RESOLUÇÃO HALLIDAY, 9ªEDIÇÃO, VOLUME 1, CAPITULO 02, Exercícios de Física

Baixe RESOLUÇÃO HALLIDAY, 9ªEDIÇÃO, VOLUME 1, CAPITULO 02 e outras Exercícios em PDF para Física, somente na Docsity! 1. A vector a can be represented in the magnitude-angle notation (a, θ), where 2 2 x ya a a= + is the magnitude and 1tan y x a a θ −= is the angle a makes with the positive x axis. (a) Given Ax = −25.0 m and Ay = 40.0 m, 2 2( 25.0 m) (40.0 m) 47.2 mA = − + = (b) Recalling that tan θ = tan (θ + 180°), tan–1 [(40.0 m)/ (– 25.0 m)] = – 58° or 122°. Noting that the vector is in the third quadrant (by the signs of its x and y components) we see that 122° is the correct answer. The graphical calculator “shortcuts” mentioned above are designed to correctly choose the right possibility. 2. The angle described by a full circle is 360° = 2π rad, which is the basis of our conversion factor. (a) ( ) 2 rad20.0 20.0 0.349 rad 360 π° = ° = ° . (b) ( ) 2 rad50.0 50.0 0.873 rad 360 π° = ° = ° . (c) ( ) 2 rad100 100 1.75 rad 360 π° = ° = ° . (d) ( ) 3600.330 rad = 0.330 rad 18.9 2 radπ ° = ° . (e) ( ) 3602.10 rad = 2.10 rad 120 2 radπ ° = ° . (f) ( ) 3607.70 rad = 7.70 rad 441 2 radπ ° = ° . 5. The vector sum of the displacements dstorm and dnew must give the same result as its originally intended displacement o ˆ(120 km)jd = where east is i , north is j . Thus, we write storm new ˆ ˆ ˆ(100 km) i , i j.d d A B= = + (a) The equation storm new od d d+ = readily yields A = –100 km and B = 120 km. The magnitude of dnew is therefore equal to 2 2 new| | 156 kmd A B= + = . (b) The direction is tan–1 (B/A) = –50.2° or 180° + ( –50.2°) = 129.8°. We choose the latter value since it indicates a vector pointing in the second quadrant, which is what we expect here. The answer can be phrased several equivalent ways: 129.8° counterclockwise from east, or 39.8° west from north, or 50.2° north from west. 6. (a) With r = 15 m and θ = 30°, the x component of r is given by rx = rcosθ = (15 m) cos 30° = 13 m. (b) Similarly, the y component is given by ry = r sinθ = (15 m) sin 30° = 7.5 m. (a) We compute the distance from one corner to the diametrically opposite corner: 2 2 2(3.00 m) (3.70 m) (4.30 m)+ + . (b) The displacement vector is along the straight line from the beginning to the end point of the trip. Since a straight line is the shortest distance between two points, the length of the path cannot be less than the magnitude of the displacement. (c) It can be greater, however. The fly might, for example, crawl along the edges of the room. Its displacement would be the same but the path length would be 11.0 m.w h+ + = (d) The path length is the same as the magnitude of the displacement if the fly flies along the displacement vector. (e) We take the x axis to be out of the page, the y axis to be to the right, and the z axis to be upward. Then the x component of the displacement is w = 3.70 m, the y component of the displacement is 4.30 m, and the z component is 3.00 m. Thus, ˆ ˆ ˆ(3.70 m) i ( 4.30 m) j (3.00 m )kd = + + . An equally correct answer is gotten by interchanging the length, width, and height. 7. The length unit meter is understood throughout the calculation. (c) The angle between the resultant and the +x axis is given by θ = tan–1(ry/rx) = tan–1 [(10 m)/( –9.0 m)] = – 48° or 132°. Since the x component of the resultant is negative and the y component is positive, characteristic of the second quadrant, we find the angle is 132° (measured counterclockwise from +x axis). 9. We write r a b= + . When not explicitly displayed, the units here are assumed to be meters. (a) The x and the y components of r are rx = ax + bx = (4.0 m) – (13 m) = –9.0 m and ry = ay + by = (3.0 m) + (7.0 m) = 10 m, respectively. Thus ˆ ˆ( 9.0m) i (10m) jr = − + . (b) The magnitude of r is 2 2 2 2| | ( 9.0 m) (10 m) 13 mx yr r r r= = + = − + = . 