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 down a ramp of length 30 cm, covering the distance in 5 seconds.
What is its average velocity?
The ball's average velocity is 6cm/s. I used the formula Average Velocity=displacement/time elapsed. aVe=30cm/5s
If the acceleration of the ball is uniform then its average velocity is equal to the average of its initial and final velocities.
If the you know the acceleration of the ball is uniform throughout the distance traveled, then Yes I do agree that the average velocity is equal to the average of its initial and final velocities. Average velocity deals in terms of displacement rather than total distance. This statement is possible, but its got me thinking. For example, in question 1 if the initial velocity is 0 and then the final velocity is 12, the average of the two would be 6cm/s, which in turn is the average velocity.
You know its average velocity, and you know the initial velocity is zero. What therefore must be the final velocity?
The final velocity of this scenario has to be 12 cm/s.
By how much did its velocity therefore change?
The velocity in this problem changed by 12.
At what average rate did its velocity change with respect to clock time?
For every second the velocity changed by 2.4.
That would be 2.4 cm/s^2. You need to use units at every step of your calculations. I've never seen a student who didn't use units who didn't make errors or tests that units calculations would have caught.
What would a graph of its velocity vs. clock time look like? Give the best description you can.
Velocity starts off at the bottom then as the seconds go by, the line for velocity increase until maximum speed. Then the ball would have leveled off at it's max velocity. So you would have seen a graph, that at first increased rapidly then leveled off once the ball reached its maximum velocity for the remainer of the 5 seconds.
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40 minutes
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I don't know if I was on the right track with this question. It shouldn't have seemed difficult for me. I was getting confused with the average velocity and velocity of the ball. I don't believe I calculated the final velocity right either, even though we have the Average velocity and the initial velocity, I didn't have a starting position or ending position. Need some feedback with this problem.
v0 = 0, vAve = 6 cm/s.
Acceleration is uniform so the v vs. t graph is linear.
The average of the initial and final values on an interval of a linear graph is the average value on that interval.
So v0 = 0 and vf must average out to 6 cm/s.
vf is therefore 12 cm/s.
In other words, your conclusion was correct. Be sure you completely understand the logic of this solution.
&#Let me know if you have questions. &#