Two-Dimensional Motion and Vectors Chapter 3 I Section Review, p. 87 Givens 2. ∆x1 = 85 m d2 = 45 m q2 = 30.0° Solutions Students should use graphical techniques. Their answers can be checked using the techniques presented in Section 3-2. Answers may vary. ∆x2 = d2(cos q2) = (45 m)(cos 30.0°) = 39 m ∆y2 = d2(sin q2) = (45 m)(sin 30.0°) = 22 m ∆xtot = ∆x1 + ∆x2 = 85 m + 39 m = 124 m ∆ytot = ∆y2 = 22 m d = (∆ m)2 +(22 m)2 xtot)2+(∆ ytot)2 = (1 24 d = 15 m2+ m2 = 15 m2 = 126 m 400 480 900 ! ! ∆ytot 22 m q = tan−1 = tan−1 = (1.0 × 101)° above the horizontal ∆xtot 124 m 3. vy, 1 = 2.50 × 102 km/h Students should use graphical techniques. v2 = 75 km/h vx,2 = v2(cos q2) = (75 km/h)[cos (−45°)] = 53 km/h q2 = −45° vy,2 = v2(sin q2) = (75 km/h)[sin (−45°)] = −53 km/h vy,tot = vy,1 + vy,2 = 2.50 × 102 km/h − 53 km/h = 197 km/h vx,tot = vs,2 = 53 km/h 2 2 v = (v vy,to )2 +(1 )2 x,to ( 3km /h 97 km /h t)+ t) = (5 Copyright © by Holt, Rinehart and Winston. All rights reserved. v = 28 00 km 2/h2+3880 0km 2/h2 = 41 600 km 2/h2 = 204 km/h ! ! v tot 204 km/h q = tan−1 y, = tan−1 = 75° north of east vx,tot 53 km/h 2.50 × 102 km/h 4. vy,1 = 2 = 125 km/h vx,2 = 53 km/h vy,2 = −53 km/h Students should use graphical techniques. vy,dr = vy,1 + vy,2 = 125 km/h − 53 km/h = 72 km/h vx,dr = vx,2 = 53 km/h 2 v = (v )2 =vy,d )2 + (72km )2 x,dr 3km /h /h r) = (5 v = 28 03km 00 km 2/h2+520 0km 2/h2 = 8. 0 1 2/h2 = 89 km/h ! ! vy dr 72 km/h q = tan−1 , = tan−1 = 54° north of east vx,dr 53 km/h Section One—Pupil’s Edition Solutions I Ch. 3–1 Practice 3A, p. 91 Givens Solutions 1. ∆x1 = 8 km east a. d = ∆x1 + ∆x2 + ∆x3 = 8 km + 3 km + 12 km = 23 km ∆x1 = 8 km I ∆x2 = 3 km west = −3km, east b. ∆xtot = ∆x1 + ∆x2 + ∆x3 = 8 km + (−3 km) + 12 km = 17 km east ∆x2 = 3 km ∆x3 = 12 km east ∆x3 = 12 km ∆y = 0 km 2. ∆x = 7.5 m ∆y = 45.0 m d= ∆ x2 + ∆ y 2 = (7 m )2 +(45 m )2 .5 .0 d = 56 m2+ m2 = 45.6 m 202 0m 2 = 20 80 Measuring direction with respect to y (north), ! ! ∆x 7.5 m q = tan−1 = tan−1 = 9.5° east of due north ∆y 45.0 m 3. ∆x = 6.0 m ∆y = 14.5 m d= ∆ ∆y2 = (6 m )2 +(14 m )2 x2+ .0 .5 d = 36 m2+ 02 m2 = 24 2.1 0×1 6m 2 = 15.7 m Measuring direction with respect to the length of the field (down the field), ! ! ∆x 6.0 m q = tan−1 = tan−1 = 22° to the side of downfield ∆y 14.5 m ∆y = –1.4 m d= ∆ ∆y2 = (1 m )2 +– (1. )2 x2+ .2 4m d = 1. m2 = 3. 4m 2+2.0 4m 2 = 1.8 m ! ! ∆y –1.4 m q = tan−1 = tan−1 = –49° = 49° below the horizontal ∆x 1.2 m Practice 3B, p. 94 1. v = 105 km/h vx = v(cos q ) = (105 km/h)(cos 25°) = 95 km/h q = 25° 2. v = 105 km/h vy = v(sin q) = (105 km/h)(sin 25°) = 44 km/h q = 25° 3. v = 22 m/s vx = v(cos q ) = (22 m/s)(cos 15°) = 21 m/s q = 15° vy = v(sin q) = (22 m/s)(sin 15°) = 5.7 m/s 4. d = 5 m q = 90° I Ch. 3–2 ∆x = d(cos q) = (5 m)(cos 90°) = 0 m ∆y = d(sin q) = (5 m)(sin 90°) = Holt Physics Solution Manual 5m Copyright © by Holt, Rinehart and Winston. All rights reserved. 4. ∆x = 1.2 m Practice 3B, p. 94 cont. Givens 5. d = 125 m q = −25° 6. d = 23.0 m q = −30.5° 7. a = 2.5 m/s2 q = −18° Solutions ∆x = d(cos q ) = (125 m)[cos (−25°)] = 1.1 × 102 m ∆y = d(sin q) = (125 m)[sin (−25°)] = −53 m I ∆x = d(cos q) = (23.0 m)[cos (−30.5°)] = 19.8 m ∆y = d(sin q) = (23.0 m)[sin (−30.5°)] = −11.7 m ax = a(cos q) = (2.5 m/s2)[cos (−18°)] = 2.4 m/s2 ay = a(sin q) = (2.5 m/s2)[sin (−18°)] = −0.77 m/s2 Practice 3C, p. 97 1. d1 = 35 m ∆x1 = d1(cos q1) = (35 m)(cos 0.0°) = 35 m q1 = 0.0° ∆y1 = d1(sin q1) = (35 m)(sin 0.0°) = 0.0 m d2 = 15 m ∆x2 = d2(cos q2 ) = (15 m)(cos 25°) = 14 m q2 = 25° ∆y2 = d2(sin q2 ) = (15 m)(sin 25°) = 6.3 m ∆xtot = ∆x1 + ∆x2 = 35 m + 14 m = 49 m ∆ytot = ∆y1 + ∆y2 = 0.0m + 6.3m = 6.3 m dtot = (∆ xtot )2 +(∆ ytot )2 = (4 )2 +(6. )2 9m 3m dtot = 24 m2+ m2 = 49 m 00 40m 2 = 24 00 ! ! Copyright © by Holt, Rinehart and Winston. All rights reserved. 