10. We label the displacement vectors A , B and C (and denote the result of their vector sum as r ). We choose east as the î direction (+x direction) and north as the ĵ direction (+y direction) All distances are understood to be in kilometers. (a) The vector diagram representing the motion is shown below: ˆ(3.1 km) j ˆ( 2.4 km) i ˆ( 5.2 km) j A B C = = − = − (b) The final point is represented by ˆ ˆ( 2.4 km )i ( 2.1 km) jr A B C= + + = − + − whose magnitude is ( ) ( )2 22.4 km 2.1 km 3.2 kmr = − + − ≈ . (c) There are two possibilities for the angle: 1 2.1 kmtan 41 ,or 221 2.4 km θ − −= = ° ° − . We choose the latter possibility since r is in the third quadrant. It should be noted that many graphical calculators have polar ↔ rectangular “shortcuts” that automatically produce the correct answer for angle (measured counterclockwise from the +x axis). We may phrase the angle, then, as 221° counterclockwise from East (a phrasing that sounds peculiar, at best) or as 41° south from west or 49° west from south. The resultant r is not shown in our sketch; it would be an arrow directed from the “tail” of A to the “head” of C . 11. We find the components and then add them (as scalars, not vectors). With d = 3.40 km and θ = 35.0° we find d cos θ + d sin θ = 4.74 km. 14. The x, y and z components of r c d= + are, respectively, (a) 7.4 m 4.4 m 12 mx x xr c d= + = + = , (b) 3.8 m 2.0 m 5.8 my y yr c d= + = − − = − , and (c) 6.1 m 3.3 m 2.8 m.z z zr c d= + = − + = − (a) Along the x axis, we have (with the centimeter unit understood) 30.0 20.0 80.0 140,xb+ − − = − which gives bx = –70.0 cm. (b) Along the y axis we have 40.0 70.0 70.0 20.0yc− + − = − which yields cy = 80.0 cm. (c) The magnitude of the final location (–140 , –20.0) is 2 2( 140) ( 20.0) 141 cm.− + − = (d) Since the displacement is in the third quadrant, the angle of the overall displacement is given by π + 1tan [( 20.0) /( 140)]− − − or 188° counterclockwise from the +x axis (172° clockwise from the +x axis). 15. Reading carefully, we see that the (x, y) specifications for each “dart” are to be interpreted as ( , )Δ Δx y descriptions of the corresponding displacement vectors. We combine the different parts of this problem into a single exposition. 16. If we wish to use Eq. 3-5 in an unmodified fashion, we should note that the angle between C and the +x axis is 180° + 20.0° = 200°. (a) The x and y components of B are given by Bx = Cx – Ax = (15.0 m) cos 200° – (12.0 m) cos 40° = –23.3 m, By =Cy – Ay = (15.0 m) sin 200° – (12.0 m) sin 40° = –12.8 m. Consequently, its magnitude is | |B = 2 2( 23.3 m) ( 12.8 m) 26.6 m− + − = . (b) The two possibilities presented by a simple calculation for the angle between B and the +x axis are tan–1[( –12.8 m)/( –23.3 m)] = 28.9°, and 180° + 28.9° = 209°. We choose the latter possibility as the correct one since it indicates that B is in the third quadrant (indicated by the signs of its components). We note, too, that the answer can be equivalently stated as 151 .− ° angle is –37.5°, which is to say that it is 37.5° clockwise from the +x axis. This is equivalent to 322.5° counterclockwise from +x. (c) We find ˆ ˆ ˆ ˆ[43.3 ( 48.3) 35.4] i [25 ( 12.9) ( 35.4)] j (127 i 2.60 j) ma b c− + = − − + − − − + − = + in unit-vector notation. The magnitude of this result is 2 2 2| | (127 m) (2.6 m) 1.30 10 m.a b c− + = + ≈ × (d) The angle between the vector described in part (c) and the +x axis is 1tan (2.6 m/127 m) 1.2− ≈ ° . (e) Using unit-vector notation, d is given by ˆ ˆ( 40.4 i 47.4 j) md a b c= + − = − + , which has a magnitude of 2 2( 40.4 m) (47.4 m) 62 m.