{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Chapter04

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

This preview shows pages 1–8. Sign up to view the full content.

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 .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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.
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.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
(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”).
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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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.
The total displacement is 123 ˆˆˆ ˆ (40.0 km)i (15.3 km)i (12.9 km) j

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 134

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

This preview shows document pages 1 - 8. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online