CHAPTER 3 - 8 - a =[(3.44m/s ) +(2.25m/s ) ] =4.11m/s . av...

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d v 0 v = a d t 0 t ; v = (4.0 m/ s 2 ) i + (3.0 m/ s 2 ) j d t 0 t = (4.0 m/ s 2 ) t i + (3.0 m/ s 2 ) t j . a av  = [(3.44 m/s 2 ) 2  + (2.25 m/s 2 ) 2 ] 1/2  =          4.11 m/s 2 . We find the direction from tan  θ  = (2.25 m/s 2 )/(3.44 m/s 2 ) = 0.654, which gives           = 33.2° north of east . ( c ) Because we do not know the distance traveled, the average speed is         unknown . 22. ( a ) For the vertical component we have a V  = (3.80 m/s 2 ) sin 30.0° =         1.90 m/s 2  down . ( b ) Because the elevation change is the vertical displacement, we find the time from the  vertical motion, taking down as the positive direction: y  =  v 0 y t  +  ! a V t 2 ; 250 m = 0 +  ! (1.90 m/s 2 ) t 2 , which gives           t  = 16.2 s . 23. The acceleration is  a  = (4.0 m/s 2 ) i  + (3.0 m/s 2 ) j . ( a ) We find the velocity by integrating: ( b ) The speed of the particle is 
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CHAPTER 3 - 8 - a =[(3.44m/s ) +(2.25m/s ) ] =4.11m/s . av...

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