6.3 m ∆ytot qtot = tan−1 = tan−1 = 7.3° to the right of downfield 49 m ∆xtot 2. d1 = 2.5 km ∆x1 = d1(cos q1) = (2.5 km)(cos 35°) = 2.0 km q1 = 35° ∆y1 = d1(sin q1) = (2.5 km)(sin 35°) = 1.4 km d2 = 5.2 km ∆x2 = d2(cos q2) = (5.2 km)(cos 22°) = 4.8 km q2 = 22° ∆y2 = d2(sin q2) = (5.2 km)(sin 22°) = 1.9 km ∆xtot = ∆x1 + ∆x2 = 2.0 km + 4.8 km = 6.8 km ∆ytot = ∆y1 + ∆y2 = 1.4 km + 1.9 km = 3.3 km xt )2 +(∆ yt )2 = (6 )2 +(3. )2 dtot = (∆ .8 km 3km ot ot dtot = 46 km 2=11 km 2 = 57 km 2 = 7.5 km ! ! ∆ytot 3.3 km qtot = tan−1 = tan−1 = 26° above the horizontal ∆xtot 6.8 km Section One—Pupil’s Edition Solutions I Ch. 3–3 Givens Solutions 3. d1 = 8.0 m I Measuring direction with respect to y = (north), q1 = 0.0° ∆x1 = d1(sin q1) = (8.0 m)(sin 0.0°) = 0.0 m d2 = 3.5 m ∆y1 = d1(cos q1) = (8.0 m)(cos 0.0°) = 8.0 m q2 = 35° ∆x2 = d2(sin q2) = (3.5 m)(sin 35°) = 2.0 m d3 = 5.0 m ∆y2 = d2(cos q2) = (3.5 m)(cos 35°) = 2.9 m q3 = 90.0° ∆x3 = d3(sin q3) = (5.0 m)(sin 90.0°) = 5.0 m ∆y3 = d3(cos q3) = (5.0 m)(cos 90.0°) = 0.0 m ∆xtot = ∆x1 + ∆x2 + ∆x3 = 0.0. m + 2.0 m + 5.0 m = 7.0 m ∆ytot = ∆y1 + ∆y2 + ∆y3 = 8.0. m + 2.9 m + 0.0 m = 10.9 m dtot = (∆ xt )2(∆ yt )2 = (7 m )2 +(10 m )2 .0 .9 ot ot dtot = 49 m2+ m2 = 16 119 8m 2 = 13.0 m ! ! ∆xtot 7.0 m qtot = tan−1 = tan−1 = 33° east of north ∆ytot 10.9 m 4. d1 = 75 km Measuring direction with respect to y (north), q1 = −30.0° ∆x1 = d1(sin q1) = (75 km)(sin − 30.0°) = −38 km d2 = 155 km ∆y1 = d1(cos q1) = (75 km)(cos − 30.0°) = 65 km q2 = 60.0° ∆x2 = d2(sin q2) = (155 km)(sin 60.0°) = 134 km ∆y2 = d2(cos q2) = (155 km)(cos 60.0°) = 78 km ∆xtot = ∆x1 + ∆x2 = −38 km + 134 km = 96 km ∆ytot = ∆y1 + ∆y2 = 65 km + 78 km = 143 km 2 2 dtot = (∆ xt )2 +( ∆ yt )2 = (9 6 km )2 +( 14 3 km )2 = 9. 2 ×1 03k m .0 ×1 04 km +2 ot ot = 2. 92 ×104km 2 = 171 km ! ! ∆xtot 96 km q = tan−1 = tan−1 = 34° east of north ∆ytot 143 km Section Review, p. 97 2. vx = 3.0 m/s vy = 5.0 m/s 2 a. v = vx2+ m/s )2 =(5. )2 v .0 0m /s y = (3 v = 9. s2 +25m s2 = 34 m2/ s2 = 5.8 m/s 0m 2/ 2/ ! ! vy 5.0 m/s q = tan−1 = tan−1 = 59° downriver from its intended path vx 3.0 m/s vx = 1.0 m/s vy = 6.0 m/s 2 b. v = vx2+ m/s )2 +(6. )2 v .0 0m /s y = (1 v = 1. s2 +36m s2 = 37 m2/ s2 = 6.1 m/s 0m 2/ 2/ ! ! v 1.0 m/s q = tan−1 x = tan−1 = 9.5° from the direction the wave is traveling vy 6.0 m/s I Ch. 3–4 Holt Physics Solution Manual Copyright © by Holt, Rinehart and Winston. All rights reserved. Givens Solutions 3. d = 10.0 km a. ∆x = d(cos q) = (10.0 km)(sin 45.0°) = 7.07 km q = 45.0° ∆y = d(sin q) = (10.0 km)(cos 45.0°) = 7.07 km a = 2.0 m/s2 b. ax = a(cos q) = (2.0 m/s2)(cos 35°) = 1.6 m/s2 q = 35° I ay = a(sin q) = (2.0 m/s2)(sin 35°) = 1.1 m/s2 4. d1 = 10.0 m ∆x1 = d1(cos q) = (10.0 m)(cos 55°) = 5.7 m q1 = 55° ∆y1 = d1(sin q) = (10.0 m)(sin 55°) = 8.2 m d2 = 5.0 m ∆x2 = d2(cos q 2) = (5.0 m)(cos 0.0°) = 5.0 m q2 = 0.0° ∆y2 = d2(sin q 2) = (5.0 m)(sin 0.0°) = 0.0 m ∆xtot = ∆x1 + ∆x2 = 5.7 m + 5.0 m = 10.7 m ∆ytot = ∆y1 + ∆y2 = 8.2 m + 0.0 m = 8.2 m dtot = (∆ xt )2 +(∆ yt )2 = (1 )2 +(8. )2 0. 7m 2m ot ot dtot = 11 4m 2+67m 2 = 18 1m 2 = 13.5 m ! ! ∆ytot 8.2 m qtot = tan−1 = tan−1 = 37° north of west ∆xtot 10.7 m Practice 3D, p. 102 1. ∆y = −0.70 m 2∆y ∆x x ∆x = 0.25 m 2 g = 9.81 m/s 2. ∆y = −1.0 m Copyright © by Holt, Rinehart and Winston. All rights reserved. "#−#g = v −g v = " #2∆#y ∆x = " #(2−#)9(.#−8#01. #7m#0/#ms#) (0.25 m) = ∆t = ∆x = 2.2 m 2 x "#−#g = v −g ∆x = (2.2 m) = v = −" "#(−29#)(.8#−11.#m#0#/ms#) 2#∆#y ∆t = 2∆y ∆x x 2 g = 9.81 m/s 3. ∆y = −5.4 m ∆x = 8.0 m "#−#g = v −g m/s v = " #2∆#y ∆x = " #(−29#)(.8#−1#5. #4#m#) (8.0 m) = g = 9.81 m/s 4. vx = 7.6 m/s ∆y = −2.7 m 2 g = 9.81 m/s 2 x "#−#g = v 2∆y (2)(−2.7 m) ∆x = " #−#g v = " #−#9.#81# #m#/s# (7.6 m/s) = ∆t = 4.9 m/s 2∆y ∆x x 2 2 x ∆t = 0.66 m/s 7.6 m/s 2∆y ∆x x x 2 5.6 m Section One—Pupil’s Edition Solutions I Ch. 3–5 Practice 3E, p. 104 Givens Solutions 1. ∆x = 4.0 m ∆x = vi (cos q)∆t q = 15° I vi = 5.0 m/s ∆x 4.0 m ∆t = = = 0.83 s vi (cos q ) (5.0 m/s)(cos 15°) ∆ymax = −2.5 m ∆y = vi (sin q)∆t − 2 g ∆t 2 = (5.0 m/s)(sin 15°)(0.83 s) − 2 (9.81 m/s2)(0.83 s)2 g = 9.81 m/s2 ∆y = 1.1 m − 3.4 m = −2.3 m 2. ∆x = 301.5 m q = 25.0° 2 g = 9.81 m/s 1 1 yes At maximum height, vy,f = 0 m/s. vy,f = vi (sin q ) − g∆t = 0 vi (sin q) ∆t = g v2 ∆x = vi (cos q)∆t = (cos q)(sin q)i g x (9.81 m/s )(301.5 m) = = 87.9 m/s " #(c#o#s #qg#)∆ (# n# q #) "## si# (cos 25.0°)(sin 25.0°) 2 vi = vy,f 2 = vi 2(sin q )2 − 2g∆ymax = 0 2 2 vi 2(sin q)2 (87.9 m/s) (sin 25.0°) = = 70.3 m ∆y max = 2 (2)(9.81 m/s ) 2g q = 25° ∆x ∆x 42.0 m ∆t = = = = 2.0 s vx vi (cos q) (23.0 m/s)(cos 25°) vi = 23.0 m/s At maximum height, vy,f = 0 m/s. 3. ∆x = 42.0 m g = 9.81 m/s2 vy,f 2 = vy,i2 − 2g∆ymax = 0 4. ∆x = 2.00 m ∆y = 0.55 m q = 32.0° g = 9.81 m/s2 I Ch. 3–6 ∆x = vi(cos q)∆t ∆x ∆t = vi(cos q) $ % $ % ∆x ∆x ∆y = vi(sin q)∆t − 12g∆t 2 = vi(sin q) − 12 g vi(cos q) vi(cos q) g∆x2 ∆y = ∆x(tan q) − 2vi2(cos q)2 g∆x2 ∆x(tan q) − ∆y = 2 2 2vi (cos q) g∆x2 2vi2(cos q)2 = ∆x(tan q) − ∆y 2 vi = "## vi = "### vi = "## "## g∆x2 2 2(cos q) [∆x(tan q) − ∆y] (9.81 m/s2)(2.00 m)2 (2)(cos 32.0°)2[(2.00 m)(tan 32.0°) − 0.55 m] (9.81 m/s2)(2.00 m)2 = (2)(cos 32.0°)2(1.25 − 0.55 m) Holt Physics Solution Manual (9.81 m/s2)(2.00 m)2 = 6.2 m/s (2)(cos 32.0°)2(0.70 m) Copyright © by Holt, Rinehart and Winston. All rights reserved. vy,i2 vi2(sin q)2 (23.0 m/s)2(sin 25°)2 = ∆ymax = = = 4.8 m (2)(9.81 m/s2) 2g 2g Givens 5. ∆x = 31.5 m q = 40.0° g = 9.81 m/s2 Solutions ∆x = vi(cos q)∆t ∆x ∆t = vi(cos q) ∆y = vi(sin q)∆t− 12 g∆t2 = 0 I ∆y = vi(sin q) − 12 g∆t = 0 g∆x ∆y = vi(sin q) − = 0 2vi(cos q) (9.81 m/s2)(31.5 m) g∆x vi = = = 17.7 m/s (2)(cos 40.0°)(sin 40.0°) 2(cos q)(sin q) "######### "## At maximum height, vy,f = 0 m/s. vy,f2 = vi2(sin q)2− 2g∆ymax = 0 2 2 vi2(sin q)2 (17.7 m/s) (sin 40.0°) = ∆ymax = = 6.60 m 2 (2)(9.81 m/s ) 2g Section Review, p. 105 3. vx = 100.0 m/s ∆y = −50.0 m g = 9.81 m/s2 4. vx = 100.0 m/s ∆x = 319 m ∆y = −50.0 m Copyright © by Holt, Rinehart and Winston. All rights reserved. g = 9.81 m/s2 1 ∆y = − 2g∆t2 ∆t = "#−#g 2∆y "## 2∆y ∆x = vx∆t = vx = (100.0 m/s) −g "(2 #−)#9(#.−85#10 #m.0#/m#s )# = 2 319 m 2 vy = vy =2g∆ m/s )2 −(2) )2(− m) = ±31.3 m/s = −31.3 m/s y = (0 (9 .8 1m /s 50 .0 ,i 2 vtot = vy2+ m/s )2 +(− m/s )2 = 1. 04 m2s2+ m2/ s2 00 .0 31 .3 00 0×1 9.8 0×102 v y = (1 vtot = 10 m2/ s2 = 104.8 m/s 980 ! ! v −31.3 m/s q = tan−1 y = tan−1 = −17.4° vx 100.0 m/s q = 17.4° below the horizontal 5. ∆y = −125 m vx = 90.0 m/s g = 9.81 m/s2 1 ∆y = − 2g∆t 2 2∆y ∆t = = −g " #−#9#.8#1 #m#/s# = (2)(−125 m) 2 5.05 s ∆x = vx ∆t = (90.0 m/s)(5.05 s) = 454 m 6. ∆t = 0.50 s ∆x = 1.5 m q = 33° 7. ∆t = 0.35 s q = 67° vi = 5.0 m/s g = 9.81 m/s2 ∆x = vi (cos q)∆t ∆x 1.5 m vi = = = 3.6 m/s (cos q )∆t (cos 33°)(0.50 s) 1 ∆y = vi (sin q )∆t − 2 g∆t 2 1 ∆y = (5.0 m/s)(sin 67°)(0.35 s) − 2 (9.81 m/s2)(0.35 s)2 ∆y = 1.6 m − 0.60 m = 1.0 m Section One—Pupil’s Edition Solutions I Ch. 3–7 Practice 3F, p. 109 Givens Solutions 1. vte = +15 m/s vbe = vbt + vte = −15 m/s + 15 m/s = 0 m/s vbt = −15 m/s I 2. vaw = +18.0 m/s vsa = −3.5 m/s 3. vfw = 2.5 m/s north vwe = 3.0 m/s east vsw = vsa + vaw = − 3.5 m/s 18.0 m/s vsw = 14.5 m/s in the direction that the aircraft carrier is moving vfe = vfw + vwe 2 v 2 = (2.5 m/s)2 + (3.0 m/s)2 vtot = vf w w+ e vtot = 6. s2 +9.0 m2/ s2 = 15 .2 m2/ s2 = 3.90 m/s 2m 2/ ! ! vfw 2.5 m/s q = tan−1 = tan = (4.0 × 101)° north of east vwe 3.0 m/s 4. vtr = 25.0 m/s north vdt = 1.75 m/s at 35.0° east of north vdr = vdt + vtr vx,tot = vx,dt = (1.75 m/s)(sin 35.0°) = 1.00 m/s vy,dt = (1.75 m/s)(cos 35.0°) = 1.43 m/s vy,tot = vtr + vy,dt = 25.0 m/s + 1.43 m/s = 26.4 m/s 2 (v 2 vtot = (v )2 +(26 m/s )2 x, to y .0 0m /s .4 t)+ ,to t) = (1 vtot = 1. m2/ s2 +697 m2/ s2 = 69 s2 = 26.4 m/s 00 8m 2/ ! ! vx, tot 1.00 m/s = tan−1 = 2.17° east of north q = tan−1 vy, tot 26.4 m/s 2. vwg = −9 m/s vbg = 1 m/s 3. vbw = 0.15 m/s north vwe = 1.50 m/s east vbw = vbg vgw = vbg − vwg = (1 m/s) – (–9 m/s) = 1 m/s + 9 m/s vbw = 10 m/s away in the oppposite direction vbe = vbw + vwe vtot = vbw )2 +(1. m/s )2 2+vwe2 = (0 .1 5m /s 50 vtot = 0. m2/s2 = 1.51 m/s 02 2m 2/s2+2.2 5m 2/s2 = 2. 27 ! ! vw 0.15 m/s q = tan−1 b = tan−1 = 5.7° north of east vwe 1.50 m/s I Ch. 3–8 Holt Physics Solution Manual Copyright © by Holt, Rinehart and Winston. All rights reserved. Section Review, p. 109 Chapter Review and Assess, pp. 113–119 Givens 6. A = 3.00 units (u) B = −4.00 units (u) Solutions Students should use graphical techniques. A2+ B 2 = (3 )2 +(− )2 a. A + B = .0 0u 4. 