problem02_14 solution

# University Physics with Modern Physics with Mastering Physics (11th Edition)

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2.14 : (a) The displacement vector is: ( 29 k j i r ˆ ) s m 0 . 3 ( ) s m 0 . 7 ( ˆ ) s m 0 . 10 ( ˆ ) s m 0 . 5 ( ) ( 2 2 t t t t t - + + - = The velocity vector is the time derivative of the displacement vector: k j i r ˆ ) ) s m 0 . 3 ( 2 s m 0 . 7 ( ˆ ) s m 0 . 10 ( ˆ ) s m 0 . 5 ( ) ( 2 t dt t d - + + - = and the acceleration vector is the time derivative of the velocity vector: k r ˆ s m 0 . 6 ) ( 2 2 2 - = dt t d At t = 5.0 s: ( 29 k j i r ˆ ) s 0 . 25 )( s m 0 . 3 ( ) s 0 . 5 )( s m 0 . 7 ( ˆ ) s 0 . 5 )( s m 0 . 10 ( ˆ s) 0 5 )( s m 0 . 5 ( ) ( 2 2 - - + + - = . t k j i ˆ ) m 0 . 40 ( ˆ ) m 0 . 50 ( ˆ ) m 0 . 25 ( - + - = k j i k j i r ˆ ) s m 0 . 23 ( ˆ ) s m 0 . 10 ( ˆ ) s m 0 . 5 ( ˆ )) s 0 . 5 )( s m 0 . 6 ( s m 0 . 7 (( ˆ ) s m 0 . 10 ( ˆ ) s m 0 . 5 ( ) ( 2 - + - = - + + - = dt t d k r ˆ s m 0 . 6 ) ( 2 2 2 - = dt t d (b) The velocity in both the
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