Phy 231
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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: 6 cm/s
Because the ball travelled 30 cm in 5 s, when can determine the change in distance divided by the change in time, which gives us the average velocity.
vAve = 30 cm / 5 s = 6 cm/s
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: 12 cm/s
Using the given equation above we can state:
6 cm/s = (initial vel + final vel) / 2
12 cm/s = 0 cm/s + final vel
12cm/s = final vel
This fits with the other definition we have learned that vAve * 2 = final velocity, and likewise final velocity / 2 = vAve
By how much did its velocity therefore change?
answer/question/discussion: 12 cm/s
The velocity went from a initial start of 0 cm/s to a final velocity of 12 cm/s. This is a change of 12 cm/s.
At what average rate did its velocity change with respect to clock time?
answer/question/discussion: 2.4 cm/s/s (or cm/s^2)
The velocity went from a initial start of 0 cm/s to a final velocity of 12 cm/s. This is a change of 12 cm/s over the course of 5 s. The change in velocity (or avg acceleration) is equal to 12 cm/s / 5 s, or 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: We are told this is constant acceleration, so the line will be straight. Since we started at clock time zero and initial velocity zero, the line will start at the origin. It will move upward and to the right in (again) a straight line. The line stops at the point (5,12). It is increasing at a constant rate.
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15 minutes
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&#Very good work. Let me know if you have questions. &#