#$&*
Phy 121
Your 'cq_1_06.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** CQ_1_06.1_labelMessages **
For each situation state which of the five quantities v0, vf, `ds, `dt and a are given, and give the value of each.
• A ball accelerates uniformly from 10 cm/s to 20 cm/s while traveling 45 cm.
answer/question/discussion: ->->->->->->->->->->->-> :
v0=10cm/s
vf=20cm/s
‘ds=45cm
‘dt=45cm/((10cm/s+20cm/s)/2)=3s
‘dv=45cm/3s=15cm/s
@&
`dv is the change in velocity.
45 cm / (3 s) is change in position / change in clock time, which is the average rate of change of position with respect to clock time, or average velocity.
What is the initial velocity?
What is the final velocity?
What therefore is the change in velocity?
*@
a=(15cm/s)/3s=5cm/s^2
#$&*
• 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: ->->->->->->->->->->->-> :
vf=50cm/s
‘dt=3s
a=10cm/s^2
‘dv=10cm/s^2*3s=30cm/s
v0=50cm/s-30cm/s=20cm/s
vAve=(20cm/s+50cm/s)/2=35cm/s
‘ds=35cm/s*3s=105cm
@&
Good
*@
#$&*
• A ball travels 30 cm along an incline, starting from rest, while accelerating at 20 cm/s^2.
answer/question/discussion: ->->->->->->->->->->->-> :
v0=0cm/s
‘ds=30cm
a=20cm/s^2
vf^2=sqrt(0cm/s)^2+2(20cm/s^2)(30cm)
sqrt(vf^2)=sqrt(1200cm^2/s^2)
vf=34.6cm/s
'dt=(34.6cm/s-0cm/s)/(20cm/s^2)=1.7s
#$&*
Then for each situation answer the following:
• Is it possible from this information to directly determine vAve?
answer/question/discussion: ->->->->->->->->->->->-> :
Yes I did so in question #2
#$&*
• Is it possible to directly determine `dv?
answer/question/discussion: ->->->->->->->->->->->-> :
Yes I did so in question #1 and #2
#$&*
** **
30 min
** **
&#Your work looks good. See my notes. Let me know if you have any questions. &#