Ch3Word - Chapter 3 CHAPTER 3 Kinematics in Two Dimensions...

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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 θ = R S / R W = (57 km)/(257 km) = 0.222, which gives θ = 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 θ = R N / R E = (2 blocks)/(10 blocks) = 0.20, which gives θ = 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 θ = V y / V x = (– 6.40)/(8.80) = – 0.727, which gives θ = 36.0° below the x -axis . 5. The resultant is 13.6 m, 18° N of E . θ D 1 S W D 2 x R D 2 y R x R y D 2 θ D 1 N E R D 2 D 3 V y V V x θ y x V R θ V 1 V 2 V 3 North East
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