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Unformatted text preview: 111 Chapter 4 1. The initial position vector r r o satisfies r r r r r r = o ∆ , which results in o ? ? ? ? (3.0j 4.0k)m (2.0i 3.0j 6.0k)m ( 2.0 m) i (6.0 m) j ( 10 m) k r r r = ∆ = + =  + +  r r r . 2. (a) The position vector, according to Eq. 41, is ? = ( 5.0 m) i + (8.0 m)j r r . (b) The magnitude is 2 2 2 2 2 2   + + ( 5.0 m) (8.0 m) (0 m) 9.4 m. r x y z = = + + = r (c) Many calculators have polar ↔ rectangular conversion capabilities which make this computation more efficient than what is shown below. Noting that the vector lies in the xy plane and using Eq. 36, we obtain: 1 8.0 m tan 58 or 122 5.0 m θ = =  ° °  where the latter possibility (122° measured counterclockwise from the + x direction) is chosen since the signs of the components imply the vector is in the second quadrant. (d) The sketch is shown on the right. The vector is 122° counterclockwise from the + x direction. (e) The displacement is r r r ′ ∆ = r r r where r r is given in part (a) and ˆ (3.0 m)i. r ′ = r Therefore, ? (8.0 m)i (8.0 m)j r ∆ = r . (f) The magnitude of the displacement is 2 2   (8.0 m) ( 8.0 m) 11 m. r ∆ = +  = r (g) The angle for the displacement, using Eq. 36, is 1 8.0 m tan = 45 or 135 8.0 m  ° °  where we choose the former possibility ( 45°, or 45° measured clockwise from + x ) since the signs of the components imply the vector is in the fourth quadrant. A sketch of r ∆ r is shown on the right. CHAPTER 4 112 3. (a) The magnitude of r r is 2 2 2   (5.0 m) ( 3.0 m) (2.0 m) 6.2 m. r = +  + = r (b) A sketch is shown. The coordinate values are in meters. 4. We choose a coordinate system with origin at the clock center and + x rightward (towards the “3:00” position) and + y upward (towards “12:00”). (a) In unitvector notation, we have 1 2 ? (10 cm)i and ( 10 cm)j. r r = =  r r Thus, Eq. 42 gives 2 1 ? ( 10 cm)i ( 10 cm)j. r r r ∆ = =  +  r r r and the magnitude is given by 2 2   ( 10 cm) ( 10 cm) 14 cm. r ∆ = +  = r (b) Using Eq. 36, the angle is 1 10 cm tan 45 or 135 . 10 cm θ = = ° °  We choose 135 ° since the desired angle is in the third quadrant. In terms of the magnitudeangle notation, one may write 2 1 ? ( 10 cm)i ( 10 cm)j (14 cm 135 ). r r r ∆ = =  +  → ∠  ° r r r (c) In this case, we have 1 2 ? ? ( 10 cm)j and (10 cm)j, and (20 cm)j. r r r =  = ∆ = r r r Thus,   20 cm. r ∆ = r (d) Using Eq. 36, the angle is given by 1 20 cm tan 90 . 0 cm θ = = ° (e) In a fullhour sweep, the hand returns to its starting position, and the displacement is zero. (f) The corresponding angle for a fullhour sweep is also zero....
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This note was uploaded on 08/09/2010 for the course PHY 202 101A2 taught by Professor Prof.yang during the Summer '10 term at National Taiwan University.
 Summer '10
 Prof.Yang
 Physics

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