71_PartUniversity Physics Solution

71_PartUniversity Physics Solution - 3 MOTION IN TWO OR...

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3-1 M OTION IN T WO OR T HREE D IMENSIONS 3.1. IDENTIFY and SET UP: Use Eq.(3.2), in component form. EXECUTE: () 21 av 5.3 m 1.1 m 1.4 m/s 3.0 s 0 x xxx v ttt Δ− == = = av 0.5 m 3.4 m 1.3 m/s 3.0 s 0 y yyy v = = EVALUATE: Our calculation gives that av v ! is in the 4th quadrant. This corresponds to increasing x and decreasing y . 3.2. IDENTIFY: Use Eq.(3.2), written in component form. The distance from the origin is the magnitude of r ! . SET UP: At time 1 t , 11 0 xy . EXECUTE: (a) av- Δ ( 3.8m s)(12.0 s) 45.6 m x xv t = and av- Δ (4.9m s)(12.0 s) 58.8 m y yv t = . (b) 22 2 2 ( 45.6 m) (58.8 m) 74.4 m. rx y =+ = + = EVALUATE: Δ r ! is in the direction of av v ! . Therefore, x Δ is negative since av- x v is negative and y Δ is positive since av- y v is positive. 3.3. (a) IDENTIFY and SET UP: From r ! we can calculate x and y for any t . Then use Eq.(3.2), in component form. EXECUTE: ( ) && 4.0 cm 2.5 cm/s 5.0 cm/s tt ⎡⎤ + ⎣⎦ ri j ! At 0, t = & 4.0 cm . = ! At 2.0 s, t = 14.0 cm 10.0 cm . j ! av 10.0 cm 5.0 cm/s. 2.0 s x x v t Δ = Δ av 10.0 cm 5.0 cm/s. 2.0 s y y v t Δ = Δ ( ) av av 1.3 m/s tan 0.9286 1.4 m/s y x v v α = 360 42.9 317 = °− °= ° ()() av av av x y vvv av (1.4 m/s) ( 1.3 m/s) 1.9 m/s v = Figure 3.1 3
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3-2 Chapter 3 ()() 22 av av av 7.1 cm/s xy vvv =+ = ( ) () av av tan 1.00 y x v v α == 45 . θ = ° Figure 3.3a EVALUATE: Both x and y increase, so av v ! is in the 1st quadrant. (b) IDENTIFY and SET UP: Calculate r ! by taking the time derivative of () . t r ! EXECUTE: 2 && 5.0 cm/s 5.0 cm/s d t dt ⎡⎤ + ⎣⎦ r vi j ! ! 0: t = 0, x v = 5.0 cm/s; y v = 5.0 cm/s v = and 90 = ° 1.0 s: t = 5.0 cm/s, x v = 5.0 cm/s; y v = 7.1 cm/s v = and 45 = ° 2.0 s: t = 10.0 cm/s, x v = 5.0 cm/s; y v = 11 cm/s v = and 27 = ° (c)
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71_PartUniversity Physics Solution - 3 MOTION IN TWO OR...

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