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: ->->->->->->->->->->->-> :
v0=0m/s
`ds=20cm
`dt=2s
20/2= 10cm/s which is vAve
So (0+vf)/2= 10 so vf= 20cm/s
A=20cm/s/2s= 10cm/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: ->->->->->->->->->->->-> :
To find the error in the time we need to subtract out 3% of the time so 2-.03= 1.97s; so vAve is 20/1.97= 10.152cm/s.
The vf can be found by multiplying the new vAve by abd subtracting v0 out= 20.304.
Now to find a we can divide `dv by `dt= 10.30cm/s^2
What is the percent error in each?
answer/question/discussion: ->->->->->->->->->->->-> : to find the percent error we will subtract the original value from the one that is 3% longer and divide by the longer value and multiply by 100 so….
vAve= 20.304-20*100= 1.5%
Acceleration: 10.3-10/10.3*100= 2.9%
If the percent error is the same for both velocity and acceleration, explain why this must be so.
answer/question/discussion: ->->->->->->->->->->->-> : the percent errors should vary because velocity only uses the time interval once whereas acceleration uses it twice which raises its percent error.
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If the percent errors are different explain why it must be so.
answer/question/discussion: ->->->->->->->->->->->-> :
They are different because although they share the same time values in their calculations, the vAve only uses the time value one time in order to find the vAve value whereas acceleration is using the time value twice. The acceleration has a greater percent error because of this.
#$&*
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20 min
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