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# Sm chapter3

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### Sm chapter3

2. 2. SOLUTIONS TO PROBLEMS Section 3.1 Coordinate Systems P3.1 x r= = ( ) = ( ) −( ) = −cos cos .θ 5 50 240 5 50 0 5 2 7. m . m .° 55 m y r= = ( ) = ( ) −( ) = −sin sinθ 5 50 240 5 50 0 8 4. m . m . 66° ..76 m P3.2 (a) x r= cosθ and y r= sinθ, therefore x1 2 50 30 0= ( ). m .cos °, y1 2 50 30 0= ( ). m .sin °, and x y1 1 2 17 1 25, . , . m( ) = ( ) x2 3 80 120= ( ) °. cosm , y2 3 80 120= ( ) °. sinm , and x y2 2 1 90 3 29, . , . m( ) = −( ) (b) d x y= + = + =( ) ( ) . . .∆ ∆2 2 2 2 4 07 2 04 4 55m m P3.3 The x distance out to the ﬂy is 2.00 m and the y distance up to the ﬂy is 1.00 m. (a) We can use the Pythagorean theorem to ﬁnd the distance from the origin to the ﬂy. distance m m m2 = + = ( ) + ( ) = =x y2 2 2 2 2 00 1 00 5 00. . . 22 24. m (b) θ = ⎛ ⎝⎜ ⎞ ⎠⎟ = °− tan .1 1 2 26 6 ; r = °2 24 26 6. , .m P3.4 We have 2 00 30 0. .= °r cos r = ° = 2 00 30 0 2 31 . cos . . and y r= = =sin sin .30 0 2 31 30 0 1 15. . .° ° . P3.5 We have r x y= +2 2 and θ = ⎛ ⎝⎜ ⎞ ⎠⎟ − tan 1 y x . (a) The radius for this new point is −( ) + = + =x y x y r 2 2 2 2 and its angle is tan− − ⎛ ⎝ ⎞ ⎠ =1 180 y x ° −θ . (b) ( ) ( )− + − =2 2 22 2 x y r This point is in the third quadrant if x y,( ) is in the ﬁrst quadrant or in the fourth quadrant if x y,( ) is in the second quadrant. It is at an angle of 180° +θ . (c) ( ) ( )3 3 32 2 x y r+ − = This point is in the fourth quadrant if x y,( ) is in the ﬁrst quadrant or in the third quadrant if x y,( ) is in the second quadrant. It is at an angle of −θ . 46 Chapter 3 13794_03_ch03_p045-064.indd 4613794_03_ch03_p045-064.indd 46 11/28/06 4:40:07 PM11/28/06 4:40:07 PM
3. 3. Vectors 47 Section 3.2 Vector and Scalar Quantities Section 3.3 Some Properties of Vectors P3.6 − =R 310 km at 57 S of W° (Scale: 1 unit = 20 km) P3.7 tan . tan . . 35 0 100 100 35 0 70 0 ° = = ( ) ° = x x m m m P3.8 Find the resultant F F1 2+ graphically by placing the tail of F2 at the head of F1. The resultant force vector F F1 2+ is of magnitude 9 5. N and at an angle of 57° abovethe axisx . FIG. P3.6 FIG. P3.7 0 1 2 3 N x y 2 1 F1 2+F F F FIG. P3.8 13794_03_ch03_p045-064.indd 4713794_03_ch03_p045-064.indd 47 11/28/06 4:40:08 PM11/28/06 4:40:08 PM
4. 4. 48 Chapter 3 P3.9 (a) d i= − =10 0 10 0. ˆ . m since the displacement is in a straight line from point A to point B. (b) The actual distance skated is not equal to the straight-line displacement. The distance follows the curved path of the semi-circle (ACB). s r= ( ) = = 1 2 2 5 15 7π π . m (c) If the circle is complete, d begins and ends at point A. Hence, d = 0 . P3.10 (a) The large majority of people are standing or sitting at this hour. Their instantaneous foot- to-head vectors have upward vertical components on the order of 1 m and randomly oriented horizontal components. The citywide sum will be ~ 105 m upward . (b) Most people are lying in bed early Saturday morning. We suppose their beds are oriented north, south, east, and west quite at random. Then the horizontal component of their total vector height is very nearly zero. If their compressed pillows give their height vectors vertical components averaging 3 cm, and if one-tenth of one percent of the population are on-duty nurses or police ofﬁcers, we estimate the total vector height as ~ . m m10 0 03 10 15 2 ( )+ ( ) ~ 103 m upward . P3.11 To ﬁnd these vector expressions graphically, we draw each set of vectors. Measurements of the results are taken using a ruler and protractor. (Scale: 1 0 5unit m= . ) (a) A + B = 5.2 m at 60° (b) A − B = 3.0 m at 330° (c) B − A = 3.0 m at 150° (d) A − 2B = 5.2 m at 300° C B A 5.00 m d FIG. P3.9 FIG. P3.11 13794_03_ch03_p045-064.indd 4813794_03_ch03_p045-064.indd 48 11/28/06 4:40:09 PM11/28/06 4:40:09 PM
