PHY 231
Your 'cq_1_06.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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For each situation state which of the five quantities v0, vf, `ds, `dt and a are given, and give the value of each.
• A ball accelerates uniformly from 10 cm/s to 20 cm/s while traveling 45 cm.
answer/question/discussion: ->->->->->->->->->->->-> :
Given:
v0 = 10 cm/s
vf = 20 cm/s
'ds = 45 cm
'dt = Not Given
a = Not Given
vAve = (( 10 cm/s + 20 cm/s ) / 2 )
vAve = (( 30 cm/s ) / 2 )
vAve = 15 cm/s ( since acceleration is uniform )
Change in Time/ 'dt:
45 cm ( distance travelled ) = 15 cm/s * x ( unknown number of seconds/ change in time )
45 cm = 15 cm/ s * x
3 s = x
'dt = 3 seconds
Change in Position/ 'ds ( Verification )
vAve = 'ds / 'dt
15 cm/s = 'ds / 3 s
45 cm = 'ds
a ( Expressed in cm/ s^2 ):
a = (( 20 cm/ s - 10 cm/s ) / 3 s )
= (( 10 cm/s ) / 3 s )
= ( 10 cm/s / 3 s ))
a = 10/3 cm/s^2
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• A ball accelerates uniformly at 10 cm/s^2 for 3 seconds, and at the end of this interval is moving at 50 cm/s.
answer/question/discussion: ->->->->->->->->->->->-> :
Given:
v0 = Not Given
vf = 50 cm/s
'ds = Not Given
'dt = 3 seconds
a = 10 cm/s ^2
v0:
v0 = 50 cm/s - ( 10 cm/s ^2 * 3 s )
v0 = 50 cm/s - ( 30 cm/s )
v0 = 20 cm/s
vAve (( Final + Initial ) / 2 ) because acceleration is uniform:
vAve = (( 20 cm/s + 50 cm/s ) / 2 )
vAve = (( 70 cm/s ) / 2 )
vAve = 35 cm/s
'ds:
'ds = vAve * 'dt
'ds = 35 cm/s * 3 s
'ds = 70 cm
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• A ball travels 30 cm along an incline, starting from rest, while accelerating at 20 cm/s^2.
answer/question/discussion: ->->->->->->->->->->->-> :
Given:
v0 = 0 cm/s
vf = Not Given
'ds = 30 cm
'dt = Not Given
a = 20 cm/s ^2
Assuming uniform acceleration, although it does not seem so since the ball is on an incline.
#### ( I am not sure if this equation is right. I have been trying to do this part of the problem for 25 minutes and I am stuck. But, I will try to solve it with the distance equation I remember from high school written below. Otherwise, I am stuck. )
Distance = 1/2 ( a )( t ^2 )
30 cm = 1/2 ( 20 cm/ s^2 )( t^2)
30 cm = ( 10 cm/ s^2 )( t^2)
3 s^2 = t^2
1.732050808 s = Time Required to Accelerate 30 cm from rest
'dt = 1.732050808 s
vf = ( 1.732050808 s * 20 cm/s ^ 2 )
vf = 34.64101615 cm/ s
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Then for each situation answer the following:
• Is it possible from this information to directly determine vAve?
answer/question/discussion: ->->->->->->->->->->->-> :
In the first situation, it is possible to determine the average velocity because both the initial and final velocities are present and the acceleration is uniform. In the second situation, it is not possible to directly derive the average velocity because we must first find the initial velocity. Once again, we can solve for the average velocity after finding the initial because the acceleration is uniform. In the third scenario, it is not possible to directly determine the average velocity because we are not given the time interval, how long it took to reach 30 cm, or the final velocity. Then, from using my distance formula, I believe I am able to derive the change in time and the final velocity. Uniform acceleration is assumed. However, not one of the three previously variables is given. Calculations are shown above.
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• Is it possible to directly determine `dv?
answer/question/discussion: ->->->->->->->->->->->-> :
In the first scenario, we can determine 'dv simply by subtracting 10 cm/s from 20 cm/s. In the second scenario, we must solve for the initial velocity prior to determining 'dv. And in the third situation, we cannot directly determine the change in velocity because the final velocity is not given. Calculations are shown above.
#$&*
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50 minutes
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Solution
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