− + = (f) The two possibilities presented by a simple calculation for the angle between the vector described in part (e) and the +x axis are 1tan (47.4 /( 40.4)) 50.0− − = − ° , and 180 ( 50.0 ) 130° + − ° = ° . We choose the latter possibility as the correct one since it indicates that d is in the second quadrant (indicated by the signs of its components). 19. Many of the operations are done efficiently on most modern graphical calculators using their built-in vector manipulation and rectangular ↔ polar “shortcuts.” In this solution, we employ the “traditional” methods (such as Eq. 3-6). Where the length unit is not displayed, the unit meter should be understood. (a) Using unit-vector notation, ˆ ˆ(50 m)cos(30 )i (50 m) sin(30 ) j ˆ ˆ(50 m)cos (195 ) i (50 m)sin (195 ) j ˆ ˆ(50 m)cos (315 ) i (50 m)sin (315 ) j ˆ ˆ(30.4 m) i (23.3 m) j. a b c a b c = ° + ° = ° + ° = ° + ° + + = − The magnitude of this result is 2 2(30.4 m) ( 23.3 m) 38 m+ − = . (b) The two possibilities presented by a simple calculation for the angle between the vector described in part (a) and the +x direction are tan–1[( –23.2 m)/(30.4 m)] = –37.5°, and 180° + ( –37.5°) = 142.5°. The former possibility is the correct answer since the vector is in the fourth quadrant (indicated by the signs of its components). Thus, the 20. Angles are given in ‘standard’ fashion, so Eq. 3-5 applies directly. We use this to write the vectors in unit-vector notation before adding them. However, a very different- looking approach using the special capabilities of most graphical calculators can be imagined. Wherever the length unit is not displayed in the solution below, the unit meter should be understood. (a) Allowing for the different angle units used in the problem statement, we arrive at E F G H E F G H = + = − = + = − + + + + = + 3 73 4 70 1 29 4 83 1 3 73 5 20 3 00 1 28 6 60 . . . . .45 . . . . . i j i j i j i j i j. (b) The magnitude of the vector sum found in part (a) is 2 2(1.28 m) (6.60 m) 6.72 m+ = . (c) Its angle measured counterclockwise from the +x axis is tan–1(6.60/1.28) = 79.0°. (d) Using the conversion factor rad = 180π ° , 79.0° = 1.38 rad. 21. (a) With i^ directed forward and j^ directed leftward, then the resultant is (5.00 i^ + 2.00 j^) m . The magnitude is given by the Pythagorean theorem: 2 2(5.00 m) (2.00 m)+ = 5.385 m ≈ 5.39 m. (b) The angle is tan−1(2.00/5.00) ≈ 21.8º (left of forward). The equation relating these is A B C D+ = + , where ˆ ˆ ˆ ˆ(1.60 m)(cos50.0 i sin50.0 j) (1.03 m)i (1.23 m)jC = ° + ° = + (a) We find ˆ ˆ(0.16 m )i ( 0.83 m ) jD A B C= + − = + − , and the magnitude is D = 0.84 m. (b) The angle is 1tan ( 0.83/ 0.16) 79− − = − ° which is interpreted to mean 79º south of east (or 11º east of south). 24. Let A → represent the first part of Beetle 1’s trip (0.50 m east or ˆ0.5 i ) and C → represent the first part of Beetle 2’s trip intended voyage (1.6 m at 50º north of east). For their respective second parts: B → is 0.80 m at 30º north of east and D → is the unknown. The final position of Beetle 1 is ˆ ˆ ˆ ˆ ˆ(0.5 m)i (0.8 m)(cos30 i sin30 j) (1.19 m) i (0.40 m) j.A B+ = + ° + ° = + 25. The resultant (along the y axis, with the same magnitude as C → ) forms (along with C → ) a side of an isosceles triangle (with B → forming the base). If the angle between C → and the y axis is 1tan (3 / 4) 36.87θ −= = ° , then it should be clear that (referring to the magnitudes of the vectors) 2 sin( / 2)B C θ= . Thus (since C = 5.0) we find B = 3.2. 