PHY 201
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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: ->->->->->->->->->->->-> :
I am given the displacement 20 cm, the initial velocity 0 cm/s, and the time interval 2 seconds, then I can solve for the rest of the variables. I am given the time interval and the displacement, so I can solve for the average velocity by: 20 cm / 2 s = 10 cm/s. Then if I know the average velocity and was given the initial velocity, then the final velocity is twice that of the average velocity: 10 m/s * 2 = 20 cm/s. Then, the change in velocity is 20 cm/s as well. Finally, if I know the change in velocity and the change in time, then I can calculate the acceleration: 20 cm/s / 2 s = 10 cm/s^2. Therefore the average velocity is: 10 m/s, the final velocity is 20 cm/s, and the acceleration is 10 cm/s^2.
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• 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 time is at an error, which it is supost to be 3 % longer than the 2 seconds, so: 2 * .03 = .06 + 2 = 2.06 seconds. Then, if the time interval is 2.06 s then the final velocity is: 20 cm/s / 2.06 s = 9.71 cm/s * 2 = 19.42 cm/s / 2.06 s = 9.43 cm/s^2. So, the final velocity is 19.42 cm/s and the acceleration is 9.43 cm/s^2.
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• What is the percent error in each?
answer/question/discussion: ->->->->->->->->->->->-> :
The percent error in the average velocity is the number without the error divided by the number with the error: 10 cm/s / 9.71 cm/s = 1.03 is the percent error of the average velocity. Then, the percent error for the acceleration is: 10 cm/s^2 / 9.43 cm/s^2 = 1.06.
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• If the percent error is the same for both velocity and acceleration, explain why this must be so.
answer/question/discussion: ->->->->->->->->->->->-> :
The percent error is not exactly the same, but approximately the same. The percent error is the same because if the error was on the time, the time affects both the average velocity and the acceleration. So, if will change both of them at the same rate.
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• If the percent errors are different explain why it must be so.
answer/question/discussion: ->->->->->->->->->->->-> :
Mine are a little off, but still very much alike. I am not sure how precise you want the percent errors to be.
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30 Minutes
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&#This looks very good. Let me know if you have any questions. &#