5. 5. Vectors 49 P3.12 The three diagrams shown below represent the graphical solutions for the three vector sums: R A B C1 = + + , R B C A2 = + + , and R C B A3 = + + . We observe that R R R1 2 3= = , illustrating that the sum of a set of vectors is not affected by the order in which the vectors are added . 2 B A R1 C C C R 3R A A B B 100 m FIG. P3.12 P3.13 The scale drawing for the graphical solution should be similar to the ﬁgure to the right. The magnitude and direction of the ﬁnal displacement from the starting point are obtained by measuring d and θ on the drawing and applying the scale factor used in making the drawing. The results should be d = = −420 3ft and θ ° . Section 3.4 Components of a Vector and Unit Vectors *P3.14 We assume the ﬂoor is level. Take the x axis in the direction of the ﬁrst displacement. If both of the 90° turns are to the right or both to the left , the displacements add like 40 0 15 0 20 0 20 0 15 0. ˆ . ˆ . ˆ . ˆ .m m mi i i+ − = +j ˆˆj( )m to give (a) displacement magnitude (202 + 152 )1ր2 m = 25.0 m at (b) tan−1 (15ր20) = 36.9° . If one turn is right and the other is left , the displacements add like 40 0 15 0 20 0 60 0 15 0. ˆ . ˆ . ˆ . ˆ .m m mi i+ + = +j i ˆˆj( )m to give (a) displacement magnitude (602 + 152 )1ր2 m = 61.8 m at (b) tan−1 (15ր60) = 14.0˚. Just two answers are possible. FIG. P3.13 13794_03_ch03_p045-064.indd 4913794_03_ch03_p045-064.indd 49 11/28/06 4:40:10 PM11/28/06 4:40:10 PM
6. 6. 50 Chapter 3 P3.15 A A A A A x y x y = − = = + = −( ) + ( ) = 25 0 40 0 25 0 40 0 42 2 2 2 . . . . 77 2. units We observe that tanφ = A A y x . So φ = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = = ( ) =− − tan tan . . tan .1 140 0 25 0 1 60 5 A A y x 88 0. °. The diagram shows that the angle from the +x axis can be found by subtracting from 180°: θ = − =180 58 122° ° ° . P3.16 The person would have to walk 3 10 1 31. 25.0 km northsin .°( ) = , and 3 10 25 0 2 81. . km eastcos .°( ) = . *P3.17 Let v represent the speed of the camper. The northward component of its velocity is v cos 8.5°. To avoid crowding the minivan we require v cos 8.5° ≥ 28 mրs. We can satisfy this requirement simply by taking v ≥ (28 mրs)րcos 8.5° = 28.3 mրs. P3.18 (a) Her net x (east-west) displacement is − + + = +3 00 0 6 00 3 00. . . blocks, while her net y (north-south) displacement is 0 4 00 0 4 00+ + = +. . blocks. The magnitude of the resultant displacement is R x y= ( ) + ( ) = ( ) + ( ) =net net 2 2 2 2 3 00 4 00 5 00. . . blocks and the angle the resultant makes with the x axis (eastward direction) is θ = ⎛ ⎝⎜ ⎞ ⎠⎟ = ( ) = °− − tan . . tan . .1 14 00 3 00 1 33 53 1 . The resultant displacement is then 5 00 53 1. .blocks at N of E° . (b) The total distance traveled is 3 00 4 00 6 00 13 0. . . .+ + = blocks . P3.19 x r= cosθ and y r= sinθ, therefore: (a) x = °12 8 150. cos , y = °12 8 150. sin , and x y, . . m( ) = − +( )111 6 40ˆ ˆi j (b) x = °3 30 60 0. cos . , y = °3 30 60 0. sin . , and x y, cm( ) = +( )1 65 2 86. ˆ . ˆi j (c) x = °22 0 215. cos , y = °22 0 215. sin , and x y, in( ) = − −( )18 0 12 6. ˆ . ˆi j P3.20 x d= = ( ) ( ) = −cos cosθ 50 0 120 25 0. m . m y d= = ( ) ( ) = = −( ) sin sin . . θ 50 0 120 43 3 25 0 . m m md ˆˆ . ˆi j+ ( )43 3 m FIG. P3.15 13794_03_ch03_p045-064.indd 5013794_03_ch03_p045-064.indd 50 11/28/06 4:40:10 PM11/28/06 4:40:10 PM