26. As a vector addition problem, we express the situation (described in the problem statement) as A → + B → = (3A) j^ , where A → = A i^ and B = 7.0 m. Since i^ ⊥ j^ we may use the Pythagorean theorem to express B in terms of the magnitudes of the other two vectors: B = (3A)2 + A2 A = 1 10 B = 2.2 m . 29. Solving the simultaneous equations yields the answers: (a) d1 → = 4 d3 → = 8 i^ + 16 j^ , and (b) d2 → = d3 → = 2 i^ + 4 j^. 30. The vector equation is R A B C D= + + + . Expressing B and D in unit-vector notation, we have ˆ ˆ(1.69i 3.63j) m+ and ˆ ˆ( 2.87i 4.10j) m− + , respectively. Where the length unit is not displayed in the solution below, the unit meter should be understood. (a) Adding corresponding components, we obtain ˆ ˆ( 3.18 m)i ( 4.72 m) jR = − + . (b) Using Eq. 3-6, the magnitude is 2 2| | ( 3.18 m) (4.72 m) 5.69 m.R = − + = (c) The angle is 1 4.72 mtan 56.0 (with axis). 3.18 m xθ −= = − ° − − If measured counterclockwise from +x-axis, the angle is then 180 56.0 124° − ° = ° . Thus, converting the result to polar coordinates, we obtain − → ∠ °318 4 72 569 124. , . .b g b g (c) If the starting point is (0, a, 0) with the corresponding position vector ̂ja , the diametrically opposite point is (a, 0, a) with the position vector ˆ ˆi ka a+ . Thus, the vector along the line is the difference ˆ ˆ ˆi j ka a a− + . (d) If the starting point is (a, a, 0) with the corresponding position vector ˆ ˆ i ja a+ , the diametrically opposite point is (0, 0, a) with the position vector k̂a . Thus, the vector along the line is the difference ˆ ˆ ˆi j ka a a− − + . (e) Consider the vector from the back lower left corner to the front upper right corner. It is ˆ ˆ ˆ i j k.a a a+ + We may think of it as the sum of the vector a i parallel to the x axis and the vector a a j k+ perpendicular to the x axis. The tangent of the angle between the vector and the x axis is the perpendicular component divided by the parallel component. Since the magnitude of the perpendicular component is 2 2 2a a a+ = and the magnitude of the parallel component is a, ( )tan 2 / 2a aθ = = . Thus θ = °54 7. . The angle between the vector and each of the other two adjacent sides (the y and z axes) is the same as is the angle between any of the other diagonal vectors and any of the cube sides adjacent to them. (f) The length of any of the diagonals is given by 2 2 2 3.a a a a+ + = 31. (a) As can be seen from Figure 3-32, the point diametrically opposite the origin (0,0,0) has position vector a a ai j k+ + and this is the vector along the “body diagonal.” (b) From the point (a, 0, 0) which corresponds to the position vector a î, the diametrically opposite point is (0, a, a) with the position vector a aj k+ . Thus, the vector along the line is the difference ˆ ˆ ˆi j ka a a− + + . 34. First, we rewrite the given expression as 4( dplane → · dcross → ) where dplane → = d1 → + d2 → and in the plane of d1 → and d2 → , and dcross → = d1 → × d2 → . Noting that dcross → is perpendicular to the plane of d1 → and d2 → , we see that the answer must be 0 (the scalar [dot] product of perpendicular vectors is zero). 35. We apply Eq. 3-30 and Eq.3-23. If a vector-capable calculator is used, this makes a good exercise for getting familiar with those features. Here we briefly sketch the method. (a) We note that ˆ ˆ ˆ8.0 i 5.0 j 6.0kb c× = − + + . Thus, ( ) = (3.0) ( 8.0) (3.0)(5.0) ( 2.0) (6.0) = 21.a b c⋅ × − + + − − (b) We note that ˆ ˆ ˆ + = 1.0 i 2.0 j + 3.0k.b c − Thus, ( ) (3.0) (1.0) (3.0) ( 2.0) ( 2.0) (3.0) 9.0.a b c⋅ + = + − + − = − (c) Finally, ˆ ˆ( + ) [(3.0)(3.0) ( 2.0)( 2.0)] i [( 2.0)(1.0) (3.0)(3.0)] j ˆ[(3.0)( 2.0) (3.0)(1.0)] k ˆ ˆ ˆ 5i 11j 9k a b c× = − − − + − − + − − = − − . We therefore obtain 2 2 (3.0)(2.0) (5.0)(4.0)ˆ 5.8. (2.0) (4.0) ba a b += ⋅ = = + 36. We apply Eq. 3-30 and Eq. 3-23. (a) ˆ = ( ) kx