CHAPTER 2 - 5 - 14 =?v/?t=(44m/s0)/(27s0)=1.6m/s . 2

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0 10 20 30 0 1 2 3 4 5 6 0 1 2 3 4 5 6 t  (s) v  (m/s) a  (m/s 2 ) acceleration velocity Æ 1 4  = ? v /? t  = (44 m/s – 0)/(27 s – 0) =         1.6 m/s 2 . Note that we cannot add the four average accelerations and divide by 4. 26. ( a ) We take the average velocity during a time interval as the instantaneous velocity at the  midpoint of the time interval: v midpoint  =  æ  =  x / t Thus for the first interval we have v 0.125 s  = (0.11 m – 0)/(0.25 s – 0) = 0.44 m/s. ( b ) We take the average acceleration during a time interval as the instantaneous acceleration at the  midpoint of the time interval: a midpoint  =  Æ  =  v / t Thus for the first interval in the velocity column we have a 0.25 s  = (1.4 m/s – 0.44 m/s)/(0.375 s – 0.125 s) = 3.8 m/s 2 . The results are presented in the following table and graph.   t   (s)     x    (m)     t   (s)     v (m/s) t   (s)     a (m/s 2 ) 0.0  0.0 0.0 0.0 0.25 0.11 0.125 0.44 0.25 3.8 0.50 0.46 0.375 1.4 0.50 4.0 0.75  1.06 0.625 2.4 0.75 4.5 1.00 1.94 0.875
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This note was uploaded on 03/08/2010 for the course PHYSICS 7A/7B taught by Professor All during the Fall '08 term at University of California, Berkeley.

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CHAPTER 2 - 5 - 14 =?v/?t=(44m/s0)/(27s0)=1.6m/s . 2

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