7. 7. Vectors 51 P3.21 Let +x be East and +y be North. x y ∑ ∑ = = = − = − 250 125 30 358 75 125 30 150 + + cos sin ° ° m 112 5 358 12 5 358 2 2 2 2 . m md x y= ( ) + ( ) = ( ) + −( ) =∑ ∑ . taan . θ θ = ( ) ( ) = − = − = − = ∑ ∑ y x 12 5 358 0 0349 2 00 35 . . ° d 88 2 00m at S of E. ° P3.22 The east and north components of the displacement from Dallas (D) to Chicago (C) are the sums of the east and north components of the displacements from Dallas to Atlanta (A) and from Atlanta to Chicago. In equation form: d d dDC east DA east AC east .= =+ −730 5 00 560cos s° iin 21 0 527. miles. DC north DA north AC ° = =d d d+ nnorth . . miles= =730 5 00 560 21 0 586sin cos° °+ By the Pythagorean theorem, d d d= + =( ) ( )DC east DC north mi2 2 788 . Then tanθ = = d d DC north DC east .111 and θ = °48 0. . Thus, Chicago is 788 miles at 48.0° northeast of Dallas . P3.23 We have B R A= − : A A x y = = − = = 150 75 0 150 120 130 cos120 cm cm ° ° . sin RR R x y = = = = 140 35 0 115 140 35 0 80 3 cos . sin . . ° ° cm cmm Therefore, B i j i j= − −( )[ ] + −[ ] = −(115 75 80 3 130 190 49 7ˆ . ˆ ˆ . ˆ)) = + = = −⎛ ⎝ ⎞− cm cmB 190 49 7 196 49 7 190 2 2 1 . tan . θ ⎠⎠ = −14 7. .° P3.24 (a) See ﬁgure to the right. (b) C A B i j i j i= + = + + − =2 00 6 00 3 00 2 00 5 00. ˆ . ˆ . ˆ . ˆ . ˆ ++ 4 00. ˆj C = + ⎛ ⎝ ⎞ ⎠ =− 25 0 16 0 6 401 . . tan .at 4 5 at 38.7° D A B i j i j= − = + − + = −2 00 6 00 3 00 2 00 1 00. ˆ . ˆ . ˆ . ˆ . ˆii j+ 8 00. ˆ D = −( ) + ( ) − ⎛ ⎝⎜ ⎞ ⎠⎟ − 1 00 8 00 8 00 1 00 2 2 1 . . tan . . at D = −( ) =8 06 180 82 9 8 06 97 2. . . .at at° ° ° x FIG. P3.23 FIG. P3.24 13794_03_ch03_p045-064.indd 5113794_03_ch03_p045-064.indd 51 11/28/06 4:40:11 PM11/28/06 4:40:11 PM
8. 8. 52 Chapter 3 P3.25 (a) A B i j i j i j+( )= −( )+ − −( )= −3 2 4 2 6ˆ ˆ ˆ ˆ ˆ ˆ (b) A B i j i j i j−( )= −( )− − −( )= +3 2 4 4 2ˆ ˆ ˆ ˆ ˆ ˆ (c) A B+ = + =2 6 6 322 2 . (d) A B− = + =4 2 4 472 2 . (e) θA B+ − = −⎛ ⎝ ⎞ ⎠ = − =tan .1 6 2 71 6 288° ° θA B− − = ⎛ ⎝ ⎞ ⎠ =tan .1 2 4 26 6° *P3.26 We take the x axis along the slope uphill. Students, get used to this choice! The y axis is perpen- dicular to the slope, at 35° to the vertical. Then the displacement of the snow makes an angle of 90° − 35° − 20° = 35° with the x axis. (a) Its component parallel to the surface is 5 m cos 35° = 4.10 m toward the top of the hill . (b) Its component perpendicular to the surface is 5 m sin 35° = 2.87 m . P3.27 d j1 3 50= −( ). ˆ m d i j i2 8 20 45 0 8 20 45 0 5 80 5= + = +. cos . ˆ . sin . ˆ . ˆ .° ° 880ˆj( ) m d i3 15 0= −( ). ˆ m R d d d i j= + + = − +( ) + −( )1 2 3 15 0 5 80 5 80 3 50. . ˆ . . ˆ == − +( )9 20 2 30. ˆ . ˆi j m (or 9.20 m west and 2.30 m north) The magnitude of the resultant displacement is R = + = −( ) + ( ) =R Rx y 2 2 2 2 9 20 2 30 9 48. . . m . The direction is θ = − ⎛ ⎝ ⎞ ⎠ =arctan . . 2 30 9 20 166° . P3.28 Refer to the sketch R A C= + + = − − + = − B i j i i 10 0 15 0 50 0 40 0 15 . ˆ . ˆ . ˆ . ˆ .. ˆ . . . / 0 40 0 15 0 42 7 2 2 1 2 j R = ( ) + −( )⎡⎣ ⎤⎦ = yards P3.29 East North x y 0 m 4.00 m 1.41 1.41 –0.500 –0.866 +0.914 4.55 R = + =x y y x 2 2 4 64at tan m at 78.6 N of1− ( ) .ր ° E A = 0.10 B = 15 0. C = 50 0. R FIG. P3.28 13794_03_ch03_p045-064.indd 5213794_03_ch03_p045-064.indd 52 11/28/06 4:40:12 PM11/28/06 4:40:12 PM
9. 9. Vectors 53 P3.30 A i j= − +8 70 15 0. ˆ . ˆ and B i j= −13 2 6 60. ˆ . ˆ A B C− + =3 0: 3 21 9 21 6 7 30 7 20 C B A i j C i j = − = − = − . ˆ . ˆ . ˆ . ˆ or Cx = 7 30. cm ; Cy = −7 20. cm P3.31 (a) F F F F i = + = ( ) + ( ) 1 2 120 60 0 120 60 0cos . ˆ sin . ˆ° ° jj i j F − ( ) + ( ) = 80 0 75 0 80 0 75 0 60 0 . cos . ˆ . sin . ˆ . ° ° ˆˆ ˆ . ˆ . ˆ . ˆ ˆi j i j i j F + − + = +( ) = 104 20 7 77 3 39 3 181 N 339 3 181 185 181 39 3 77 8 2 2 1 . tan . . + = = ⎛ ⎝ ⎞ ⎠ =− N θ ° (b) F F i j3 39 3 181= − = − −( ). ˆ ˆ N P3.32 A = 3 00. m, θA = °30 0. B = 3 00. m, θB = °90 0. A Ax A= = ° =cos . cos . .θ 3 00 30 0 2 60 m A Ay A= = =sin . sin . .θ 3 00 30 0 1 50° m A i j i j= + = +( )A Ax y ˆ ˆ . ˆ . ˆ2 60 1 50 m Bx = 0, By = 3 00. m so B j= 3 00. ˆ m A B i j j i j+ = +( )+ = +(2 60 1 50 3 00 2 60 4 50. ˆ . ˆ . ˆ . ˆ . ˆ)) m P3.33 B i j k i j k B = + + = + + = B B Bx y z ˆ ˆ ˆ . ˆ . ˆ . ˆ . 4 00 6 00 3 00 4 000 6 00 3 00 7 81 4 00 7 81 59 2 2 2 1 + + = = ⎛ ⎝ ⎞ ⎠ =− . . . cos . . α .. cos . 