#$&*
course Phy 231
1-30-11 about 9:00pm
3 cm / sec to 11 cm / sec in 9 seconds, then at what average rate is the velocity changing?A ball rolling from rest down a constant incline requires 8.6 seconds to roll the 68 centimeter length of the incline.
• What is its average velocity?
An object which accelerates uniformly from rest will attain a final velocity which is double its average velocity.
• What therefore is the final velocity of this ball?
• What average rate is the velocity of the ball therefore changing?
An automobile accelerates uniformly down a constant incline, starting from rest. It requires 15 seconds to cover a distance of 172 meters. At what average rate is the velocity of the automobile therefore changing?
(11cm/sec – 3cm/sec)/9sec= (8cm/sec)/9sec= .89cm/sec/sec
68cm/8.6sec= 7.9cm/sec=ave_v
v_f= 7.9cm/sec * 2= 15.8cm/sec
(15.8cm/sec – 7.9cm/sec)/8.6sec= (7.9cm/sec)/8.6sec= .92cm/sec/sec
@& For your numerator you have subtracted average velocity from final velocity.
Change in velocity is final velocity - initial velocity, not final velocity - average velocity.*@
ave_v= 172m/15sec= 11.47m/sec
v_f= 11.47cm/sec * 2= 22.94cm/sec
(22.94cm/sec – 11.47cm/sec)/15sec= (11.47cm/sec)/15sec= .76cm/sec/sec
@& You made the same error on the last question as on the one preceding it.
It's easily fixed, but very important. I expect that you will easily understand how your calculation is not consistent with the definintion of average rate, but if you're not absolutely clear on this let me know.
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