CHAPTER 2 - 8

# CHAPTER 2 - 8 - x =x(65m/s(6.0s(9.7m/s(6.0s =x 565m 6 0 0 2...

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D a y  = 0 v 0  = 0 L TRAIN x 6  =  x 0  + (65 m/s)(6.0 s) +  !  (9.7 m/s 2 )(6.0 s) 2  =  x 0  + 565 m. Thus the distance moved is x 6  –  x 2  = 565 m – 149 m = 416 m =       4.2 × 10 2  m . 41. We use a coordinate system with the origin at the  initial position of the front of the train.  We can  find the acceleration of the train from the motion  up to the point where the front of the train passes  the worker: v 1 2  =  v 0 2  + 2 a ( D  – 0);    (25 m/s) 2  = 0 + 2 a (140 m – 0),  which gives   a  = 2.23 m/s 2 . Now we consider the motion of the last car, which starts at –  L , to the point where it passes the worker: v 2 2 v 0 2  + 2 a [ D  – (–  L )]    = 0 + 2(2.23 m/s 2 )(140 m + 75 m), which gives  v 2  =        31 m/s . 42. With the origin at the beginning of the reaction, the location when the brakes are applied is

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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 Berkeley.

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CHAPTER 2 - 8 - x =x(65m/s(6.0s(9.7m/s(6.0s =x 565m 6 0 0 2...

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