Chapter 03 - 3.1 a s m 4 1 s 3 m 1 1 m 3 5 ave = = x v s m...

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Unformatted text preview: 3.1 : a) , s m 4 . 1 ) s . 3 ( ) m 1 . 1 ( ) m 3 . 5 ( ave , =- = x v . s m 3 . 1 ) s . 3 ( ) m 4 . 3 ( ) m 5 . ( ave ,- =-- = y v b) s m 9 . 1 or , s m 91 . 1 ) s m 3 . 1 ( ) s m (1.4 2 2 ave =- + = v to two significant figures, ( 29 °- = =- 43 arctan 1.4 1.3 θ . 3.2 : a) and m 6 . 45 ) s . 12 )( s m 8 . 3 ( Δ ) ( ave ,- =- = = t v x x m. 8 . 58 ) s . 12 )( s m 9 . 4 ( Δ ) ( ave , = = = t v y y b) m. 4 . 74 ) m 8 . 58 ( ) m 6 . 45 ( 2 2 2 2 = +- = + = y x r 3.3: The position is given by j i r ˆ ) s cm . 5 ( ˆ ] ) s cm 5 . 2 ( cm . 4 [ 2 2 t t + + = . (a) i ˆ ] cm . 4 [ ) ( = r , and j i j i ˆ ) cm . 10 ( ˆ ) cm . 14 ( ˆ s) 2 )( s cm . 5 ( ˆ ] s) 2 )( s cm (2.5 cm . 4 [ ) s 2 ( 2 2 + = + + = r . Then using the definition of average velocity, . j i v j i ˆ ) s cm 5 ( ˆ ) s cm 5 ( s 2 ˆ ) cm 10 ( ˆ ) cm 4 cm 14 ( ave + = =- +- s cm 1 . 7 ave = v at an angle of ° 45 . b) j i j i v ˆ ) s cm 5 ( ˆ ) s cm 5 ( ˆ ) s cm 5 ( ˆ ) s cm 5 . 2 )( 2 ( + = + = = t t dt r d . Substituting for s 1 , = t , and 2 s, gives: j i v j v ˆ ) s cm 5 ( ˆ ) s cm 5 ( s) 1 ( , ˆ ) s cm 5 ( ) ( + = = , and j i v ˆ ) s cm 5 ( ˆ ) s cm 10 ( s) 2 ( + = . The magnitude and direction of v at each time therefore are: s cm . 5 : = t at ° 90 ; s cm 1 . 7 : 05 . 1 = t at s cm 11 : 05 . 2 ; 45 = ° t at ° 27 . c) 3.4: j i v ˆ 3 ˆ 2 2 ct bt + = . This vector will make a ° 45 -angle with both axes when the x- and y-components are equal; in terms of the parameters, this time is c b 3 2 . 3.5: a) b) , s m 7 . 8 ) s . 30 ( ) s m 90 ( ) s m 170 ( 2 ave ,- =-- = x a . s m 3 . 2 ) s . 30 ( ) s m 110 ( ) s m 40 ( 2 ave ,- =- = y a c) ( 29 . 195 180 8 . 14 arctan , s m . 9 ) s m 3 2 ( ) s m 7 . 8 ( 7 . 8 3 . 2 2 2 2 2 2 ° = ° + ° = =- +--- . 3.6: a) 2 2 2 2 s m 23 . . 31 sin ) s m 45 . ( , s m 39 . . 31 cos ) s m 45 . ( = ° = = ° = y x a a , so s m 5 . 6 ) s . 10 )( s m 39 . ( s m 6 . 2 2 = + = x v and s m 52 . ) s . 10 )( s m 23 . ( s m 8 . 1 2 = +- = y v . b) s m 48 . 6 ) s m 52 . ( ) s m 5 . 6 ( 2 2 = + = v , at an angle of ( 29 ° = 6 . 4 arctan 5 . 6 52 . above the horizontal. c) 3.7: a) b) . ˆ ) s m 4 . 2 ( ˆ 2 ˆ ] ) s m 4 . 2 [( ˆ ) s m 4 . 2 ( ˆ 2 ˆ 2 2 j j a j i j i v- =- =- =- = β β α t t c) At s . 2 = t , the velocity is j i v ˆ ) s m 8 . 4 ( ˆ ) s m 4 . 2 (- = ; the magnitude is s m 4 . 5 ) s m 8 . 4 ( ) s m 4 . 2 ( 2 2 =- + , and the direction is ( 29 °- =- 63 arctan 4 . 2 8 . 4 . The acceleration is constant, with magnitude 2 s m 4 . 2 in the y--direction. d) The velocity vector has a component parallel to the acceleration, so the bird is speeding up. The bird is turning toward the y--direction, which would be to the bird’s right (taking the z +- direction to be vertical)....
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This homework help was uploaded on 02/26/2008 for the course PHYS 11 and 21 taught by Professor Hickman during the Spring '08 term at Lehigh University .

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Chapter 03 - 3.1 a s m 4 1 s 3 m 1 1 m 3 5 ave = = x v s m...

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