00 u A + B = 9. 00 u2+16. 0u2 = 25 .0 u2 = 5.00 units ! I ! B − 4.00 u q = tan−1 = tan−1 = 53.1° below the positive x-axis A 3.00 u A2+ )2 = (3 )2 +(4. )2 b. A − B = (−B .0 0u 00 u A − B = 9. 00 u2+16. 0u2 = 25 .0 u2 = 5.00 units ! ! −B 4.00 u q = tan−1 = tan−1 = 53.1° above the positive x-axis A 3.00 u A2+ )2 = (3 )2 +(− )2 c. A + 2B = (2B .0 0u 8. 00 u A + 2B = 9. 00 u2+64. 0u2 = 73 .0 u2 = 8.54 units ! ! 2B −8.00 u q = tan−1 = tan−1 = 69.4° below the positive x-axis A 3.00 u B 2+ )2 = (− )2 +(− )2 = 5.00 units d. B − A = (−A 40 0u 3. 00 u ! ! B − 4.00 u q = tan−1 = tan−1 = 53.1° below the negative x-axis −A −3.00 u or 127° clockwise from the positive x-axis Copyright © by Holt, Rinehart and Winston. All rights reserved. 7. A = 3.00 m Students should use graphical techniques. B = 3.00 m Ax = A(cos q) = (3.00 m)(cos 30.0°) = 2.60 m q = 30.0° Ay = A(sin q) = (3.00 m)(sin 30.0°) = 1.50 m 2 + (A + B)2 = (2.60 m)2 + (4.50 m)2 a. A + B = A y x A + B = 6. m2+ m2 = 5.20 m 76 20. 2m 2 = 27 .0 ! ! Ay + B 4.50 m q = tan−1 = tan−1 = 60.0° above the positive x-axis Ax 2.60 m 2 + (A − B)2 = (2.60 m)2 + (−1.50 m)2 b. A − B = A y x A − B = 6. m2+ m2 = 3.00 m 76 2.2 5m 2 = 9. 01 ! ! Ay − B −1.50 m q = tan−1 = tan−1 = 30.0° below the positive x-axis Ax 2.60 m Section One—Pupil’s Edition Solutions I Ch. 3–9 Chapter Review and Assess, pp. 113–119 continued Givens Solutions 2 (−A )2 = (1.50 m)2 + (−2.60 m)2 c. B − A = (B A − y)+ x B − A = 3.00 m I ! ! B − Ay 1.50 m q = tan−1 = tan−1 = 30.0° above the negative x-axis −Ax −2.60 m or 150° counterclockwise from the positive x-axis 2 + (A − 2B)2 = (2.60)2 + (−4.50)2 = 5.20 m d. A − 2B = A y x ! ! Ay − 2B − 4.50 m q = tan−1 = tan−1 = 60.0° below the positive x-axis Ax 2.60 m 8. ∆y1 = −3.50 m Students should use graphical techniques. d2 = 8.20 m ∆x2 = d2 (cos q2 ) = (8.20 m)(cos 30.0°) = 7.10 m q2 = 30.0° ∆y2 = d2 (sin q2 ) = (8.20 m)(sin 30.0°) = 4.10 m ∆x3 = 15.0 m ∆xtot = ∆x2 + ∆x3 = 7.10 m − 15.0 m = −7.9 m ∆ytot = ∆y1 + ∆y2 = −3.50 m + 4.10 m = 0.60 m d = (∆ xt )2 +(∆ yt )2 = (− )2 +(0. m )2 7. 9m 60 ot ot d = 62 m2+ 0.3 6m 2 = 62 m 2= 7.9 m ! ! ∆ytot 0.60 m q = tan−1 = tan−1 = 4.3° north of west ∆xtot −7.9 m ∆y = 13.0 m Students should use graphical techniques. d= ∆ x2 + ∆ y 2 = (− m )2 +(13 m )2 8. 00 .0 d = 64 m2+ m2 = 23 .0 169 3m 2 = 15.3 m ! ! ∆y 13.0 m q = tan−1 = tan−1 = 58.4° south of east ∆x −8.00 m 22. ∆x1 = 3 blocks west = −3 blocks east ∆y = 4 blocks north a. ∆xtot = ∆x1 + ∆x2 = −3 blocks + 6 blocks = 3 blocks ∆ytot = ∆y = 4 blocks d = (∆ xtot )2 +(∆ ytot )2 = (3 blo ck s) 2+(4blo ck s) 2 ∆x2 = 6 blocks east d = 9b s2 +16b s2 = 25 s2 = 5 blocks lo ck lo ck blo ck ! ! ∆ytot 4 blocks q = tan−1 = tan−1 = 53° north of east ∆xtot 3 blocks b. distance traveled = 3 blocks + 4 blocks + 6 blocks = 13 blocks I Ch. 3–10 Holt Physics Solution Manual Copyright © by Holt, Rinehart and Winston. All rights reserved. 9. ∆x = −8.00 m Givens Solutions 23. ∆x = 6.00 m d= ∆ ∆y2 = (6 )2 + (−5. m )2 x2+ .0 0m 40 ∆y = −5.40 m d = 36 m2+ m2 = 8.07 m .0 29. 2m 2 = 65 .2 ! ! ∆y −5.40 m q = tan−1 = tan−1 = 42.0° south of east ∆x 6.00 m 24. ∆y1 = −10.0 yards I ∆ytot = ∆y1 + ∆y2 = −10.0 yards + 50.0 yards = 40.0 yards ∆x = 15.0 yards ∆xtot = ∆x = 15.0 yards ∆y2 = 50.0 yards d = (∆ xt )2 +(∆ yt )2 = (1 5. 0ya rd s) 2+(40 .0 yar ds) 2 ot ot d = 22 s2 +1.6 03yar s2 = 18 s2 = 42.7 yards 5ya rd 0×1 d 20 yar d 25. ∆y1 = −40.0 m Case 1: ∆ytot = ∆y1 + ∆y2 = −40.0 m − 20.0 m = −60.0 m ∆x = ±15.0 m ∆xtot = ∆x = +15.0 m ∆y2 = ±20.0 m d = (∆ m)2 +(15 m)2 ytot)2+(∆ xtot)2 = (− 60 .0 .0 d = 3. 