# CH_3 - Chapter 3 CHAPTER 3 Kinematics in Two Dimensions...

This preview shows pages 1–3. Sign up to view the full content.

Chapter 3 Page 1 CHAPTER 3 - Kinematics in Two Dimensions; Vectors 1. We choose the west and south coordinate system shown. For the components of the resultant we have R W = D 1 + D 2 cos 45° = (200 km) + (80 km) cos 45° = 257 km; R S = 0 + D 2 = 0 + (80 km) sin 45° = 57 km. We find the resultant displacement from R = ( R W 2 + R S 2 ) 1/2 = [(257 km) 2 + (57 km) 2 ] 1/2 = 263 km ; tan = R S / R W = (57 km)/(257 km) = 0.222, which gives = 13° S of W . 2. We choose the north and east coordinate system shown. For the components of the resultant we have R E = D 2 = 10 blocks; R N = D 1 D 3 = 18 blocks – 16 blocks = 2 blocks. We find the resultant displacement from R = ( R E 2 + R N 2 ) 1/2 = [(10 blocks) 2 + (2 blocks) 2 ] 1/2 = 10 blocks ; tan = R N / R E = (2 blocks)/(10 blocks) = 0.20, which gives = 11° N of E . 3. From Fig. 3–6c, if we write the equivalent vector addition, we have V 1 + V wrong = V 2 , or V wrong = V 2 V 1 . 4. We find the vector from V = ( V x 2 + V y 2 ) 1/2 = [(8.80) 2 + (– 6.40) 2 ] 1/2 = 10.9 ; tan = V y / V x = (– 6.40)/(8.80) = – 0.727, which gives = 36.0° below the x -axis . 5. The resultant is 13.6 m, 18° N of E . D 1 S W D 2 x R D 2 y R x R y D 2 D 1 N E R D 2 D 3 V y V V x y x V R V 1 V 2 V 3 North East

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

View Full Document
Chapter 3 Page 2 6. ( a ) V 1 x = – 6.0, V 1 y = 0 ; V 2 x = V 2 cos 45° = 4.5 cos 45° = 3.18 = 3.2 , V 2 y = V 2 sin 45° = 4.5 sin 45° = 3.18 = 3.2 . ( b ) For the components of the sum we have R x = V 1 x + V 2 x = – 6.0 + 3.18 = – 2.82; R y = V 1 y + V 2 y = 0 + 3.18 = 3.18. We find the resultant from R = ( R x 2 + R y 2 ) 1/2 = [(– 2.82) 2 + (3.18) 2 ] 1/2 = 4.3 ; tan = R y / R x = (3.18)/(2.82) = 1.13, which gives = 48° above – x -axis . Note that we have used the magnitude of R x for the angle indicated on the diagram. 7. ( a ) ( b ) For the components of the vector we have V x = – V cos = – 14.3 cos 34.8° = – 11.7 ; V y = V sin = 14.3 sin 34.8° = 8.16 . ( c ) We find the vector from V = ( V x 2 + V y 2 ) 1/2 = [(– 11.7) 2 + (8.16) 2 ] 1/2 = 14.3 ; tan = V y / V x = (8.16)/(11.7) = 1.42, which gives = 34.9° above – x -axis . This is within significant figures. Note that we have used the magnitude of V x for the angle indicated on the diagram. 8. Because the vectors are parallel, the direction can be indicated by the sign. ( a ) C = A + B = 6.8 + (– 5.5) = 1.3 in the + x -direction . ( b ) C = A B = 6.8 – (– 5.5) = 12.3 in the + x -direction . ( c ) C = B A = – 5.5 – (6.8) = – 12.3 in the + x -direction or 12.3 in the x -direction . 9. ( a ) Using the given angle, we find the components from V N = V cos 38.5° = (635 km/h) cos 41.5° = 476 km/h ; V W = V sin 38.5° = (635 km/h) sin 41.5° = 421 km/h . ( b
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 04/08/2008 for the course PHY 53 taught by Professor Mueller during the Fall '08 term at Duke.

### Page1 / 24

CH_3 - Chapter 3 CHAPTER 3 Kinematics in Two Dimensions...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online