2 6 01 ° is the angle with the axisx β = − 00 7 81 39 8 . .⎛ ⎝ ⎞ ⎠ = ° is the angle with the axy iis is the angle wγ = ⎛ ⎝ ⎞ ⎠ =− cos . . .1 3 00 7 81 67 4° iith the axisz P3.34 (a) D A B C i j= + + = −2 2ˆ ˆ D = + = =2 2 2 83 3152 2 . m at θ ° (b) E A B C i j= − − + = − +6 12ˆ ˆ E = + = =6 12 13 4 1172 2 . m at θ ° P3.35 (a) C A B i j k= + = − −( )5 00 1 00 3 00. ˆ . ˆ . ˆ m C = ( ) + ( ) + ( ) =5 00 1 00 3 00 5 92 2 2 2 . . . .m m (b) D A B i j k= − = − +( )2 4 00 11 0 15 0. ˆ . ˆ . ˆ m D = ( ) + ( ) + ( ) =4 00 11 0 15 0 19 0 2 2 2 . . . .m m 13794_03_ch03_p045-064.indd 913794_03_ch03_p045-064.indd 9 12/7/06 6:01:55 PM12/7/06 6:01:55 PM
10. 10. P3.36 Let the positive x-direction be eastward, the positive y-direction be vertically upward, and the positive z-direction be southward. The total displacement is then d i j j k= +( ) + −( ) =4 80 4 80 3 70 3 70 4. ˆ . ˆ . ˆ . ˆ .cm cm 880 8 50 3 70ˆ . ˆ . ˆi j k+ −( ) cm. (a) The magnitude is d = ( ) + ( ) + −( ) =4 80 8 50 3 70 10 4 2 2 2 . . . .cm cm . (b) Its angle with the y axis follows from cos . . θ = 8 50 10 4 , giving θ = 35 5. ° . P3.37 (a) A i j k= + −8 00 12 0 4 00. ˆ . ˆ . ˆ (b) B A i j k= = + − 4 2 00 3 00 1 00. ˆ . ˆ . ˆ (c) C A i j k= − = − − +3 24 0 36 0 12 0. ˆ . ˆ . ˆ P3.38 The y coordinate of the airplane is constant and equal to 7 60 103 . × m whereas the x coordinate is given by x ti= v where vi is the constant speed in the horizontal direction. At t = 30 0. s we have x = ×8 04 103 . , so vi = =8 40 2680 m 30 s m s. The position vector as a function of time is P i j= ( ) + ×( )268 7 60 103 m s mtˆ . ˆ. At t = 45 0. s, P i j= × + ×⎡ ⎣ ⎤ ⎦1 21 10 7 60 104 3 . ˆ . ˆ m. The magnitude is P = ×( ) + ×( ) = ×1 21 10 7 60 10 1 43 104 2 3 2 4 . . .m m and the direction is θ = × × ⎛ ⎝⎜ ⎞ ⎠⎟ =arctan . . . 7 60 10 1 21 10 32 2 3 4 ° above tthe horizontal . P3.39 The position vector from radar station to ship is S i j i= +( ) = −17 3 136 17 3 136 12 0. sin ˆ . cos ˆ . ˆ° ° km 112 4. ˆj( ) km. From station to plane, the position vector is P i j k= + +( )19 6 153 19 6 153 2 20. sin ˆ . cos ˆ . ˆ° ° km, or P i j k= − +( )8 90 17 5 2 20. ˆ . ˆ . ˆ km. (a) To ﬂy to the ship, the plane must undergo displacement D S P i j k= − = + −( )3 12 5 02 2 20. ˆ . ˆ . ˆ km . (b) The distance the plane must travel is D = = ( ) + ( ) + ( ) =D 3 12 5 02 2 20 6 31 2 2 2 . . . .km km . 54 Chapter 354 Chapter 3 13794_03_ch03_p045-064.indd 1013794_03_ch03_p045-064.indd 10 12/7/06 2:08:46 PM12/7/06 2:08:46 PM
11. 11. Vectors 55 55 P3.40 (a) E i j= ( ) + ( )17 0 27 0 17 0 27 0. cos . ˆ . sin . ˆcm cm° ° E i j= +( )15 1 7 72. ˆ . ˆ cm (b) F i j= −( ) + ( )17 0 27 0 17 0 27 0. sin . ˆ . cos . ˆcm cm° ° F i j= − +( )7 72 15 1. ˆ . ˆ cm Note that we do not need to explicitly identify the angle with the positive x axis. (c) G i j= +( ) + ( )17 0 27 0 17 0 27 0. sin . ˆ . cos . ˆcm cm° ° G i j= + +( )7 72 15 1. ˆ . ˆ cm P3.41 Ax = −3 00. , Ay = 2 00. (a) A i j i j= + = − +A Ax y ˆ ˆ . ˆ . ˆ3 00 2 00 (b) A = + = −( ) + ( ) =A Ax y 2 2 2 2 3 00 2 00 3 61. . . tan . . .θ = = −( ) = − A A y x 2 00 3 00 0 667, tan . .− −( ) = − °1 0 667 33 7 θ is in the 2nd quadrant, so θ = −( ) =180 33 7 146° ° °+ . . (c) Rx = 0, Ry = −4 00. , R A B= + thus B R A= − and B R Ax x x= − = − −( ) =0 3 00 3 00. . , B R Ay y y= − = − − = −4 00 2 00 6 00. . . . Therefore, B i j= −3 00 6 00. ˆ . ˆ . P3.42 The hurricane’s ﬁrst displacement is 41 0 3 00 . . km h h ⎛ ⎝⎜ ⎞ ⎠⎟ ( ) at 60 0. ° N of W, and its second displacement is 25 0 1 50 . . km h h ⎛ ⎝⎜ ⎞ ⎠⎟ ( ) due North. With ˆi representing east and ˆj representing north, its total displacement is: 41 0 60 0 3 00 41 0. cos . . ˆ . sin km h h km h °⎛ ⎝ ⎞ ⎠ ( ) −( )+i 660 0 3 00 25 0 1 50. . ˆ . . ˆ°⎛ ⎝ ⎞ ⎠ ( ) + ⎛ ⎝ ⎞ ⎠ ( ) = h km h hj j 661 5 144. ˆ ˆkm km−( )+i j with magnitude 61 5 144 157 2 2 . km km km( ) + ( ) = . P3.43 (a) R R x y = + = = 40 0 45 0 30 0 45 0 49 5 40 0 . cos . . cos . . . sin ° ° 445 0 30 0 45 0 20 0 27 1 49 5 27 1 . . sin . . . . ˆ . ° °− + = = +R i ˆˆj (b) R = ( ) + ( ) = = ⎛ ⎝ ⎞ ⎠ =− 49 5 27 1 56 4 27 1 49 5 2 2 1 . . . tan . . θ 228 7. ° F y x 27.0º G 27.0º E 27.0º FIG. P3.40 A y x B 45º C 45º O FIG. P3.43 13794_03_ch03_p045-064.indd 1113794_03_ch03_p045-064.indd 11 12/7/06 2:08:47 PM12/7/06 2:08:47 PM