y y xa b a b a b× − since all other terms vanish, due to the fact that neither a nor b have any z components. Consequently, we obtain ˆ ˆ[(3.0)(4.0) (5.0)(2.0)]k 2.0k− = . (b) = x x y ya b a b a b⋅ + yields (3.0)(2.0) + (5.0)(4.0) = 26. (c) ˆ ˆ (3.0 2.0) i (5.0 4.0) j a b+ = + + + ( + ) = (5.0) (2.0) + (9.0) (4.0) = 46a b b⋅ . (d) Several approaches are available. In this solution, we will construct a b unit-vector and “dot” it (take the scalar product of it) with a . In this case, we make the desired unit- vector by 2 2 ˆ ˆ2.0 i 4.0 jˆ . | | (2.0) (4.0) bb b += = + 39. Since ab cos φ = axbx + ayby + azbz, cos .φ = + +a b a b a b ab x x y y z z The magnitudes of the vectors given in the problem are 2 2 2 2 2 2 | | (3.00) (3.00) (3.00) 5.20 | | (2.00) (1.00) (3.00) 3.74. a a b b = = + + = = = + + = The angle between them is found from (3.00) (2.00) (3.00) (1.00) (3.00) (3.00)cos 0.926. (5.20) (3.74) φ + += = The angle is φ = 22°. 40. Using the fact that ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆi j k, j k i, k i j× = × = × = we obtain ( ) ( )ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ2 2 2.00i 3.00j 4.00k 3.00i 4.00j 2.00k 44.0i 16.0j 34.0k.A B× = + − × − + + = + + Next, making use of ˆ ˆ ˆ ˆ ˆ ˆi i = j j = k k = 1 ˆ ˆ ˆ ˆ ˆ ˆi j = j k = k i = 0 ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ we have ( ) ( ) ( )ˆ ˆ ˆ ˆ ˆ3 2 3 7.00 i 8.00 j 44.0 i 16.0 j 34.0 k 3[(7.00) (44.0)+( 8.00) (16.0) (0) (34.0)] 540. C A B⋅ × = − ⋅ + + = − + = 41. From the definition of the dot product between A and B , cosA B AB θ⋅ = , we have cos A B AB θ ⋅= With 6.00A = , 7.00B = and 14.0A B⋅ = , cos 0.333θ = , or 70.5 .θ = ° 44. The two vectors are written as, in unit of meters, 1 1 1 2 2 2 ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ4.0 i+5.0 j i j, 3.0 i+4.0 j i jx y x yd d d d d d= = + = − = + (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 kx y y xd d d d d d× = − = − − (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 yd d d d d d⋅ = + = − + = (c) 2 2 2 1 2 2 1 2 2( ) 8.0 ( 3.0) (4.0) 33.d d d d d d+ ⋅ = ⋅ + = + − + = (d) Note that the magnitude of the d1 vector is 16+25 = 6.4. Now, the dot product is (6.4)(5.0)cosθ = 8. Dividing both sides by 32 and taking the inverse cosine yields θ = 75.5°. Therefore the component of the d1 vector along the direction of the d2 vector is 6.4cosθ ≈ 1.6. 45. Although we think of this as a three-dimensional movement, it is rendered effectively two-dimensional by referring measurements to its well-defined plane of the fault. (a) The magnitude of the net displacement is 2 2 2 2| | | | | | (17.0 m) (22.0 m) 27.8m.AB AD AC → = + = + = (b) The magnitude of the vertical component of AB → is |AD| sin 52.0° = 13.4 m. cos (3.2) (0.50) (1.6) (4.5) (3.58) (4.53) cos x x y ya b a b a b ab φ φ ⋅ = + = + = which leads to φ = 57° (the inverse cosine is double-valued as is the inverse tangent, but we know this is the right solution since both vectors are in the same quadrant). (b) Since the angle (measured from +x) for a is tan–1(1.6/3.2) = 26.6°, we know the angle for c is 26.6° –90° = –63.4° (the other possibility, 26.6° + 90° would lead to a cx < 0). Therefore, cx = c cos (–63.4° )= (5.0)(0.45) = 2.2 m. (c) Also, cy = c sin (–63.4°) = (5.0)( –0.89) = – 4.5 m. (d) And we know the angle for d to be 26.6° + 90° = 116.6°, which leads to dx = d cos(116.6°) = (5.0)( –0.45) = –2.2 m. (e) Finally, dy = d sin 116.6° = (5.0)(0.89) = 4.5 m. 46. Where the length unit is not displayed, the unit meter is understood. (a) We first note that 2 2| | (3.2) (1.6) 3.58a a= = + = and 2 2| | (0.50) (4.5) 4.53b b= = + = . Now, (b) The y-component of d1 is d1y = d1 sin θ1 = –2.83 m. (c) The x-component