03 m2+ m2 = 38 m2 = 61.8 m 60 ×1 225 20 ! ! ∆ytot −60.0 m q = tan−1 = tan−1 = 76.0° south of east ∆xtot 15.0 m Case 2: ∆ytot = ∆y1 + ∆y2 ! −40.0 m+ 20.0 m+ −20.0 m ∆xtot = ∆x = +15.0 m d = (∆ yt )2 +)∆ xt )2 = (− m )2 +(15 m )2 20 .0 .0 ot ot d = 4. m2 = (6 m)2 = 25.0 m 00 ×102+225 25 ! ! ∆ytot −20.0 m ∆ = tan−1 = tan−1 = 53.1° south of east ∆tot 15.0 m Case 3: ∆ytot = ∆y1 + ∆y2 = −40.0 m − 20.0 m = −60.0 m Copyright © by Holt, Rinehart and Winston. All rights reserved. ∆xtot = ∆x = −15.0 m d = (∆ yt )2 +(∆ xt )2 = (− m )2 +(− m )2 60 .0 15 .0 ot ot d = 61.8 m ! ! ∆ytot −60.0 m q = tan−1 = tan−1 = 76.0° south of west ∆xtot −15.0 m Case 4: ∆ytot = ∆y1 + ∆y2 = −40.0 m + 20.0 m = −20.0 m ∆xtot = ∆x = −15.0 m d = (∆ yt )2 +(∆ xt )2 = (− m )2 +(− m )2 20 .0 15 .0 ot ot d = 25.0 m ! ! ∆ytot −20.0 m q = tan−1 = tan−1 = 53.1° south of west ∆xtot −15.0 m 26. d = 110.0 m ∆ = −10.0° ∆x = d(cos q) = (110.0 m)[cos(−10.0°)] = 108 m ∆x = d(sin q) = (110.0 m)[sin(−10.0°)] = −19.1 m Section One—Pupil’s Edition Solutions I Ch. 3–11 Givens Solutions 27. q = 25.0° ∆x = d(cos q) = (3.10 km)(cos 25.0°) = 2.81 km east d = 3.10 km I 28. d = 41.1 m ∆y = d(sin q) = (3.10 km)(sin 25.0°) = 1.31 km north ∆x = d(cos q) = (41.1 m)(cos 40.0°) = 31.5 m q = 40.0° ∆y = d(sin q) = (41.1 m)(sin 40.0°) = 26.4 m 29. d1 = 100.0 m ∆x1 = d1(cos q1) = (100.0 m)(cos 0.00°) = 100.0 m q1 = 0.00° east = 0.00° ∆y1 = d1(sin q1) = (100.0 m)(sin 0.00°) = 0.000 m d2 = 300.0 m ∆x2 = d2(cos q2) = (300.0 m)[cos (−90.0°)] = 10.00 m q2 = 90.0° south = −90.0° ∆y2 = d2(sin q2) = (300.0 m)[sin (−90.0°)] = −300.0 m d3 = 150.0 m ∆x3 = d3(cos q3) = (150.0 m)[cos (−150°)] = −129.9 m q3 = 30.0° south of west ∆y3 = d3(sin q3) = (150.0 m)[sin (−150°)] = −75.00 m = 180.0° − 30.0° south of east = −150° d4 = 200.0 m q4 = 60.0 ° north of west = 180° − 60.0° north of east = 120° ∆x4 = d4(cos q4) = (200.0 m)(cos (120°)] = −100.0 m ∆y4 = d4(sin q4) = (200.0 m)[sin (120°)] = 173.2 m ∆xtot = ∆x1 + ∆x2 + ∆x3 + ∆x4 = (100.0 m) + (0.00 m) + (−129.9 m) + (−100.0 m) = −129.9 m ∆ytot = ∆y1 + ∆y2 + ∆y3 + ∆y4 = (0.000 m) + (−300.0 m) + (−75.00 m) + (173.2 m) = −201.8 m d = (∆ xtot)2+(∆ ytot)2 = (− 12 9. 9m )2+(−20 1. 8m )2 d = 16 m2+ m2 = 240.0 m 870 4072 0m 2 = 57 590 ! ! 34. ∆y = −0.809 m ∆x = 18.3 m 2 g = 9.81 m/s ∆x ∆t = vx ! −g m/s v = " #2∆#y ∆x = " #(2−#)#(9−.8#01#.8 #0#9 #m#) (18.3 m) = 1 1 ∆x ∆y = − 2g∆t 2 = − 2 g vx 2 2 x 35. vi = 1.70 × 103 m/s q = 55.0° g = 9.81 m/s2 1 45.1 m/s, or 162 km/h 1 a. ∆y = vi(sin q)∆t − 2g∆t2 = vi(sin q ) − 2g∆t = 0 2vi(sin q) (2)(1.70 × 103 m/s)(sin 55.0°) = ∆t = = 284 s 9.81 m/s2 g ∆x = vi(cos q)∆t = (1.70 × 103 m/s)(cos 55.0°)(284 s) = 2.77 × 105 m b. ∆t = 284 s (See a.) I Ch. 3–12 Holt Physics Solution Manual Copyright © by Holt, Rinehart and Winston. All rights reserved. ∆ytot −201.8 m = tan−1 = 57.23° south of west q = tan−1 ∆xtot −129.9 m Givens Solutions 36. vx = 18 m/s ∆y = − 2 g∆t2 1 ∆y = −52 m g = 9.81 m/s2 ∆t = "#−#g = " #−#9.#81# #m#/#s = 2∆y (2)(−52 m) 2 3.3 s When the stone hits the water, I vy = −g∆t = (−9.81 m/s)(3.3 s) = −32 m/s 2 vtot = vx2+ )2+(− m/s )2 v 8m /s 32 y = (1 vtot = 32 m2/s2 = 36 m/s 0m 2/s2+100 0m 2/s2 = 13 00 37. vx,s = 15 m/s vx,o = 26 m/s ∆y = −5.0 m g = 9.81 m/s2 1 ∆y = − 2g∆t2 2∆y ∆t = = −g "## " #−#9.#81# ##/s# = 1.0 s m 2(−5.0 m) 2 ∆xs = vx,s ∆t = (15 m/s)(1.0 s) = 15 m ∆xo = vx,o ∆t = (26 m/s)(1.0 s) = 26 m ∆xo − ∆xs = 26 m − 15 m = 11 m 38. ∆x = 36.0 m vi = 20.0 m/s q = 53° a. ∆x = vi(cos q)∆t ∆x 36.0 m ∆t = = = 3.0 s vi(cos q) (20.0 m/s)(cos 53°) 1 1 ∆ybar = 3.05 m ∆y = vi(sin q)∆t − 2g∆t2 = (20.0 m/s)(sin 53°)(3.0 s) − 2(9.81 m/s2)(3.0 s)2 g = 9.81 m/s2 ∆y = 48 m − 44 m = 4 m ∆y = ∆ybar = 4 m − 3.05 m = 1 m The ball clears the goal by 1 m. b. vy,f = vi(sin q) − g∆t = (20.0 m/s)(sin 53°) − (9.81 m/s2)(3.0 s) Copyright © by Holt, Rinehart and Winston. All rights reserved. vx,f = 16 m/s − 29 m/s = −13 m/s The velocity of the ball as it passes over the crossbar is negative; therefore, the ball is falling. 