#$&*
phy 231
Your 'cq_1_09.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** **
cq_1_091
#$&*
phy 231
Your 'cq_1_09.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** **
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: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> :
We know our given values:
`Ds = 20cm v0 = 0 `dt = 2s
Use `ds/`dt to find vf
20cm / 2s = 10cm/s = vAve
(vAve * 2)- v0 = vf vf = (10 cm/s * 2) - 0 = 20 cm/s
`a = (vf-v0)/`dt
`a = (20 cm/s - 0)/ 2s = 10 cm/s^2
You could also use reasoning and say a object moving uniformly from rest will have 2 times the vf as the acceleration.
#$&*
If the time interval is in error so that it is 3% longer than the actual time interval, then what are the actual values of the final velocity and acceleration?
answer/question/discussion: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> :
You then use the same equations, but replace 2 with 2.06s and 1.94s:
(2.06) Use `ds/`dt to find vf
20cm / 2.06s = 9.71cm/s = vAve
(vAve * 2)- v0 = vf (9.71 cm/s * 2) - 0 = 19.42 cm/s
`a = (vf-v0)/`dt
`a = (19.42 cm/s - 0)/ 2.06s = 9.43 cm/s^2
Now 1.94 seconds:
vAve = 20 cm/1.94s = 10.31 cm/s
vf = (10.31 cm/s * 2) - 0 = 20.62 cm/s
`a = (20.62 cm/s - 0)/1.94 s = 10.63 cm/s^2
#$&*
What is the percent error in each?
answer/question/discussion: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> :
For 2.06 s:
`a = (9.43s - 10) / 10 * 100% = -5.7%
vf = (19.42 - 20) / 20 * 100% = -5.8%
For 1.94 s:
`a = (10.63 - 10)/10 *100% = 6.3%
Vf = (20.62 - 20 )/10 * 100% = 6.2%
#$&*
If the percent error is the same for both velocity and acceleration, explain why this must be so.
answer/question/discussion: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> :
They are almost the same, because both are experiencing a 3% error. However, `a % error is just slightly larger than vf.
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
MORE COMPLETE ANSWER:
All the calculations are correct; I kept my velocities and accelerations to two decimal points which make a slight difference in the % error calculation. I also needed to calculate the percent error for vAve. It’s with this percent error that you can see the doubling effect.
% error with vAve = (9.71 - 10)/10 = -2.9% If I drop the .01 from vAve to 9.7 cm/s then the % error is exactly 3%.
Likewise with acceleration: if I drop .03 from the value to 10.6 cm/s/s, then the % error is exactly 6%.
Either 2.9% to 6.3% or 3% to 6%, you see the doubling effect due to acceleration and time both having the 3% uncertainty.
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
#$&*
*#&!
&#Your work looks very good. Let me know if you have any questions. &#