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CHAPTER 2 - 6

# CHAPTER 2 - 6 - (c v=dx/dt=A 18Bt =A 18B(5.0s =A(450s)B...

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( c ) For the given time we have v  = d x /d t  =  A  + 18 Bt 2  =  A  + 18 B (5.0 s) 2  =        A  + (450 s 2 ) B ; a  = d v /d t  = 36 Bt  = 36 B (5.0 s) =        (180 s) B . ( d ) We find the velocity by differentiating: v  = d x /d t  =  A  – 3 Bt –4 . 29. We find the acceleration from v  =  v 0   +   a ( t  –  t 0 ); 21 m/s = 12 m/s +  a (6.0 s), which gives        a  = 1.5 m/s 2 . We find the distance traveled from     x   ! ( v  +  v 0 ) t     ! (21 m/s + 12 m/s)(6.0 s) =        99 m . 30. We find the acceleration (assumed constant) from v 2  =  v 0 2  + 2 a ( x 2  –  x 1 );  0 = (25 m/s) 2  + 2 a (75 m), which gives        = – 4.2 m/s 2 . 31. We find the length of the runway from v 2  =  v 0 2  + 2 aL (32 m/s) 2  = 0 + 2(3.0 m/s 2 ) L , which gives        L  = 1.7 × 10 2  m . 32. We find the average acceleration from v 2  =  v 0 2  + 2 Æ ( x 2  –  x 1 );  (44 m/s) 2  = 0 + 2 Æ (3.5 m), which gives

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CHAPTER 2 - 6 - (c v=dx/dt=A 18Bt =A 18B(5.0s =A(450s)B...

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