39. vi = 25.0 m/s q = 45.0° ∆x = 50.0 m 2 g = 9.81 m/s ∆x = vi (cos q)∆t ∆x ∆t = vi (cos q) $ % $ % ∆x ∆x 1 1 ∆y = vi (sin q)∆t − 2g∆t 2 = vi (sin q) − 2g vi(cos q) vi(cos q) 2 g∆x2 (9.81 m/s2)(50.0 m)2 ∆y = ∆x(tan q) − = (50.0 m)(tan 45.0°) − 2vi2(cos q)2 (2)(25.0 m/s)(cos 45.0)2 ∆y = 50.0 m − 39.2 m = 10.8 m Section One—Pupil’s Edition Solutions I Ch. 3–13 Givens Solutions 40. ∆y = −1.00 m Find the initial velocity of the water when shot at rest horizontally 1 m above the ground. ∆x = 5.00 m 1 v = 2.00 m/s ∆y = − 2g∆t2 2∆ y ∆t = −g ∆t = 0.329 s ∆x = vx∆t q = 45.0° I "## g = 9.81 m/s2 ∆x 5.00 m ∆x vx = = = = 11.1 m/s y 2 ∆ (2 )(−1.00 m) ∆t −g −9.81 m/s2 "## "######## Find how far the water will go if it is shot horizontally 1 m above the ground while the child is sliding down the slide. vx, tot = vx + v(cos q) ∆x = vx, tot∆t = [vx + v(cos q)]∆t = 11.1 m/s + (2.00 m/s)(cos 45.0°)](0.329 s) ∆x = [11.1 m/s + 1.41 m/s](0.329 s) = (12.5 m/s)(0.329 s) = 4.11 m 41. ∆x1 = 2.50 × 103 m ∆x2 = 6.10 × 102 m ∆ymountain = 1.80 × 103 m 2 vi = 2.50 × 10 m/s q = 75.0° g = 9.81 m/s2 For projectile’s full flight, ∆x ∆t = vi(cos q) 1 1 ∆y = vi (sin q)∆t − 2g∆t 2 = vi(sin q) − 2g∆t = 0 $ % ∆x 1 vi (sin q) − 2g = 0 vi(cos q) 2vi2(sin q)(cos q) (2)(2.50 × 102 m/s)2(sin 75.0°)(cos 75.0°) = ∆x = = 3190 m 9.81 m/s2 g Distance between projectile and ship = ∆x − ∆x1 − ∆x2 = 3190 m − 2.50 × 103 m − 6.10 × 102 m = 80 m $ % $ % ∆x1 ∆x1 1 1 ∆y = vi(sin q)∆t′ − 2g∆t′2 = vi(sin q) − g vi(cos q) 2 vi(cos q) 2 g∆x12 ∆y = ∆x1(tan q) − 2 2 2vi (cos q) ∆y = (2.50 × 103 m)(tan 75.0°) (9.81 m/s2)(2.50 × 103 m)2 − (2)(2.50 × 102 m/s)2 (cos 75.0°)2 ∆y = 9330 m − 7320 = 2010 m distance above peak = ∆y − ∆ymountain = 2010 m − 1.80 × 103 m = 47. vap = 165 km/h south = −165 km/h north vpe = 145 km/ north I Ch. 3–14 vae = vap + vpe vae = −165 km/h + 145 km/h = −20 km/h north = 20 km/h south Holt Physics Solution Manual 210 m Copyright © by Holt, Rinehart and Winston. All rights reserved. For projectile’s flight to the mountain, ∆xi ∆t′ = vi (cos q) Givens Solutions 48. vre = 1.50 m/s east a. vbe = vbr + vre vbr = 10.0 m/s north 2 2 vbe ! vb vr )2 =(1. m/s )2 0. 0m /s 50 r e = (1 vbe! 1. 02 m2/ s2 +2.2 s2 = 10 s2 = 10.1 m/s 00 ×1 5m 2/ 2m 2/ ! ! v 1.50 m/s = tan−1 = 8.53° east of north q = tan−1 re vbr 10.0 m/s I ∆x 325 m b. ∆t = = = 32.5 s vbr 10.0 m/s ∆x = 325 m 49. vwe = 50.0 km/h south vaw = 205 km/h vae is directed due west ∆y = vre ∆t = (1.50 m/s)(32.5 s) = 48.8 m a. vaw = vae + (−vwe) ! ! v we = sin q vaw v 50.0 km/h q = sin−1 we = sin−1 = 14.1° north of west vaw 205 km/h b. vaw2 = vae2 + vwe2 2 vae = va )2 −(50 )2 vwe2 = (2 05 km /h .0 km /h w− vae = 4. 04km 03km 20 ×1 2/h2−2.5 0×1 2/h2 vae = 3. 04km 95 ×1 2/h2 = 1.99 km/h 50. ∆x = 1.5 km vre = 5.0 km/h vbr = 12 km/h 51. vre = 3.75 m/s downstream Copyright © by Holt, Rinehart and Winston. All rights reserved. vsr = 9.50 m/s vse is directed across the river The boat’s velocity in the x direction is greatest when the boat moves directly across the river with respect to the river. 1.5 km ∆x ∆tmin = = = 7.5 min vbr (12 km/h)(1 h/60 min) vre = sin q vsr a. vsr = vse + (−vre) ! 3.75 m/s q = sin−1 = 23.2° upstream from straight across 9.50 m/s b. vsr2 = vse2 + vre2 2 vse = vs −vre )2 −(3. m/s )2 2 = (9 .5 0m /s 75 r vse = 90 m2/ s2 −14. s2 = 76 m2/ s2 = 8.72 m/s .2 1m 2/ .1 vse = 8.72 m/s directly across the river 52. vbr = 12.0 m/s east vre = 3.5 m/s south a. vbe = vbr + vre 2 vbe = vb +vre 2 = (1 2. 0m /s )2+(3. 5m /s )2 r vbe = 14 s2 +12m s2 = 15 s2 = 12.5 m/s 4m 2/ 2/ 6m 2/ ! ! v 3.5 = tan−1 = 16° south of east q = tan−1 re vbr 12.0 ∆x = 1360 m ∆x 1360 m b. ∆t = = = 113 s vbr 12.0 m/s Section One—Pupil’s Edition Solutions I Ch. 3–15 Givens Solutions 53. ∆x = 130.0 m a. ∆x = vi(cos q)∆t q = 35.0° ∆y = 21.0 m − 1.0 m = 20.0 m I g = 9.81 m/s2 ∆x ∆t = vi(cos q ) $ % $ ∆x ∆x 1 1 ∆y = vi(sin q)∆t − 2 g∆t2 = vi(sin q) − 2 g vi(cos q) vi(cos q) % 2 2 g∆x ∆y = ∆x(tan q) − 2vi2(cos q)2 2vi2(cos q)2[∆x(tan q) − ∆y] = g∆x2 vi = "## g∆x2 2 2(cos q) [∆x(tan q) − ∆y] "### (9.81 m/s )(130.0 m) = 41.7 m/s v = "## (2)(cos 35.0°)(91.0 m − 20.0 m) (9.81 m/s2)(130.0 m)2 (2)(cos 35.0)2[(130.0 m)(tan 35.0°) − 20.0 m] vi = 2 2 i 130.0 m ∆x b. ∆t = = = 3.81 s vi(cos q) (41.7 m/s)(cos 35.0°) c. vy,f = vi(sin q) − g∆t = (41.7 m/s)(sin 35.0°) − (9.81 m/s2)(3.81 s) vy,f = 23.9 m/s − 37.4 m/s = −13.5 m/s vx,f = vx = vi(cos q) = (41.7 m/s)(cos 35.0°) = 34.2 m/s 2 2 vf = (v vy )2 +(− m/s )2 x, ( 4. 