Chapter04 - 1. The initial position vector ro satisfies r -...

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1. The initial position vector G r o satisfies G G G rr r −= o Δ , which results in o ˆˆ ˆ ˆ ˆ ˆ (3.0j 4.0k)m (2.0i 3.0j 6.0k)m ( 2.0 m)i (6.0 m) j ( 10 m)k rr r =− Δ= + + +− GG G .
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2. (a) The position vector, according to Eq. 4-1, is ˆˆ = ( 5.0 m) i + (8.0 m)j r G . (b) The magnitude is 222 2 2 2 | | + + ( 5.0 m) (8.0 m) (0 m) 9.4 m. rx y z == + + = G (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. 3-6, we obtain: 1 8.0 m tan 58 or 122 5.0 m θ §· ° ° ¨¸ ©¹ 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 rrr Δ= − GGG where G r is given in part (a) and ˆ (3.0 m)i. r ′ = G Therefore, (8.0 m)i (8.0 m)j r Δ= G . (f) The magnitude of the displacement is 22 | | (8.0 m) ( 8.0 m) 11 m. r +− = G (g) The angle for the displacement, using Eq. 3-6, 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 Δ G is shown on the right.
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3. (a) The magnitude of G r is 22 2 | | (5.0 m) ( 3.0 m) (2.0 m) 6.2 m. r =+ + = G (b) A sketch is shown. The coordinate values are in meters.
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(a) In unit-vector notation, we have 12 ˆˆ (10 cm)i and ( 10 cm)j. rr == GG Thus, Eq. 4-2 gives 21 (1 0 cm ) i 0 ) j . rrr Δ= − =− +− GGG and the magnitude is given by 22 | | ( 10 cm) 14 cm. r Δ= − = G (b) Using Eq. 3-6, the angle is 1 10 cm tan 45 or 135 . θ §· ° ° ¨¸ ©¹ We choose 135 −° since the desired angle is in the third quadrant. In terms of the magnitude-angle notation, one may write 0 ) i 0 ) j ( 1 4 1 3 5) . rr r ∠− ° GG G (c) In this case, we have ˆ ( 10 cm)j and (10 cm)j, and (20 cm)j. r =− = Δ= G Thus, || 2 0 c m . r G (d) Using Eq. 3-6, the angle is given by 1 20cm 90 . 0cm ° (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. 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”).
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5. Using Eq. 4-3 and Eq. 4-8, we have avg ˆˆ ˆ ˆ ˆ ˆ ( 2.0i + 8.0j 2.0k) m (5.0i 6.0j + 2.0k) m ˆ ( 0.70i +1.40j 0.40k) m/s. 10 s v −− == G
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6. To emphasize the fact that the velocity is a function of time, we adopt the notation v ( t ) for /. dx dt (a) Eq. 4-10 leads to 2 ˆˆ ˆ ˆ ˆ ( ) (3.00 i 4.00 j + 2.00k) (3.00 m/s)i (8.00 m/s) j d vt t t t dt =− = (b) Evaluating this result at t = 2.00 s produces = (3.00i 16.0j) m/s. v G (c) The speed at t = 2.00 s is 22 | | (3.00 m/s) ( 16.0 m/s) 16.3 m/s. vv == +− = G (d) The angle of G v at that moment is 1 16.0 m/s tan 79.4 or 101 3.00 m/s §· ° ° ¨¸ ©¹ where we choose the first possibility (79.4° measured clockwise from the + x direction, or 281° counterclockwise from + x ) since the signs of the components imply the vector is in the fourth quadrant.
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The total displacement is 123 ˆˆˆ ˆ (40.0 km)i (15.3 km)i (12.9 km) j
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This note was uploaded on 05/11/2010 for the course PHYS 2101,2102 taught by Professor Giammanco during the Spring '10 term at LSU.

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Chapter04 - 1. The initial position vector ro satisfies r -...

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