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ch04 - Chapter 4 1(a The magnitude of r is | r |(5.0 m)2...

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107 Chapter 4 1. (a) The magnitude of r is 2 2 2 | | (5.0 m) ( 3.0 m) (2.0 m) 6.2 m. r (b) A sketch is shown. The coordinate values are in meters. 2. (a) The position vector, according to Eq. 4-1, is ? = ( 5.0 m) i + (9.0 m)j r . (b) The magnitude is 2 2 2 2 2 2 | | + + ( 5.0 m) (9.0 m) (0 m) 10 m. r x y z (c) Many calculators have polar rectangular conversion capabilities that make this computation more efficient than what is shown below. Noting that the vector lies in the xy plane and using Eq. 3-6, we obtain: 1 9.0 m tan 61 or 119 5.0 m   where the latter possibility (119° 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 to the right. The vector is 119° counterclockwise from the + x direction. (e) The displacement is r r r where r is given in part (a) and ˆ (3.0 m)i. r   Therefore, ? (8.0 m)i (9.0 m)j r . (f) The magnitude of the displacement is 2 2 | | (8.0 m) ( 9.0 m) 12 m. r (g) The angle for the displacement, using Eq. 3-6, is 1 8.0 m tan = 42 or 138 9.0 m 119 0 (-5, 9) (-5, 9)
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CHAPTER 4 108 where we choose the former possibility ( 42°, or 42° measured clockwise from + x ) since the signs of the components imply the vector is in the fourth quadrant. A sketch of r is shown on the right. 3. The initial position vector r o satisfies 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       . 4. We choose a coordinate system with origin at the clock center and + x rightward (toward the “3:00” position) and + y upward (toward “12:00”). (a) In unit-vector notation, we have 1 2 ? (12 cm)i and ( 12 cm)j. r r Thus, Eq. 4-2 gives 2 1 ? ( 12 cm)i ( 12 cm)j. r r r     The magnitude is given by 2 2 | | ( 12 cm) ( 12 cm) 17 cm. r   (b) Using Eq. 3-6, the angle is 1 12 cm tan 45 or 135 . 12 cm We choose 135 since the desired angle is in the third quadrant. In terms of the magnitude-angle notation, one may write 2 1 ? ( 12 cm)i ( 12 cm)j. r r r     (c) In this case, we have 1 2 ? ? ( 12 cm)j and (12 cm)j, and (24 cm)j. r r r Thus, | | 24 cm. r (d) Using Eq. 3-6, the angle is given by 1 24 cm tan 90 . 0 cm (e) In a full-hour sweep, the hand returns to its starting position, and the displacement is zero. (f) The corresponding angle for a full-hour sweep is also zero.
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109 5. The average velocity of the entire trip is given by Eq. 4-8: avg / , v r t   where the total displacement 1 2 3 r r r r       r r r r is the sum of three displacements (each result of a constant velocity during a given time), and 1 2 3 t t t t        is the total amount of time for the trip. We use a coordinate system with +
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