2m /s 13 .5 f )+ ,f) = (3 vf = 11 m/s m2/ s2 = 13 m2/ s2 = 36.7 m/s 70 2+182 50 54. ∆x = 12 m q = 15° 2 g = 9.81 m/s ∆x ∆t = vi(cos q) 1 1 ∆y = vi(sin q)∆t − 2 g∆t2 = vi(sin q) − 2 g∆t = 0 $ % ∆x 1 vi(sin q) − 2 g = 0 vi(cos q) 2vi2(sin q)(cos q) = g∆x vi = "# "## g∆x = 2(sin q)(cos q) (9.81 m/s2)(12 m) = 15 m/s (2)(sin 15°)(cos 15°) 12 m ∆x b. ∆t = = = 0.83 s (15 m/s)(cos 15°) vi(cos q) vy,f = vi(sin q) − g∆t = (15 m/s)(sin 15°) − (9.81 m/s2)(0.83 s) vy,f = 3.9 m/s − 8.1 m/s = −4.2 m/s I Ch. 3–16 Holt Physics Solution Manual Copyright © by Holt, Rinehart and Winston. All rights reserved. a. ∆x = vi(cos q )∆t Givens Solutions vx,f = vx = vi(cos q ) = (15 m/s)(cos 15°) = 14 m/s 2 vf = (v x,f)2+(vy 4m /s )2+(−4. 2m /s )2 ,f) = (1 vf = 2. 02 m2/ s2 +18m s2 = 22 s2 = 15 m/s 0×1 2/ 0m 2/ See solution to Chapter 3 Review and Assess problem 53 for a derivation of the following equation. 55. ∆x = 10.0 m q = 45.0° ∆y = 3.05 m − 2.00 m = 1.05 m vi = "## "### vi = = = 10.5 m/s "## (2)(cos 45.0°) (10.0 m − 1.05 m) "## (2)(cos 45.0°) (8.95 m) g = 9.81 m/s2 g∆x2 = 2 2(cos q) [∆x(tan q) − ∆y] I (9.81 m/s2)(10.0 m)2 (2)(cos 45.0°)2[(10.0 m)(tan 45.0°) − 1.05 m] (9.81 m/s2)(10.0 m)2 (9.81 m/s2)(10.0 m)2 2 2 1 ∆y = vi(sin q)∆t − 2 g∆t2 = vi(sin q) − g∆t = 0 56. ∆t = 3.00 s q = 30.0° 2 g = 9.81 m/s 57. ∆x = 20.0 m ∆t = 50.0 s vpe = ±0.500 m/s g∆t (9.81 m/s2)(3.00 s) vi = = = 29.4 m/s 2(sin q ) (2)(sin 30.0°) ∆x 20.0 m veg = = = 0.400 m/s ∆t 50.0 s vpg = vpe + veg a. Going up: vpg = vpe + veg = 0.500 m/s + 0.400 m/s = 0.900 m/s 20.0 m ∆x ∆tup = = = 22.2 s vpg 0.900 m/s b. Going down: vpg = −vpe + veg = −0.500 m/s + 0.400 m/s = −0.100 m/s Copyright © by Holt, Rinehart and Winston. All rights reserved. −20.0 m −∆x ∆tdown = = = 2.00 × 102 s vpg −0.100 m/s 58. ∆y = −1.00 m ∆x = 1.20 m g = 9.81 m/s2 ∆x ∆t = vx a. ∆x = vx ∆t 1 1 ∆x ∆y = − 2 g∆t2 = − 2 g vx vx = ! 2 g∆x2 = 2vx2 "# "## −g∆x2 = 2y −(9.81 m/s2)(1.20 m)2 = 2.66 m/s (2)(−1.00 m) b. The ball’s velocity vector makes a 45° angle with the horizontal when vx = vy . vx vx = vy,f = −g∆t ∆t = g vx 2 vx 2 1 1 2 ∆y = − 2 g∆t = − 2 g = − 2g g ! 2 (2.66 m/s) ∆y = − = − 0.361 m (2)(9.81 m/s2) h = 1.00 m − 0.361 m = 0.64 m Section One—Pupil’s Edition Solutions I Ch. 3–17 Givens Solutions 59. v1 = 40.0 km/h For lead car: v2 = 60.0 km/h ∆xi = 125 m ∆xtot = v1∆t + ∆xi For chasing car: ∆xtot = v2∆t I v2∆t = v1∆t + ∆xi (125 m)(10−3 km/m) ∆xi ∆t = = v2 − v1 (60.0 km/h − 40.0 km/h)(1 h/3600 s) 125 × 10−3 km ∆t = = 22.5 s (20.0 km/h)(1 h/3600 s) 60. q = 60.0° d1 = v1∆t = (41.0 km/h)(3.00 h) = 123 km v1 = 41.0 km/h ∆x1 = d1(cos q) = (123 km)(cos 60.0°) = 61.5 km v2 = 25.0 km/h ∆y1 = d1(sin q) = (123 km)(sin 60.0°) = 107 km ∆t1 = 3.00 h ∆t2 = ∆t − ∆t1 = 4.50 h − 3.00 h = 1.50 h ∆t = 4.50 h ∆y2 = v2∆t2 = (25.0 km/h)(1.50 h) = 37.5 km ∆xtot = ∆x1 = 61.5 km ∆ytot = ∆y1 + ∆y2 = 107 km + 37.5 km = 144 km d = (∆ xtot )2 +(∆ ytot )2 = (6 )2 +(14 )2 1. 5km 4km d = 37 80 km 2+2070 0km 2 = 24 500 km 2 = 157 km 61. q = −24.0° a = 4.00 m/s2 d = 50.0 m 1 a. d = 2 a∆t2 ∆t1 = = 5.00 s "#a = "# 4.00 m/s 2d (2)(50.0 m) 2 ∆y = −30.0 m vi = a∆t1 = (4.00 m/s2)(5.00 s) = 20.0 m/s g = 9.81 m/s2 2 vy,f = vi ( si n q )2 −2 g∆y = (2 ]2 −(2) m) 0. 0m /s 2)[s in (− 24 .0 °) (9 .8 1m /s 2)(− 30 .0 vy,f = 66 m2/ s2 =589 m2/ s2 = 65 s2 = ±25.6 m/s = −25.6 m/s .2 5m 2/ vy,f = vi(sin q) − g∆t2 vy,f − vi(sin q) −25.6 m/s − (20.0 m/s)(sin −24.0°) ∆t2 = = −9.81 m/s2 −g −25.6 m/s + 8.13 m/s −17.5 m/s = 2 = 1.78 s ∆t2 = −9.81 m/s2 −9.81 m/s ∆x = vi(cos q)∆t2 = (20.0 m/s)[cos(−24.0°)](1.78 s) = 32.5 m b. ∆t2 = 1.78 s (See a.) 62. vbw = ±7.5 m/s vwe = 1.5 m/s ∆x d = 250 m vbe = vbw + vwe Going downstream: vbe,d = 7.5 m/s + 1.5 m/s = 9.0 m/s ∆x u = −250 m Going upstream: vbe,u = −7.5 m/s + 1.5 m/s = −6.0 m/s ∆xd ∆xu 250 m −250 m ∆t = + = + = 28 s + 42 s = 7.0 × 101 s vbe,d vbe,u 9.0 m/s − 6.0 m/s I Ch. 3–18 Holt Physics Solution Manual Copyright © by Holt, Rinehart and Winston. All rights reserved. Givens Solutions 63. q = 34° a. ∆x = vi (cos q )∆t ∆x ∆t = vi (cos q ) ∆x = 240 m g = 9.81 m/s2 1 1 ∆y = vi (sin q)∆t − 2 g∆t 2 = vi (sin q) − 2 g∆t = 0 $ % ∆x g ∆x 1 vi (sin q) − 2 g = vi2(sin q) − = 0 vi(cos q) 2(cos q) vi = I "######### "## g ∆x = 2(cos q )(sin q ) (9.81 m/s2)(240 m) = 5.0 × 101 m/s (2)(cos 34°)(sin 34°) vy,f 2 − vy,i2 b. ∆ymax = −2g Because vy,f = 0 m/s, −vi2(sin q)2 (5.0 × 101 m/s)2(sin 34°)2 ∆ymax = = −2g (2)(9.81 m/s2) ∆ymax = 4.0 × 101 m $ % $ ∆x ∆x 1 vi(sin q) − g 2vi(cos q) 2 2vi(cos q) % 2 2 ∆x g∆x ∆ymax = (tan q) − 2 8 vi2(cos q)2 (240 m)(tan 34°) (9.81 m/s2)(240 m)2 ∆ymax = − 2 (8)(5.0 × 101 m/s)2(cos 34°) ∆ymax = 81 m − 41 m = 4.0 × 101 m 64. vwe = −0.500 m/s ∆x = 0.560 m Copyright © by Holt, Rinehart and Winston. All rights reserved. ∆tupstream = 0.800 s a. vse = vsw + vwe Going upstream: ∆x 0.560 m vse = = = 0.700 m/s ∆tupstream 0.800 s 0.700 m/s = vsw + (−0.500 m/s) vsw = 0.700 m/s + 0.500 m/s = 1.20 m/s Going downstream: vse = −0.500 m/s (same as the water) −0.500 m/s = vsw + (−0.500 m/s) vsw = −0.500 m/s + 0.500 m/s = 0.00 m/s b. d = vsw ∆t = (1.200 m/s)(0.800 s) = 0.960 m −∆x − 0.560 m c. ∆tdownstream = = = 1.12 s vse − 0.500 m/s ∆ttotal = ∆tupstream + ∆tdownstream = 0.800 s + 1.12 s = 1.92 s d 0.960 m vsw, avg = = = 0.500 m/s ∆ttotal 1.92 s Section One—Pupil’s Edition Solutions I Ch. 3–19 Givens Solutions 65. vce = 50.0 km/h east a. vce = vrc (sin q ) v 50.0 km/h = = 57.7 km/h vrc = ce (sin q) (sin 60.0°) q = 60.0° vrc = 57.7 km/h at 60.0° west of the vertical I b. vre = vrc (cos q) = (57.7 km/h)(cos 60.0°) = 28.8 km/h vre = 28.8 km/h straight down 66. ∆twalk = 30.0 s ∆tstand = 20.0 s L L vpe = = ∆twalk 30.0 s L L veg = = ∆tstand 20.0 s vpg = vpe + veg vpg = vpe + veg L L 2L + 3L 5L vpg = + = = 30.0 s 20.0 s 60.0 s 60.0 s L 5L = ∆t 60.0 s 60.0 s ∆t = = 12.0 s 5 67. ∆x Earth = 3.0 m 2 g = 9.81 m/s 1 1 ∆y = vi (sin q)∆t − 2 g∆t 2 = vi (sin q) − 2 g∆t = 0 2vi (sin q) ∆t = g $ % 2vi (sin q) ∆xEarth = vi (cos q)∆t = vi (cos q) g 2vi2(cos q)(sin q) Because vi and q are the same for all locations, k ∆xEarth = , where k = 2vi2(cos q )(sin q) g ! g k = g∆xEarth = ∆xmoon = (0.38g)∆xMars 6 ∆xmoon = 6∆xEarth = (6)(3.0 m) = 18 m ∆xEarth 3.0 m ∆xMars = = = 7.9 m 0.38 0.38 68. vx = 10.0 m/s q = 60.0° g = 9.81 m/s2 The observer on the ground sees the ball rise vertically, which indicates that the x-component of the ball’s velocity is equal and opposite the velocity of the train. vx = vi(cos q ) 10.0 m/s vx vi = = = 20.0 m/s (cos q) (cos 60.0°) At maximum height, vy = 0, so vy,f 2 − vy,i2 vi2(sin q)2 ∆ymax = = −2g 2g (20.0 m/s)2(sin 60.0°)2 ∆ymax = = 15.3 m (2)(9.81 m/s2) I Ch. 3–20 Holt Physics Solution Manual Copyright © by Holt, Rinehart and Winston. All rights reserved. ∆xEarth = g Givens Solutions 69. vi = 18.0 m/s ∆y = vi (sin q)∆t − 2 g∆t 2 = vi (sin q) − 2 g∆t = 0 q = 35.0° 1 1 ∆xi = 18.0 m 2vi (sin q) 2(18.0 m/s)(sin 35.0°) = ∆t = = 2.10 s g 9.81 m/s2 g = 9.81 m/s2 ∆x = vi (cos q)∆t = (18.0 m/s)(cos 35.0°)(2.10 s) = 31.0 m I ∆xrun = ∆x − ∆xi = 31.0 m − 18.0 m = 13.0 m ∆x un 13.0 m vrun = r = = 6.19 m/s downfield ∆t 2.10 s ay = a(sin q) = (25 m/s2)(sin 53°) = 2.0 × 101 m/s2 70. q = 53° 1 1 vi = 75 m/s ∆y = vi (sin q)∆t + 2 ay ∆t 2 = (75 m/s)(sin 53°)(25 s) + 2 (2.0 × 101 m/s2)(25 s)2 ∆t = 25 s ∆y = 1500 m + 6200 m = 7700 m 2 a = 25 m/s vf = vi + a∆t = 75 m/s + (25 m/s2)(25 s) = 75 m/s + 620 m/s = 7.0 × 102 m/s For the motion of the rocket after the boosters quit: vi = vf = 7.0 × 102 m/s q = 53° g = 9.81 m/s2 vy,f = vi (sin q ) − g∆t = 0 2 vi (sin q) (7.0 × 10 m/s)(sin 53°) = ∆t = = 57 s 9.81 m/s2 g 1 1 ∆y = vi (sin q)∆t − 2 g∆t 2 = (7.0 × 102 m/s)(sin 53°)(57 s) − 2 (9.81 m/s2)(57 s)2 ∆y = 32 000 m − 16 000 m = 16 000 m a. ∆ytotal = 7700 m + 16 000 m = 2.4 × 104 m 1 b. ∆y = − 2 g∆t 2 ∆t = = 7.0 × 10 s "#−g# = "## −9.81 m/s 2∆y (2)(−24 000 m) 2 1 Copyright © by Holt, Rinehart and Winston. All rights reserved. ∆ttotal = 25 s + 57 s + 7.0 × 101 s = 152 s c. ax = a(cos q) = (25 m/s2)(cos 53°) = 15 m/s2 1 1 ∆x = vi (cos q)∆t + 2 a∆t 2 = (75 m/s)(cos 53°)(25 s) + 2 (75 m/s2)(25 s)2 ∆x = 1.1 × 103 m + 2.3 × 104 m = 2.4 × 104 m vi = 7.0 × 102 m/s After the rockets quit: q = 53° ∆t = 57 s + 7.0 × 101 s = 127 s ∆x = vi (cos q)∆t = (7.0 × 101 m/s)(cos 53°)(127 s) = 5.4 × 104 m ∆xtot = 2.4 × 104 m + 5.4 × 104 m = 7.8 × 104 m Section One—Pupil’s Edition Solutions I Ch. 3–21