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HW_2_solutions

# HW_2_solutions - P112 Spring 2007 Homework#2 Solutions Note...

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P112 Spring 2007 Homework #2 Solutions Note: Solutions to boxed problems are given in detail, “unboxed” problem solutions will just contain a short answer. 1.( unboxed) Speed at the first point is 5.00 m/s, acceleration is 1.43 m/s 2 2.( unboxed) Deceleration 7.11 m/s 2 , stopping time is 3.75 s. 3. (unboxed) (a) | r A | = A = A x 2 + A y 2 = (4.00) 2 + (3.00) 2 = 5.00 . 39 . 5 0 . 29 ) 00 . 2 ( ) 00 . 5 ( | | 2 2 2 2 = = ! + = + = = y x B B B B r . (b) j i j i B A ˆ )) 00 . 2 ( 00 . 3 ( ˆ ) 00 . 5 00 . 4 ( ˆ ) ( ˆ ) ( ! ! + ! = ! + ! = ! y y x x B A B A r r . j i B A ˆ 00 . 5 ˆ 00 . 1 + ! = ! r r . (c) 10 . 5 0 . 26 ) 00 . 5 ( ) 00 . 1 ( | | 2 2 = = + ! = ! B A r r . ° ! = " # \$ % & ! = " " # \$ % % & ! ! = ! ! ! 7 . 78 00 . 1 00 . 5 tan ) ( ) ( tan 1 1 x x y y B A B A B A r r ( . Since 0 ) ( < ! x x B A and 0 ) ( > ! y y B A , the vector j i B A ˆ 00 . 5 ˆ 00 . 1 + ! = ! r r points 78.7 ° above the x ! direction or at an angle of 101.3 ° with respect to the x + direction. 4. 2 ) ( t t v x ! " # = , where m/s 00 . 4 = ! and 3 m/s 00 . 2 = ! . At 0 = t , 0 0 = x . (a) 3 ) ( ) ( ) ( 3 0 2 0 0 t t t d t t d t v x t x x t t x ! " ! " # = \$ \$ # = \$ \$ = # = % & & . 3 ) ( 3 0 t t x t x ! " # + = . t dt t dv t a x x ! 2 ) ( ) ( " = = . (b) Maxima of ) ( t x occur when 0 ) ( = t v x and 0 ) ( ) ( < = dt t dv t a x x . We are looking for the maximum positive value of ) ( t x . Setting ) ( t v x equal to zero, 0 2 = ! t " # , and solving for t yields two values: !

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