12. 12. 56 Chapter 3 *P3.44 (a) Taking components along ˆi and ˆj, we get two equations: 6 00 8 00 26 0 0. . .a b− + = and − + + =8 00 3 00 19 0 0. . .a b . We solve simultaneously by substituting a = 1.33 b − 4.33 to ﬁnd −8(1.33 b − 4.33) + 3 b + 19 = 0 or 7.67b = 53.67 so b = 7.00 and a = 1.33(7) − 4.33. Thus a b= =5 00 7 00. , . . Therefore, 5 00 7 00 0. .A B C+ + = . (b) In order for vectors to be equal, all of their components must be equal. A vector equa- tion contains more information than a scalar equation. *P3.45 The displacement from the start to the ﬁnish is 16 ˆi + 12 ˆj − (5 ˆi + 3 ˆj) = (11 ˆi + 9 ˆj) meters. The displacement from the starting point to A is f(11 ˆi + 9 ˆj) meters. (a) The position vector of point A is 5 ˆi + 3 ˆj + f(11 ˆi + 9 ˆj) = (5 + 11f ) ˆi + (3 + 9f ) ˆj meters . (b) For f = 0 we have the position vector (5 + 0) ˆi + (3 + 0) ˆj meters. This is reasonable because it is the location of the starting point, 5 ˆi + 3 ˆj meters. (c) For f = 1 = 100%, we have position vector (5 + 11) ˆi + (3 + 9) ˆj meters = 16 ˆi + 12 ˆj meters . This is reasonable because we have completed the trip and this is the position vector of the endpoint. *P3.46 We note that − =ˆi west and − =ˆj south. The given mathematical representation of the trip can be written as 6 3 4 3. b west b at 40 south of west b at+ +° 550 south of east b south° + 5 . (a) (b) The total odometer distance is the sum of the magnitudes of the four displacements: 6 3 4 3 5 18 3. .b b b b b+ + + = . (c) R i= − − +( ) + − − −( )6 3 3 06 1 93 2 57 2 30 5. . . ˆ . . ˆb b jj i j= − − = ( ) + ( ) 7 44 9 87 7 44 9 87 2 2 . ˆ . ˆ . . b b b b at south of west b at tan . . . − = 1 9 87 7 44 12 4 553.0 south of west b at 233 counter ° °= 12 4. cclockwise from east R = 1 block EW S N f i 13794_03_ch03_p045-064.indd 5613794_03_ch03_p045-064.indd 56 11/28/06 4:40:16 PM11/28/06 4:40:16 PM
13. 13. Vectors 57 Additional Problems P3.47 Let θ represent the angle between the directions of A and B. Since A and B have the same magnitudes, A, B, and R A B= + form an isosceles triangle in which the angles are 180° −θ, θ 2 , and θ 2 . The magnitude of R is then R A= ⎛ ⎝⎜ ⎞ ⎠⎟2 2 cos θ . This can be seen from applying the law of cosines to the isosceles triangle and using the fact that B A= . Again, A, −B, and D A B= − form an isosceles triangle with apex angle θ. Applying the law of cosines and the identity 1 2 2 2 −( ) = ⎛ ⎝⎜ ⎞ ⎠⎟cos sinθ θ gives the magnitude of D as D A= ⎛ ⎝⎜ ⎞ ⎠⎟2 2 sin θ . The problem requires that R D= 100 . Thus, 2 2 200 2 A Acos sin θ θ⎛ ⎝⎜ ⎞ ⎠⎟ = ⎛ ⎝⎜ ⎞ ⎠⎟ . This gives tan . θ 2 0 010 ⎛ ⎝⎜ ⎞ ⎠⎟ = and θ = 1 15. ° . P3.48 Let θ represent the angle between the directions of A and B. Since A and B have the same magnitudes, A, B, and R A B= + form an isosceles triangle in which the angles are 180° −θ, θ 2 , and θ 2 . The magnitude of R is then R A= ⎛ ⎝⎜ ⎞ ⎠⎟2 2 cos θ . This can be seen by applying the law of cosines to the isosceles triangle and using the fact that B A= . Again, A, –B, and D A B= − form an isosceles triangle with apex angle θ. Applying the law of cosines and the identity 1 2 2 2 −( ) = ⎛ ⎝⎜ ⎞ ⎠⎟cos sinθ θ gives the magnitude of D as D A= ⎛ ⎝⎜ ⎞ ⎠⎟2 2 sin θ . The problem requires that R nD= or cos sin θ θ 2 2 ⎛ ⎝⎜ ⎞ ⎠⎟ = ⎛ ⎝⎜ ⎞ ⎠⎟n giving θ = ⎛ ⎝⎜ ⎞ ⎠⎟ − 2 