of d2 is d2x = d2 cos θ2 = 5.00 m. (d) The y-component of d2 is d2y = d2 sin θ2 = 0. (e) The x-component of d3 is d3x = d3 cos θ3 = 3.00 m. (f) The y-component of d3 is d3y = d3 sin θ3 = 5.20 m. (g) The sum of x-components is dx = d1x + d2x + d3x = –2.83 m + 5.00 m + 3.00 m = 5.17 m. (h) The sum of y-components is dy = d1y + d2y + d3y = –2.83 m + 0 + 5.20 m = 2.37 m. (i) The magnitude of the resultant displacement is 2 2 2 2(5.17 m) (2.37 m) 5.69 m.x yd d d= + = + = (j) And its angle is θ = tan–1 (2.37/5.17) = 24.6° which (recalling our coordinate choices) means it points at about 25° north of east. (k) and (l) This new displacement (the direct line home) when vectorially added to the previous (net) displacement must give zero. Thus, the new displacement is the negative, or opposite, of the previous (net) displacement. That is, it has the same magnitude (5.69 m) but points in the opposite direction (25° south of west). 49. We choose +x east and +y north and measure all angles in the “standard” way (positive ones are counterclockwise from +x). Thus, vector d1 has magnitude d1 = 4.00 m (with the unit meter) and direction θ1 = 225°. Also, d2 has magnitude d2 = 5.00 m and direction θ2 = 0°, and vector d3 has magnitude d3 = 6.00 m and direction θ3 = 60°. (a) The x-component of d1 is d1x = d1 cos θ1 = –2.83 m. 50. From the figure, it is clear that a→ + b→ + c→ = 0, where a→ ⊥ b→ . (a) a→ · b→ = 0 since the angle between them is 90º. (b) a → · c→ = a→ · (− a→ − b→ ) = −| a→ |2 = − 16 . (c) Similarly, b → · c→ = − 9.0 . 51. Let A → represent the first part of his actual voyage (50.0 km east) and C → represent the intended voyage (90.0 km north). We are looking for a vector B → such that A → + B → = C → . (a) The Pythagorean theorem yields 2 2(50.0 km) (90.0 km) 103 km.B = + = (b) The direction is 1tan (50.0 km / 90.0 km) 29.1− = ° west of north (which is equivalent to 60.9° north of due west). 2 277 m ( 3.2 m) 8.2 m.d⊥ = − − = This gives the magnitude of the perpendicular component (and is consistent with what one would get using Eq. 3-27), but if more information (such as the direction, or a full specification in terms of unit vectors) is sought then more computation is needed. 54. The three vectors are 1 2 3 ˆ ˆ ˆ4.0 i 5.0 j 6.0k ˆ ˆ ˆ1.0 i 2.0 j+3.0k ˆ ˆ ˆ4.0 i 3.0 j+2.0k d d d = + − = − + = + (a) 1 2 3 ˆ ˆ ˆ(9.0 m)i (6.0 m) j ( 7.0 m)kr d d d= − + = + + − . (b) The magnitude of r→ is 2 2 2| | (9.0 m) (6.0 m) ( 7.0 m) 12.9 mr = + + − = . The angle between r and the z-axis is given by k̂ 7.0 mcos 0.543 | | 12.9 m r r θ ⋅ −= = = − which implies 123 .θ = ° (c) The component of 1d along the direction of 2d is given by 1 1û= cosd d d ϕ= ⋅ where ϕ is the angle between 1d and 2d , and û is the unit vector in the direction of 2d . Using the properties of the scalar (dot) product, we have 1 2 1 2 1 2 2 2 1 2 2 (4.0)( 1.0) (5.0)(2.0) ( 6.0)(3.0) 12= 3.2 m. 14( 1.0) (2.0) (3.0) d d d dd d d d d ⋅ ⋅ − + + − −= = = = − − + + (d) Now we are looking for d⊥ such that 2 2 2 2 2 2 1 (4.0) (5.0) ( 6.0) 77d d d⊥= + + − = = + . From (c), we have 55. The two vectors are given by ˆ ˆ ˆ ˆ8.00(cos130 i sin130 j) 5.14 i 6.13 j ˆ ˆ ˆ ˆi j 7.72 i 9.20 j.x y A B B B = ° + ° = − + = + = − − (a) The dot product of 5A B⋅ is ˆ ˆ ˆ ˆ5 5( 5.14 i 6.13 j) ( 7.72 i 9.20 j) 5[( 5.14)( 7.72) (6.13)( 9.20)] 83.4. A B⋅ = − + ⋅ − − = − − + − = − (b) In unit vector notation 3ˆ ˆ ˆ ˆ ˆ ˆ4 3 12 12( 5.14 i 6.13 j) ( 7.72 i 9.20 j) 12(94.6k) 1.14 10 kA B A B× = × = − + × − − = = × (c) We note that the azimuthal angle is undefined for a vector along the z axis. Thus, our result is “1.14×103, θ not defined, and φ = 0°.” (d) Since A → is in the xy plane, and A B× is perpendicular to that plane, then the answer is 90°. (e) Clearly, A → + 3.00 k^ = –5.14 i^ + 6.13 j^ + 3.00 k^ . (f) The Pythagorean theorem yields magnitude 2 2 2(5.14) (6.13) (3.00) 8.54A = + + = . The azimuthal angle is θ = 130°, just as it was in the problem statement ( A → is the projection onto to the xy plane of the new vector created in part (e)). The angle measured from the +z axis is φ = cos−1(3.00/8.54) = 69.4°. (c) 1 2 0d d⋅ = since ˆ ˆi j = 0.