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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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?
vAve = ‘ds/’dt
vAve = 30cm/5s
vAve = 6cm/s
average velocity is determined by displacement divided by time. In this case the displacement is 30cm and the time is 5 seconds
answer/question/discussion:
• If the acceleration of the ball is uniform then its average velocity is equal to the average of its initial and final velocities.
Thus the average velocity could also by found by dividing the initial plus the final velocity by 2.
answer/question/discussion:
• You know its average velocity, and you know the initial velocity is zero. What therefore must be the final velocity?
12cm/s
This is determined by setting 6cm/s = (0cm/s + final) /2
muliply both sides by 2 to get the result of 12cm/s and subtract 0 cm/s from both sides to get final = 12cm/s
answer/question/discussion:
• By how much did its velocity therefore change?
The velocity changed by 12cm/s...final velocity minus initial velocity
answer/question/discussion:
• At what average rate did its velocity change with respect to clock time?
rate = 'dv/'dt
rate = 12cm/s / 5s
rate = 2.4cm/(s^2)
rate is the change in velocity divided by the time interval.
answer/question/discussion:
• What would a graph of its velocity vs. clock time look like? Give the best description you can.
the graph would show a linear line with a slope of 2.4
answer/question/discussion:
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.5 hours
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Would like to have feedback. Thanks, Joshua
&#This looks very good. Let me know if you have any questions. &#