Phy 121
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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: ->->->->->->->->->->->-> : Average velocity is 10 cm/s. Final velocity is 20 cm/s. Acceleration is 10 cm/s^2.
`ds = 20 cm
V0 = 0 cm/s
`dt = 2 s
`ds = vAve * `dt
20 cm = vAve * 2 s
20 cm/2s = vAve
vAve = 10 cm/s
vAve = (v0 + vf) / 2
10 cm/s = (0 cm/s + vf) / 2
10 cm/s = vf / 2
Vf = 20 cm/s
Vf = v0 + a * `dt
20 cm/s = 0 cm/s + a * 2 s
20 cm/s = a * 2 s
20 cm/s / 2 s = a
10 cm / s^2 = a
• 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 final velocity would then be 20.6 cm/s and the acceleration would then be 10.3 cm/s^2.
`dt * 1.03 = 2 s
`dt = 2 s / 1.03
`dt = 1.94
• What is the percent error in each?
answer/question/discussion: ->->->->->->->->->->->-> : Approximately 3% error in each
10.3 – 10 = .3
.3 / 10.3 = .29
20.6 – 20 = .6
.6 / 20.6 = .29
• If the percent error is the same for both velocity and acceleration, explain why this must be so.
answer/question/discussion: ->->->->->->->->->->->-> : The 3% error is never squared. The time is off by 3% so when that is multiplied
• If the percent errors are different explain why it must be so.
answer/question/discussion: ->->->->->->->->->->->-> :
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10 minutes
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Solution
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