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HW_2_solutions

# HW_2_solutions - P1112 Fall 2008 Homework#2 Solutions 1 For...

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P1112 Fall 2008 Homework #2 Solutions 1. For the car to stop 0 ) ( = stop t v . Thus using the equation for constant acceleration ( ) at v t v + = 0 we get a v t t a v stop stop 0 0 0 = ! " = . Now if 2 / a a ! and 0 0 2 v v ! then stop stop t a v t 4 2 2 0 = ! 2. Data ( ) ( ) j i t v j i t v s m s m s m s m ˆ 40 ˆ 170 , ˆ 110 ˆ 90 2 1 + ! = + = r r (a) The two vectors differ by ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) j i j i v j t v t v i t v t v t v t v v s m s m s m s m s m s m y y x x ˆ 70 ˆ 260 ˆ 110 40 ˆ 90 170 ˆ ˆ 1 2 1 2 1 2 ! ! = " # \$ % & ! + " # \$ % & ! " # \$ % & ! = ( ) ! + ! = ! = ( r r (b) By definition t v a avg ! ! = r r and s t 30 = ! so r a avg = " 260 m ˆ i " 70 m ˆ j 30 s = " 8.67 ˆ i " 2.33 ˆ j (c) The magnitude a avg = x , avg 2 + a y , avg 2 = 8.67 m 2 + 2.33 m 2 = 9 m The angle is given by ( ) ! " # \$ % & == = x y x y a a a a arctan tan ( As will be the case here it is important to remember that arc tangent is well define up to a factor of o 180 . arctan 2.33 8.67 " # \$ % & = 15 o of course both the x and y components are negative so the angle is really " = 15 o + 180 o = 195 o 3. Going back to a formula for constant acceleration ( ) ( ) 2 0 v t v t v avg + = . The Car is moving at constant velocity so t v x C C ! = ! . In order for the motorcycle to

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HW_2_solutions - P1112 Fall 2008 Homework#2 Solutions 1 For...

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