⋅ The two vectors are perpendicular to each other. (d) 1 2 1 2( / 4) ( ) / 4 0d d d d⋅ = ⋅ = , as in part (c). (e) 1 2 1 2 1 2ˆ ˆ ˆ( j i) = kd d d d d d× = − × , in the +z-direction. (f) 2 1 2 1 1 2ˆ ˆ ˆ(i j) = kd d d d d d× = − × − , in the −z-direction. (g) The magnitude of the vector in (e) is 1 2d d . (h) The magnitude of the vector in (f) is 1 2d d . (i) Since 1 2 1 2 ˆ( / 4) ( / 4)kd d d d× = , the magnitude is 1 2 / 4.d d (j) The direction of 1 2 1 2 ˆ( / 4) ( / 4)kd d d d× = is in the +z-direction. 56. The two vectors 1d and 2d are given by 1 1 2 2ˆ ˆj and i.d d d d= − = (a) The vector 2 2 ˆ/ 4 ( / 4) id d= points in the +x direction. The ¼ factor does not affect the result. (b) The vector 1 1 ˆ/( 4) ( / 4) jd d− = points in the +y direction. The minus sign (with the “−4”) does affect the direction: −(–y) = + y. (h) Also, | | | | a b b a ab× = × = . (i) With d > 0, we find that ( / )a b d× has magnitude ab/d. (j) The vector ( / )a b d× points in the +z direction. 59. The vectors can be written as ˆ ˆi and ja a b b= = where , 0.a b > (a) We are asked to consider b d b d = FHG I KJ j in the case d > 0. Since the coefficient of j is positive, then the vector points in the +y direction. (b) If, however, d < 0, then the coefficient is negative and the vector points in the –y direction. (c) Since cos 90° = 0, then 0a b⋅ = , using Eq. 3-20. (d) Since b d/ is along the y axis, then (by the same reasoning as in the previous part) ( / ) 0a b d⋅ = . (e) By the right-hand rule, a b× points in the +z-direction. (f) By the same rule, b a× points in the –z-direction. We note that b a a b× = − × is true in this case and quite generally. (g) Since sin 90° = 1, Eq. 3-27 gives | |a b ab× = where a is the magnitude of a . 60. The vector can be written as ˆ(2.5 m)jd = , where we have taken ĵ to be the unit vector pointing north. (a) The magnitude of the vector 4.0a d= is (4.0)(2.5 m) = 10 m. (b) The direction of the vector a d= 4.0 is the same as the direction of d (north). (c) The magnitude of the vector = 3.0c d− is (3.0)(2.5 m) = 7.5 m. (d) The direction of the vector = 3.0c d− is the opposite of the direction of d . Thus, the direction of c is south. 61. We note that the set of choices for unit vector directions has correct orientation (for a right-handed coordinate system). Students sometimes confuse “north” with “up”, so it might be necessary to emphasize that these are being treated as the mutually perpendicular directions of our real world, not just some “on the paper” or “on the blackboard” representation of it. Once the terminology is clear, these questions are basic to the definitions of the scalar (dot) and vector (cross) products. (a) ˆ ˆi k=0⋅ since ˆ ˆi k⊥ (b) ˆ ˆ( k) ( j)=0− ⋅ − since ˆ ˆk j⊥ . (c) ˆ ˆj ( j)= 1.⋅ − − (d) ˆ ˆ ˆk j= i (west).× − (e) ˆ ˆ ˆ( i) ( j)= k (upward).− × − + (f) ˆ ˆ ˆ( k) ( j)= i (west).