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Phy 121
Your 'cq_1_02.1' 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_02.1_labelMessages.txt **
The problem:
A ball starts with velocity 4 cm/sec and ends with a velocity of 10 cm/sec.
• What is your best guess about the ball's average velocity?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
Since no time is given, I add v_0 + v_f and divide by 2.
(v_0 + v_f)/2=(4cm/s+10cm/s)/2=(14cm/s)/2=7cm/s=average velocity
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• Without further information, why is this just a guess?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
The equation we are given to find average velocity is the change in velocity with respect to time is the change in velocity divided by the change in time. We only know half the data required the change in velocity. We do not know anything about the time.
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• If it takes 3 seconds to get from the first velocity to the second, then what is your best guess about how far it traveled during that time?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
To find how far an object traveled in a time interval we need to know vAve and ‘dt.
‘ds=vAve*’dt.=7cm/s*3s=21cm
This is also a best guess because 7cm/s is a best guess.
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• At what average rate did its velocity change with respect to clock time during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
The average rate of velocity change with respect to clock time is
aAve=’dv/’dt=(v_f - v_0)/(t_1 - t_0)=(10cm/s-4cm/s)/(3s)=(14cm/s)/(3s)=4.67cm/s^2
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20 minutes
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&#Very good responses. Let me know if you have questions. &#