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A ball accelerates at 8 cm/s^2 for 3 seconds, starting with velocity 12 cm/s.
• What will be its velocity after the 3 seconds has elapsed?
8cm/s per second for 3 secs, 8 * 3 = 24cm/s
12cm/s + 24cm/s = 36cm/s
• Assuming that acceleration is constant, what will be its average velocity during this interval?
vAve = ( 36cm/s – 12cm/s ) / 2 = 12cm/s
This cannot be the average velocity.
The initial velocity on this interval is 12 cm/s, and the final is 36 cm/s.
The average velocity will lie between the initial and final velocities.
The average velocity is not equal to 12 cm/s or to any lesser velocity.
Subtraction is not used in the process of averaging two numbers. Subtraction is used, in both the numerator and denominator, when finding the average rate of change of one quantity with respect to another.
• How far will it travel during this interval?
36cm/s * 3 secs = 108 cm
36 cm/s is the maximum velocity on this interval. Your result indicates how far it would travel in 3 seconds, if it traveled at its maximum velocity.
However reaches its maximum velocity only at the end of the interval, so your result is an overestimate.
What quantity would you multiply by the time interval to get the displacement, or change in position?
This question can be answered by using the definition of velocity.
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5 mins
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I'm not sure that I did this correctly.
You made a couple of errors.
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