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Solutions 3

Solutions 3 - 3.1 IDENTIFY and SET UP Use Eq(3.2 in...

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3.1. I DENTIFY and S ET U P : Use Eq.(3.2), in component form. E XECUTE : ( ) 2 1 av 2 1 5.3 m 1.1 m 1.4 m/s 3.0 s 0 x x x x v t t t = = = = ( ) 2 1 av 2 1 0.5 m 3.4 m 1.3 m/s 3.0 s 0 y y y y v t t t = = = = − E VALUATE : Our calculation gives that av v r is in the 4th quadrant. This corresponds to increasing x and decreasing y . 3.6. I DENTIFY : Use Eq.(3.8), written in component form. S ET U P : 2 2 2 2 (0.45m s )cos31.0 0.39m s , (0.45m s )sin31.0 0.23m s x y a a = ° = = ° = E XECUTE : (a) av- x x v a t = and 2 2.6 m s (0.39 m s )(10.0 s) 6.5 m s x v = + = . av- y y v a t = and 2 1.8 m s (0.23 m s )(10.0 s) 0.52 m s y v = − + = . (b) 2 2 (6.5m s) (0.52m s) 6.48m s v = + = , at an angle of 0.52 arctan 4.6 6.5 = ° above the horizontal. (c) The velocity vectors 1 v r and 2 v r are sketched in Figure 3.6. The two velocity vectors differ in magnitude and direction. E VALUATE : 1 v r is at an angle of 35 ° below the -axis x + and has magnitude 1 3.2 m/s v = , so 2 1 v v > and the direction of 2 v r is rotated counterclockwise from the direction of 1 v r . Figure 3.6 ( ) ( ) av av 1.3 m/s tan 0.9286 1.4 m/s y x v v α = = = − 360 42.9 317 α = ° − ° = ° ( ) ( ) 2 2 av av av x y v v v = + 2 2 av (1.4 m/s) ( 1.3 m/s) 1.9 m/s v = + − = Figure 3.1

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3.9. I DENTIFY : The book moves in projectile motion once it leaves the table top. Its initial velocity is horizontal.
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Solutions 3 - 3.1 IDENTIFY and SET UP Use Eq(3.2 in...

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