CHAPTER 2 - 7 - 35....

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35. For the constant acceleration the average speed is  ! ( v  +  v 0 ), thus  x   ! ( v  +  v 0 ) t : ! (0 + 22.0 m/s)(5.00 s) =        55.0 m . 36. We find the speed of the car from v 2  =  v 0 2  + 2 a ( x 1  –  x 0 );  0  =  v 0 2  + 2(– 7.00 m/s 2 )(75 m), which gives         v 0  = 32 m/s . 37. We convert the units for the speed:  (55 km/h)/(3.6 ks/h) = 15.3 m/s. ( a ) We find the distance the car travels before stopping from v 2  =  v 0 2  + 2 a ( x 1  –  x 0 );   0  = (15.3 m/s) 2  + 2(– 0.50 m/s 2 )( x 1  –  x 0 ), which gives         x 1  –  x 0  = 2.3 × 10 2  m . ( b ) We find the time it takes to stop the car from v  =  v 0   +   at  ; 0 = 15.3 m/s + (– 0.50 m/s 2 ) t , which gives        t  = 31 s . ( c ) With the origin at the beginning of the coast, we find the position at a time  t  from   x    =  v
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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 - 7 - 35....

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