Ch 3Word - Chapter 3 CHAPTER 3 Kinematics in Two Dimensions...

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Chapter  3 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 . Page 1
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Chapter  3 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  θ  =  R y / R x  = (3.18)/(2.82) = 1.13, which gives   θ  = 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  θ  = – 14.3 cos 34.8 °  =         – 11.7 ; V y V  sin  θ  =  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  θ  =  V y / V x  = (8.16)/(11.7) = 1.42, which gives   θ  = 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.
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