#$&*
Phy 241
Your 'cq_1_09.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** CQ_1_09.1_labelMessages **
A ball accelerates uniformly as it rolls 20 cm down a ramp, starting from rest, in 2 seconds.
What are its average velocity, final velocity and acceleration?
answer/question/discussion: ->->->->->->->->->->->-> :
v_0 = 0
'ds = 20 cm
'dt = 2 seconds
v_ave = 'ds/'dt
= 20 cm / 2 seconds
= 10 cm/s
'ds = v_0 * 'dt + .5 * a * 'dt^2
a = 2 * ['ds - (v_0 * 'dt)] / 'dt^2
= 2 * [2 m] / 4 s^2
@&
`ds is 20 cm, not 2 m.
*@
= 1 m/s^2
v_f = v_0 + a * 'dt
= 0 + 1 m/s^2 * 2 seconds
= 2 m/s
#$&*
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: ->->->->->->->->->->->-> :
2 s * .03 = 0.6
2 + 0.06 = 2.06 is corrected time interval
a = 2 * ['ds - (v_0 * 'dt)] / 'dt^2
= 2 * [2.06 s] / 4.2436 s^2
@&
2.06 s is not a value of `ds.
*@
= 0.97 m/s^2
v_f = v_0 + a * 'dt
= 0 + 0.97 m/s^2 * 2.06
= 2 m/s
#$&*
What is the percent error in each?
answer/question/discussion: ->->->->->->->->->->->-> :
3 % error in acceleration.
0 % error in v_f
#$&*
If the percent error is the same for both velocity and acceleration, explain why this must be so.
answer/question/discussion: ->->->->->->->->->->->-> :
They aren't. I got 3% error for acceleration and 0 % error for final velocity. If the were the same, then I guess that would be because they both rely on each other. In order to solve for V_f, you have to find acceleration first.
#$&*
If the percent errors are different explain why it must be so.
answer/question/discussion: ->->->->->->->->->->->-> :
The reason they are different is becuase if an object is accelerating, it will reach a time destination quicker and velocity. If the object doesn't get to the time destination till alittle bit later, it's velocity will still be the same but the acceleration slightly slower.
#$&*
cq_1_091
#$&*
Phy 241
Your 'cq_1_09.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** CQ_1_09.1_labelMessages **
A ball accelerates uniformly as it rolls 20 cm down a ramp, starting from rest, in 2 seconds.
What are its average velocity, final velocity and acceleration?
answer/question/discussion: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> :
v_0 = 0
'ds = 20 cm
'dt = 2 seconds
v_ave = 'ds/'dt
= 20 cm / 2 seconds
= 10 cm/s
'ds = v_0 * 'dt + .5 * a * 'dt^2
a = 2 * ['ds - (v_0 * 'dt)] / 'dt^2
= 2 * [2 m] / 4 s^2
= 1 m/s^2
v_f = v_0 + a * 'dt
= 0 + 1 m/s^2 * 2 seconds
= 2 m/s
@&
&&&&
v_0 = 0
'ds = 20 cm
'dt = 2 seconds
v_ave = 'ds/'dt
= 20 cm / 2 seconds
= 10 cm/s
'ds = v_0 * 'dt + .5 * a * 'dt^2
a = 2 * ['ds - (v_0 * 'dt)] / 'dt^2
= 2 * [0.2 m] / 4 s^2
= 0.1 m/s^2
v_f = v_0 + a * 'dt
= 0 + 0.1 m/s^2 * 2 seconds
= 0.2 m/s
&&&&
Having found vAve, which is 10 cm/s, you can use the uniformity of velocity and the initial velocity to reason out vf, which is easily found to
be 20 cm/s.
Your 2 m/s result for vf is not consistent with your result for vAve. An initial velocity of 0 and a final velocity of 2 m/s implies an average
velocity of 1 m/s, not 10 cm/s.
You error occurs on two levels. On the first level your results fail the test of consistency. On the second level, you use the equations correctly
but 20 cm is not 2 meters; this is more of a clerical error that should have been caught by straight reasoning.
When solving a problem, especially on a test where it counts, if a quantity can be reasoned out easily, it should be. And if a quantity can be
obtained using equations it should be. If the two approaches don't agree, you know something is amiss.
*@
#$&*
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: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> :
2 s * .03 = 0.06
2 + 0.6 = 2.06 is corrected time interval
a = 2 * ['ds - (v_0 * 'dt)] / 'dt^2
= 2 * [2.06 s] / 4.2436 s^2
= 0.97 m/s^2
&&&&
2 s * .03 = 0.06 s
2 s + 0.6 = 2.06 s is corrected time interval
a = 2 * ['ds - (v_0 * 'dt)] / 'dt^2
= 2 * [0.2 m / (2.06 s)^2]
= 0.09425 m/2s
&&&&
@&
The units of this result would be s^-1, not m/s^2. This shows that either your equation is dimensionally inconsistent,
or you have used the wrong dimensions for at least one quantity.
Most of your subsequent results are based on this erroneous value of the acceleration and need to be revised.
*@
@&
The acceleration is easily reasoned out based on the definitions.
You can get the average velocity, then the final velocity, then the acceleration without resorting to equations.
You should, of course, also solve using the equations and resolve any inconsistencies.
*@
v_f = v_0 + a * 'dt
= 0 + 0.97 m/s^2 * 2.06
= 2 m/s
#$&*
What is the percent error in each?
answer/question/discussion: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> :
3 % error in acceleration.
0 % error in v_f
#$&*
If the percent error is the same for both velocity and acceleration, explain why this must be so.
answer/question/discussion: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> :
They aren't. I got 3% error for acceleration and 0 % error for final velocity. If the were the same, then I guess that would be because they both rely on each other. In order to solve for V_f, you have to find acceleration first.
#$&*
If the percent errors are different explain why it must be so.
answer/question/discussion: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> :
The reason they are different is becuase if an object is accelerating, it will reach a time destination quicker and velocity. If the object doesn't get to the time destination till alittle bit later, it's velocity will still be the same but the acceleration slightly slower.
#$&*
*#&!
@&
You've made a couple of errors regarding `ds for this motion.
Your conclusions are not correct.
A longer time interval results in a lesser average velocity, a lesser final velocity and a lesser change in velocity, therefore in a lesser acceleration.
&#Please see my notes and submit a copy of this document with revisions, comments and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes. &#
*@