cq_1_031Phy 201
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: average velocity = displacement/ time elapsed so average velocity is 30cm/5sec=6cm/sec
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: I dont really know this answer but I think that the average velocity plus the initial velocity would be the final velocity so 6 + 0 = 6cm/sec
#### The initial velocity of the ball is 0 and acceleration is uniform therefore the final velocity = vf=2*vAve = 2(6cm/sec = 12cm/s
the ave velocity would be between the initial and final velocities, so final velocity couldn't be ave velocity + initial velocity
however you're on the right track in your overall approach
By how much did its velocity therefore change?
answer/question/discussion: The velocity did not change
#### The velocity of the ball changed from its initial velocity v0 to its final velocity vf = 12cm/s so dv =vf-v0 = 12cm/s 0cm/s = 12cm/s
you claim different initial and final velocities, and this constitutes a change in velocity
At what average rate did its velocity change with respect to clock time?
answer/question/discussion: it stayed the same.
#### ave rate = change in velocity/ change in clock time = 12cm/s / 5s = 2.4 cm/s^2
What would a graph of its velocity vs. clock time look like? Give the best description you can.
answer/question/discussion: it would be a straight line because it is constant speed it would rise up to the 30 cm mark on the y axis and it would go to the left on the x axis to the 5 sec mark.
#### v vs t graph with the slope being constant the graph of v vs t rises along a straight line from v = 0 to v = 30 cm/s while clock time t changes by 5 sec. The rise of the graph between these points is 30 cm/s, the run is 5 s , and the slope of the line is rise/run = 30 cm/s / 5s = 6cm/s^2. The slope is identical to the uniform acceleration.
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30 min
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Very good revisions.