− × − − 64. The point P is displaced vertically by 2R, where R is the radius of the wheel. It is displaced horizontally by half the circumference of the wheel, or πR. Since R = 0.450 m, the horizontal component of the displacement is 1.414 m and the vertical component of the displacement is 0.900 m. If the x axis is horizontal and the y axis is vertical, the vector displacement (in meters) is ( )ˆ ˆ1.414 i + 0.900 j .r = The displacement has a magnitude of ( ) ( )2 2 22 4 1.68mr R R R= π + = π + = and an angle of 1 12 2tan tan 32.5R Rπ π − −= = ° above the floor. In physics there are no “exact” measurements, yet that angle computation seemed to yield something exact. However, there has to be some uncertainty in the observation that the wheel rolled half of a revolution, which introduces some indefiniteness in our result. 65. Reference to Figure 3-18 (and the accompanying material in that section) is helpful. If we convert B to the magnitude-angle notation (as A already is) we have B = ∠ °14 4 337. .b g (appropriate notation especially if we are using a vector capable calculator in polar mode). Where the length unit is not displayed in the solution, the unit meter should be understood. In the magnitude-angle notation, rotating the axis by +20° amounts to subtracting that angle from the angles previously specified. Thus, A = ∠ ° ′12 0 40 0. .b g and B = ∠ ° ′( . . )14 4 137 , where the ‘prime’ notation indicates that the description is in terms of the new coordinates. Converting these results to (x, y) representations, we obtain (a) ˆ ˆ(9.19 m) i (7.71 m) j .A ′ ′= + (b) Similarly, ˆ ˆ(14.0 m) i (3.41 m) jB ′ ′= + . 1 1 216.5 mtan tan 36 . 300 m y x r r − −= = ° The direction is 36° north of due east. (c) The total distance walked is d = 250 m + 175 m = 425 m. (d) The total distance walked is greater than the magnitude of the resultant displacement. The diagram shows why: A and B are not collinear. 66. The diagram shows the displacement vectors for the two segments of her walk, labeled A and B , and the total (“final”) displacement vector, labeled r . We take east to be the +x direction and north to be the +y direction. We observe that the angle between A and the x axis is 60°. Where the units are not explicitly shown, the distances are understood to be in meters. Thus, the components of A are Ax = 250 cos60° = 125 and Ay = 250 sin60° = 216.5. The components of B are Bx = 175 and By = 0. The components of the total displacement are rx = Ax + Bx = 125 + 175 = 300 ry = Ay + By = 216.5 + 0 = 216.5. (a) The magnitude of the resultant displacement is 2 2 2 2| | (300 m) (216.5 m) 370m.x yr r r= + = + = (b) The angle the resultant displacement makes with the +x axis is 68. The two vectors can be found be solving the simultaneous equations. (a) If we add the equations, we obtain 2 6a c= , which leads to ˆ ˆ3 9 i 12 ja c= = + . (b) Plugging this result back in, we find b c= = +3 4i j . 69. (a) This is one example of an answer: (−40 i^ – 20 j^ + 25 k^) m, with i^ directed anti- parallel to the first path, j^ directed anti-parallel to the second path and k^ directed upward (in order to have a right-handed coordinate system). Other examples are (40 i^ + 20 j^ + 25 k^ ) m and (40i^ – 20 j^ – 25 k^ ) m (with slightly different interpretations for the unit vectors). Note that the product of the components is positive in each example. (b) Using Pythagorean theorem, we have 2 2(40 m) (20 m)+ = 44.7 m ≈ 45 m. (b) The new direction of 5d is the same as the old: south. The vector 2.0d− can be written as ˆ2.0 (6.0 m) j.d− = (c) The absolute value of the scalar factor (|−2.0| = 2.0) affects the magnitude. The new magnitude is 6.0 m. (d) The minus sign carried by this scalar factor reverses the direction, so the new direction is ĵ+ , or north. 70. The vector d (measured in meters) can be represented as ˆ(3.0 m)( j)d = − , where ĵ− is the unit vector pointing south. Therefore, ˆ ˆ5.0 5.0( 3.0 m j) ( 15 m) j.d = − = − (a) The positive scalar factor (5.0) affects the magnitude but not the direction. The magnitude of 5.0d is 15 m.