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Problem_Set_2_Solutions

Problem_Set_2_Solutions - 22 The desired result is the...

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22. The desired result is the displacement vector, in units of km, A = (5.6 km), 90º (measured counterclockwise from the + x axis), or ˆ (5.6 km)j A = G , where ˆ j is the unit vector along the positive y axis (north). This consists of the sum of two displacements: during the whiteout, (7.8 km), 50 B = ° G , or ˆ ˆ ˆ ˆ (7.8 km)(cos50 i sin50 j) (5.01 km)i (5.98 km)j B = ° + ° = + G and the unknown C G . Thus, A B C = + G G G . (a) The desired displacement is given by ˆ ˆ ( 5.01 km) i (0.38 km) j C A B = = − G G G . The magnitude is 2 2 ( 5.01 km) ( 0.38 km) 5.0 km. + − = (b) The angle is 1 tan [( 0.38 km) /( 5.01 km)] 4.3 , = ° south of due west.
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44. The two vectors are written as, in unit of meters, 1 1 1 2 2 2 ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ 4.0i+5.0 j i j, 3.0i+4.0 j i j x y x y d d d d d d = = + = − = + G G (a) The vector (cross) product gives 1 2 1 2 1 2 ˆ ˆ ˆ ( )k [(4.0)(4.0) (5.0)( 3.0)]k=31 k x y y x d d d d d d × = = G G (b) The scalar (dot) product gives 1 2 1 2 1 2 (4.0)( 3.0) (5.0)(4.0) 8.0. x x y y d d d d d d = + = + = G G (c) 2 2 2 1 2 2 1 2 2 ( ) 8.0 ( 3.0) (4.0) 33. d d d d d d + = + = + − + = G G G G G (d) Note that the magnitude of the d 1 vector is 16+25 = 6.4. Now, the dot product is (6.4)(5.0)cos θ = 8. Dividing both sides by 32 and taking the inverse cosine yields θ = 75.5 ° . Therefore the component of the d 1 vector along the direction of the d 2 vector is 6.4cos θ 1.6.
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(h) The y component of the net displacement net d G is , 1 2 3 ( 0.28 m) (0 m) (0.52 m) 0.24 m. net y y y y d d d d = + + = − + + = (i) The magnitude of the net displacement is 2 2 2 2 , , (0.52 m) (0.24 m) 0.57 m. net net x net y d d d = + = + = (j) The direction of the net displacement is , 1 1 , 0.24 m tan tan 25 (north of east) 0.52 m net y net x d d θ § · § · = = = ° ¨ ¸ ¨ ¸ ¨ ¸ © ¹ © ¹ If the ant has to return directly to the starting point, the displacement would be net d G . (k) The distance the ant has to travel is | | 0.57 m.
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