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phy121
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.
** CQ_1_06.1_labelMessages **
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: ->->->->->->->->->->->-> :
v0= 10cm/s
vf= 20cm/s
‘ds= 45cm
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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: ->->->->->->->->->->->-> :
a= 10cm/s^2
‘dt= 3 seconds
Vf= 50cm/s
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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: ->->->->->->->->->->->-> :
‘ds= 30cm
V0= 0cm/s
A= 20cm/s^2
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Then for each situation answer the following:
• Is it possible from this information to directly determine vAve?
answer/question/discussion: ->->->->->->->->->->->-> :
First question: yes it is possible to determine vAve by adding the initial and final velocities and divide by 2.
(10cm/s+20cm/s)/2= 30cm/s /2= 15cm/s
Question 2: yes it is possible, by using the formula a= ‘dv/’dt, this will find the initial velocity, then do the same for the first question.
10cm/s^2= (50cm/s-v0)/3s
10cm/s^2*3s= 50cm/s-v0
30cm/s-50cm/s= -v0
V0= 20cm/s
vAve= (50cm/s+20cm/s)/2= 70cm/s /2= 35cm/s
Question 3: yes it is possible, by using the formula vf^2 = v0^2 + 2 a `ds, to find the final velocity then finding the average.
Vf^2= 0cm/s^2+ 2*20cm/s^2*30cm
Vf^2= 1200cm/s
Vf= sqrt1200cm/s
Vf= 34.64cm/s
vAve= (34.64cm/s+0cm/s)/2= 17.32cm/s
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• Is it possible to directly determine `dv?
answer/question/discussion: ->->->->->->->->->->->-> :
For all the questions, it is possible to determine ‘dv, since we found the initial and final velocities for each question in the previous question to find vAve. Therefore:
Question 1: 20cm/s-10cm/s= 10cm/s
Question 2: 50cm/s-20cm/s= 30cm/s
Question 3: 34.64cm/s-0cm/s= 34.64cm/s
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&#Very good responses. Let me know if you have questions. &#