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Assignment 9 Seed Question 1
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:
To find the vf, we can use the equation ‘ds = vAve * ‘dt. We know ‘ds= 20 cm and ‘dt = 2 sec, so we can plug these in. We also know v0= 0 cm/s, so we plug this in for the equation vAve = (vf + v0) / 2. We can now solve for vf, and when we do so, we get vf= 20 cm/s. To get vAve, we plug in (20cm/s + 0cm/s) / 2 = 10cm/s. Now, we can solve for a by dividing ‘dv/’dt = (20cm/s-0cm/s) /2sec = 10 cm/s/s.
Good. Note that you could have found vAve directly by dividing 20 cm by 2 seconds. Nothing wrong with what you did but be sure you understand it both ways.
• 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: The actual values of the final velocity and acceleration would be between some intervals if this were true. Therefore, the time interval would be between 1.94 and 2.06 sec, found by multiplying 2 * ,03 to get the 3 % of 2 sec, and getting .06 sec from this tells us that we have a percent error of +- 3% for our time interval. We would need to solve for vf and a differently for such a case. The vf would then be found with ‘ds= vAve * ‘dt still, but with two different values for ‘dt. If we plug in 1.94 sec for ‘dt, we get a vf of 20.62 cm/s. If we plug in 2.06 sec for ‘dt, we get a vf of 19.42 cm/s. We now solve for acceleration using these new vf values and get a= (20.62 cm/s – 0cm/s) / 1.94 sec = 20.63 cm/s/s. We do the same thing for the upper time interval and find an a value of (19.420 cm/s – 0 cm/s) / 2.06 sec = 9.43 cm/s/s.
• What is the percent error in each?
answer/question/discussion: The percent error in each is +- 3%, because this was the percent error of the time interval.
(20.62 cm/s – 0cm/s) / 1.94 sec is not equal to 20.63 cm/s/s. Closer to 21.2 cm/s. Better check your calculations. Your 9.43 cm/s is not correct either, for a very similar reason.
21.2 cm/s differs from the original 20 cm/s by 1.2 cm/s, which is more than 3% of that result.
What therefore is the percent error?
Note that you divided by the time interval to find the average velocity, and thus the final velocity.
You then divided the change in velocity by the time interval.
So your calculation involved two separate divisions by the time interval.
The result is that the error in the time interval is compounded twice, so that the percent error in the final result is greater than the percent error in the time interval.
• If the percent error is the same for both velocity and acceleration, explain why this must be so.
answer/question/discussion: The percent error for velocity and acceleration is the same because they are both dependent on time, which has a percent error of +- 3 %.
• If the percent errors are different explain why it must be so.
answer/question/discussion: The percent errors are not different.
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It took me about 15 minutes to answer these questions.
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Good work for the most part. The percent errors for the acceleration, however, are not calculated correctly, and the percent errors are in fact different.
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