11 tan n . The larger R is to be compared to D, the smaller the angle between A and B becomes. P3.49 The position vector from the ground under the controller of the ﬁrst airplane is r i j1 19 2 25 19 2 25= ( )( ) + ( )( ). cos ˆ . sin ˆkm km° ° ++ ( ) = + +( ) 0 8 17 4 8 11 0 8 . ˆ . ˆ . ˆ . ˆ . km km k i j k The second is at r i j2 17 6 20 17 6 20= ( )( ) + ( )( ). cos ˆ . sin ˆkm km° ° ++ ( ) = + +( ) 1 16 5 6 02 1 1 .1 km km ˆ . ˆ . ˆ . ˆ . k i j k Now the displacement from the ﬁrst plane to the second is r r i j k2 1 0 863 2 09 0 3− = − − +( ). ˆ . ˆ . ˆ km with magnitude 0 863 2 09 0 3 2 29 2 2 2 . . . .( ) + ( ) + ( ) = km . A B R θ θ θ /2 A D -B FIG. P3.47 FIG. P3.48 13794_03_ch03_p045-064.indd 5713794_03_ch03_p045-064.indd 57 11/28/06 4:40:17 PM11/28/06 4:40:17 PM
14. 14. 58 Chapter 3 P3.50 Take the x axis along the tail section of the snake. The displacement from tail to head is 240 420 240 180 105 180m m mˆ cos ˆ sii i+ −( ) −( ) −° ° nn ˆ ˆ ˆ75 287°j i j= −m 174 m . Its magnitude is 287 174 335 2 2 ( ) + ( ) =m m. From v = distance ∆t , the time for each child’s run is Inge: distance m h km s ∆t = = ( )( )( ) v 335 1 3 600 12 km m h s Olaf: m s 3. ( )( )( ) = = ⋅ 1 000 1 101 420 ∆t 333 m s= 126 . Inge wins by 126 101 25 4− = . s . P3.51 Let A represent the distance from island 2 to island 3. The displacement is A = A at 159°. Represent the displacement from 3 to 1 as B = B at 298°. We have 4.76 km at 37 0° + + =A B . For x components 4 76 37 159 298 0 3 80 . cos cos cos . km km ( ) + + = − ° ° °A B 00 934 0 469 0 8 10 1 99 . . . . A B B A + = = − +km For y components 4 76 37 159 298 0 2 86 . sin sin sin . km km ( ) + + = + ° ° °A B 00 358 0 883 0. .A B− = (a) We solve by eliminating B by substitution: 2 86 0 358 0 883 8 10 1 99 0 2 86 . . . . . . km km+ − − +( ) =A A kkm km km + + − = = = 0 358 7 15 1 76 0 10 0 1 40 7 . . . . . . A A A A 117 km (b) B = − + ( ) =8 10 1 99 7 17 6 15. . . .km km km P3.52 (a) Rx = 2 00. , Ry = 1 00. , Rz = 3 00. (b) R = + + = + + = =R R Rx y z 2 2 2 4 00 1 00 9 00 14 0 3 74. . . . . (c) cos cos .θ θx x x xR R = ⇒ = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ =− R R f1 57 7° rrom + x cos cos .θ θy y y yR R = ⇒ = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ =− R R f1 74 5° rrom + y cos cos .θ θz z z zR R = ⇒ = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ =− R R f1 36 7° rrom + z N B28º A C 69º 37º 1 2 3 E FIG. P3.51 13794_03_ch03_p045-064.indd 5813794_03_ch03_p045-064.indd 58 11/28/06 4:40:19 PM11/28/06 4:40:19 PM
15. 15. Vectors 59 P3.53 v i j i= + = +( ) +v vx y ˆ ˆ cos . ˆ sin .300 100 30 0 100 30 0° °°( ) = +( ) = ˆ ˆ . ˆ j v i j v 387 50 0 390 mi h mi h at 7.337 N of E° *P3.54 (a) A j i j= − = +60 80 80cm and B cos sinθ θ(( ) cm so A B i j+ = + −( )80 80 60cos sinθ θ centimeters and A B+ = ( ) + −( )⎡⎣ ⎤⎦ =80 80 60 80 2 2 cos sin coθ θ2 1/2 cm ss sin ( )( )cos2 2 2 2 80 2 80 60 60θ θ θ+ − +⎡⎣ ⎤⎦ 1/2 cm Now sin2 θ + cos2 θ = 1 for all θ, so we have A B+ = + −⎡⎣ ⎤⎦80 60 2 80 60 102 2 ( )( )cosθ 1/2 cm = 0000 600 cm 1/2 −[ ]9 cosθ (b) For θ = 270° , cosθ = −1 and the expression takes on its maximum value, [10 000 + 9 600] 1ր2 cm = 140 cm . (c) For θ = 90° , cosθ = +1 and the expression takes on its minimum value, [10 000 − 9 600] 1ր2 cm = 20.0 cm . (d) They do make sense. The maximum value is attained when A and B are in the same direction, and it is 60 cm + 80 cm. The minimum value is attained when A and B are in opposite direc- tions, and it is 80 cm − 60 cm. *P3.55 ∆r i j= −( ) =∫ 1.2 m s m s 1.2 s ˆ . ˆ . ր ր9 8 0 0 380 t dt 22 m s m s 1.2 s 2 s t tˆ . ˆ ˆ . . i jր ր 0 0 38 2 0 0 38 9 8 2 − = ii jm s s m s s2 ր ր( ) −( )− ( ) −⎛ 0 38 0 9 8 0 38 0 2 2 . . ˆ . ⎝⎝⎜ ⎞ ⎠⎟ = −0 456 0 708. ˆ . ˆi jm m P3.56 Choose the + x axis in the direction of the ﬁrst force, and the y axis at 90° counterclockwise from the x axis. Then each force will have only one nonzero component. The total force, in newtons, is then 12 0 31 0 8 40 24 0 3 60 7 00. ˆ . ˆ . ˆ . ˆ . ˆ . ˆi j i j i j+ − − = ( )+ (( ) N . The magnitude of the total force is 3 60 7 00 7 87 2 2 . . .