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: ->->->->->->->->->->->-> :
With a ball rolling down a ramp we can assume that acceleration is 9.8 m/s^2.
If it was in free fall we could make that assumption, but not on a ramp.
Average velocity = (0m/s + 9.8m/s) / 2
Average velocity = 9.8m/s / 2
Average velocity = 4.9 m/s
`ds = 2sec * 4.9 m/s
`ds = 9.8m
Vf= sqrt(v0^2 + 2a`ds)
Vf = sqrt(0 + 2 (9.8)(20cm))
Vf = sqrt(392)
Vf = 19.8 m/s
• 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 interval would be 2.03 seconds. The acceleration would remain at 9.8 m/s^2.
If the time interval changes the final velocity would be:
Vf= sqrt(0 + 2(9.947) 20)
Vf= sqrt(397.88)
Vf = 19.9
• What is the percent error in each?
answer/question/discussion: ->->->->->->->->->->->-> :
There is a .01% error between the two and there is a 0% error for acceleration.
• If the percent error is the same for both velocity and acceleration, explain why this must be so.
answer/question/discussion: ->->->->->->->->->->->-> :
It is not the same.
• If the percent errors are different explain why it must be so.
answer/question/discussion: ->->->->->->->->->->->-> :
The acceleration remained constant with gravity at 9.8 m/s^2. The velocity changed by .1.
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30 minutes
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Solution
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