CHAPTER 3 - 1 - CHAPTER 3 ­ Kinematics in Two Dimensions;...

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Unformatted text preview: CHAPTER 3 ­ Kinematics in Two Dimensions; Vectors D2x W D1 D 2y D 2 θ R S Ry Rx 1. We choose the west and south coordinate system shown. For the components of the resultant we have R = D + D cos 45° = (200 km) + (80 km) cos 45° = 257 km; R = 0 + D S 2 = 0 + (80 km) sin 45° = 57 km. We find the resultant displacement from R = (R 2 W 1 2 tan θ = R /R = (57 km)/(257 km) = 0.222, which gives θ = 13° S of W. S W W + R ) = [(257 km) + (57 km) ] = 263 km; S 2 1/2 2 2 1/2 N D2 D1 D3 θ 2. We choose the north and east coordinate system shown. For the components of the resultant we have R = D = 10 blocks; R E R E We find the resultant displacement from 2 2 1/2 E N N = D – D = 18 blocks – 16 blocks = 2 blocks. 1 3 2 2 1/2 2 R = (R + R ) = [(10 blocks) + (2 blocks) ] = 10 blocks; tan θ = R /R = (2 blocks)/(10 blocks) = 0.20, which gives θ = 11° N of E. N E 3. From Fig. 3–6c, if we write the equivalent vector addition, we have V + V 1 wrong = V , or V 2 wrong = V – V . 2 1 y Vx x Vy θ V 4. We find the vector from V = (V + V ) = [(8.80) + (– 6.40) ] = 10.9; x y 2 2 1/2 2 2 1/2 tan θ = V /V = (– 6.40)/(8.80) = – 0.727, which gives y x θ = 36.0° below the x­axis. 5. The resultant is 13.6 m, 18° N of E. North V2 V1 VR V3 θ East ...
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This note was uploaded on 03/08/2010 for the course PHYSICS 7A/7B taught by Professor All during the Fall '08 term at Berkeley.

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CHAPTER 3 - 1 - CHAPTER 3 ­ Kinematics in Two Dimensions;...

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