( ) + ( ) =N N and the angle it makes with our + x axis is given by tan . . θ = ( ) ( ) 7 00 3 60 , θ = °62 8. . Thus, its angle counterclockwise from the horizontal is 35 0 62 8 97 8. . .° ° °+ = . R 35.0º y 24 N horizontal 31 N 8.4 N 12 N x FIG. P3.56 13794_03_ch03_p045-064.indd 5913794_03_ch03_p045-064.indd 59 11/28/06 4:40:19 PM11/28/06 4:40:19 PM
16. 16. 60 Chapter 3 P3.57 d i d j d i 1 2 3 100 300 150 30 0 150 = = − = − ( ) − ˆ ˆ cos . ˆ° ssin . ˆ ˆ . ˆ cos . 30 0 130 75 0 200 60 04 ° ° ( ) = − − = − j i j d (( ) + ( ) = − + = + ˆ sin . ˆ ˆ ˆi j i j R d d 200 60 0 100 173 1 ° 22 3 4 2 130 202 130 202 + + = − −( ) = −( ) + −( d d i j R ˆ ˆ m )) = = ⎛ ⎝ ⎞ ⎠ = = − 2 1 240 202 130 57 2 180 m φ θ tan . ° ++ =φ 237° P3.58 d dt d t dt r i j j j= + −( ) = + − = −( ) 4 3 2 0 0 2 2 00 ˆ ˆ ˆ ˆ . ˆm s jj The position vector at t = 0 is 4 3ˆ ˆi j+ . At t = 1 s, the position is 4 1ˆ ˆi j+ , and so on. The object is moving straight downward at 2 mրs, so d dt r represents its velocity vector . P3.59 (a) You start at point A: r r i j1 30 0 20 0= = −( )A . ˆ . ˆ m. The displacement to B is r r i j i j iB A− = + − + = +60 0 80 0 30 0 20 0 30 0. ˆ . ˆ . ˆ . ˆ . ˆ 1100ˆj. You cover half of this, 15 0 50 0. ˆ . ˆi j+( ) to move to r i j i j i2 30 0 20 0 15 0 50 0 45 0 30 0= − + + = +. ˆ . ˆ . ˆ . ˆ . ˆ . ˆˆj. Now the displacement from your current position to C is r r i j i jC − = − − − − = −2 10 0 10 0 45 0 30 0 55 0. ˆ . ˆ . ˆ . ˆ . ˆii j− 40 0. ˆ . You cover one-third, moving to r r r i j i3 2 23 45 0 30 0 1 3 55 0 40 0= + = + + − −∆ . ˆ . ˆ . ˆ . ˆjj i j( )= +26 7 16 7. ˆ . ˆ . The displacement from where you are to D is r r i j i j iD − = − − − = −3 40 0 30 0 26 7 16 7 13 3. ˆ . ˆ . ˆ . ˆ . ˆ 446 7. ˆj. You traverse one-quarter of it, moving to r r r r i j i4 3 3 1 4 26 7 16 7 1 4 13 3= + −( ) = + + −D . ˆ . ˆ . ˆ 446 7 30 0 5 00. ˆ . ˆ . ˆj i j( )= + . The displacement from your new location to E is r r i j i j iE − = − + − − = −4 70 0 60 0 30 0 5 00 100. ˆ . ˆ . ˆ . ˆ ˆ ++ 55 0. ˆj of which you cover one-ﬁfth the distance, − +20 0 11 0. ˆ . ˆi j, moving to r r i j i j4 45 30 0 5 00 20 0 11 0 10 0+ = + − + =∆ . ˆ . ˆ . ˆ . ˆ . ˆii j+16 0. ˆ . The treasure is at 10 0. m, 16.0 m( ) . FIG. P3.57 continued on next page 13794_03_ch03_p045-064.indd 6013794_03_ch03_p045-064.indd 60 11/28/06 4:40:20 PM11/28/06 4:40:20 PM
17. 17. Vectors 61 (b) Following the directions brings you to the average position of the trees. The steps we took numerically in part (a) bring you to r r r r r A B A A B + −( ) = +⎛ ⎝⎜ ⎞ ⎠⎟ 1 2 2 then to r r r r r r r rA B C A B A B C +( ) + − +( ) = + + 2 2 3 3 ր then to r r r r r r r r rA B C D A B C A B + +( ) + − + +( ) = + 3 3 4 ր ++ +r rC D 4 and last to r r r r r r r r rA B C D E A B C D+ + +( ) + − + + +( ) 4 4 5 ր == + + + +r r r r rA B C D E 5 . This center of mass of the tree distribution is the same location whatever order we take the trees in. P3.60 (a) Let T represent the force exerted by each child. The x component of the resultant force is T T T T T Tcos cos cos . .0 120 240 1 0 5 0 5+ + = ( )+ −( )+ −(° ° )) = 0 The y component is T T T T Tsin sin sin . .0 120 240 0 0 866 0 866 0+ + = + − = . Thus, ∑ =F 0 (b) If the total force is not zero, it must point in some direction. When each child moves one space clockwise, the whole set of forces acting on the tire turns clockwise by that angle so the total force must turn clockwise by that angle, 360° N . Because each child exerts the same force, the new situation is identical to the old and the net force on the tire must still point in the original direction. But the force cannot have two different directions. The contradiction indicates that we were wrong in supposing that the total force is not zero. The total force must be zero. FIG. P3.60 13794_03_ch03_p045-064.indd 1713794_03_ch03_p045-064.indd 17 12/7/06 6:33:00 PM12/7/06 6:33:00 PM
