Phy 231
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A ball accelerates uniformly as it rolls 20 cm down a ramp, starting from rest, in 2 seconds.
What are its average velocity, final velocity and acceleration?
answer/question/discussion: vAve = 20 cm / 2 s = 10 cm/s ; vf = vAve *2 – v0, which is 20 cm/s = 10 cm/s *2 ; a = (20 cm/s)/ 2 s = 10 cm/s^2.
If the time interval is in error so that it is 3% longer than the actual time interval, then what are the actual values of the final velocity and acceleration?
answer/question/discussion: All we have to do is subsitute 3% less of a time interval into each of the above equations, which gives us vAve = 20 cm/ 1.97 s = 10.152 cm/s ; vf = 10.152 * 2 – v0 which gives us vAve 20.304 = 10.152 *2; a = 20.304 / 1.97 s = 10.30 cm/s^2
What is the percent error in each?
answer/question/discussion: for vAVe percent error is [(20.304 – 20)/20.304] 100 = 1.50 % error ; for a percent error is [(10.30 – 10)/10.3] * 100 = 2.91 % error
If the percent error is the same for both velocity and acceleration, explain why this must be so.
answer/question/discussion:Not the same for reasons below.
If the percent errors are different explain why it must be so.
answer/question/discussion: Not the same, because it is true that both are affected by the difference in time; however, the average velocity is only calculated by the time value in one instance, while in the acceleration calculations the time interval value is used twice, which means that the is a greater discrepancy in the acceleration value than in the velocity as seen by the higher percent error.
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20 minutes
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&#Very good work. Let me know if you have questions. &#