18. 18. 62 Chapter 3 P3.61 Since A B j+ = 6 00. ˆ, we have A B A Bx x y y+( ) + +( ) = +ˆ ˆ ˆ . ˆi j i j0 6 00 giving A Bx x+ = 0 or A Bx x= − [1] and A By y+ = 6 00. . [2] Since both vectors have a magnitude of 5.00, we also have A A B Bx y x y 2 2 2 2 2 5 00+ = + = . . From A Bx x= − , it is seen that A Bx x 2 2 = . Therefore, A A B Bx y x y 2 2 2 2 + = + gives A By y 2 2 = . Then, A By y= and Equation [2] gives A By y= = 3 00. . Deﬁning θ as the angle between either A or B and the y axis, it is seen that cos . . .θ = = = = A A B B y y 3 00 5 00 0 600 and θ = °53 1. . The angle between A and B is then φ θ= =2 106° . P3.62 (a) From the picture, R i j1 = +a bˆ ˆ and R1 2 2 = +a b . (b) R i j k2 = + +a b cˆ ˆ ˆ ; its magnitude is R1 2 2 2 2 2 + = + +c a b c . ANSWERS TO EVEN PROBLEMS P3.2 (a) 2 17 1 25. , .m m( ); −( )1 90 3 29. , .m m (b) 4.55 m P3.4 y = 1.15; r = 2.31 P3.6 310 km at 57° S of W P3.8 9.5 N at 57° P3.10 (a) ~105 m vertically upward (b) ~103 m vertically upward P3.12 See the solution; the sum of a set of vectors is not affected by the order in which the vectors are added. FIG. P3.62 FIG. P3.61 13794_03_ch03_p045-064.indd 6213794_03_ch03_p045-064.indd 62 11/28/06 4:40:21 PM11/28/06 4:40:21 PM
19. 19. Vectors 63 P3.14 We assume that the shopping cart stays on the level ﬂoor. There are two possibilities. If both of the turns are right or both left, the net displacement is (a) 25.0 m (b) at 36.9°. If one turn is right and one is left, we have (a) 61.8 m (b) at 14.0°. P3.16 1.31 km north; 2.81 km east P3.18 (a) 5.00 blocks at 53.1° N of E (b) 13.0 blocks P3.20 − +25 0 43 3. ˆ . ˆm mi j P3.22 788 48 0mi at north of east. ° P3.24 (a) see the solution (b) 5.00ˆi + 4.00ˆj, 6.40 at 38.7°, –1.00ˆi + 8.00ˆj, 8.06 at 97.2° P3.26 (a) 4.10 m toward the top of the hill (b) 2.87 m P3.28 42.7 yards P3.30 C i j= −7 30 7 20. ˆ . ˆcm cm P3.32 A B+ = (2.60 ˆi + 4.50 ˆj)m P3.34 (a) 2.83 m at θ = 315° (b) 13.4 m at θ = 117° P3.36 (a) 10.4 cm; (b) 35.5° P3.38 1 43 104 . × m at 32.2° above the horizontal P3.40 (a) 15 1 7 72. ˆ . ˆi j+( ) cm (b) − +( )7 72 15 1. ˆ . ˆi j cm (c) + +( )7 72 15 1. ˆ . ˆi j cm P3.42 157 km P3.44 (a) a = 5.00 and b = 7.00 (b) For vectors to be equal, all of their components must be equal. A vector equation contains more information than a scalar equation. P3.46 (a) see the solution (b) 18.3 b (c) 12.4 b at 233° counterclockwise from east P3.48 2 11 tan− ⎛ ⎝ ⎞ ⎠n P3.50 25.4 s P3.52 (a) 2.00, 1.00, 3.00 (b) 3.74 (c) θx = 57.7°, θy = 74.5°, θz = 36.7° P3.54 (a) (10 000 − 9 600 cos θ )1ր2 cm (b) 270° ; 140 cm (c) 90° ; 20.0 cm (d) They do make sense. The maximum value is attained when A and B are in the same direction, and it is 60 cm + 80 cm. The minimum value is attained when A and B are in opposite directions, and it is 80 cm − 60 cm. P3.56 We choose the x axis to the right at 35° above the horizontal and the y axis at 90° counterclock- wise from the x axis. Then each vector has only a single nonzero component. The resultant is 7.87 N at 97.8° counterclockwise from a horizontal line to the right. P3.58 −( )2 00. ˆmրs j; its velocity vector P3.60 (a) zero (b) see the solution P3.62 (a) R i j1 = +a bˆ ˆ; R1 2 2 = +a b (b) R i j k2 = + +a b cˆ ˆ ˆ; R2 2 2 2 = + +a b c 13794_03_ch03_p045-064.indd 6313794_03_ch03_p045-064.indd 63 11/28/06 4:40:22 PM11/28/06 4:40:22 PM
20. 20. 13794_03_ch03_p045-064.indd 2013794_03_ch03_p045-064.indd 20 12/7/06 2:08:57 PM12/7/06 2:08:57 PM