Conceptual Physics 11th Edition 7 Solutions Manual

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Conceptual physics 12th edition ebook hewitt conceptual physics 12th edition pdf conceptual physics 12th edition online conceptual physics hewitt 12th edition pdf free download 12th Edition. In Conceptual Physics, Eleventh Edition Paul Hewitt shows how a compelling text and the most advanced media can be integrated to empower professors as they bring. Instructor Solutions for Lab Manual (Download only) for Conceptual Physics, 11th Edition. For Conceptual Physics, 11th Edition. Instructor Solutions for Problem Solving Book.

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Edition: 11TH 10
Copyright: 2010
Publisher: Addison-Wesley Longman, Inc.
Published: 2010
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ISBN13: 9780321568090

Edition: 11TH 10

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Brief Description: Since defining this course 30 years ago, Paul Hewitt's best-selling text continues to be the benchmark book that two-thirds of professors use and by which all others are judged. InConceptual Physics, Eleventh EditionPaul Hewitt shows how a compelling text and the most advanced media can be integrated to empower professors as they bring physics to life for non-science majors, both in and out of class. For theEleventh Edition,Hewitt helps students connect physics to their everyday experiences and the world around them, and provides additional help on solving mathematical problems. Key Topics: About Science, Newton's First Law of Motion: Inertia, Linear Motion, Newton's Second Law of Motion: Force and Acceleration Newton's Third Law of Motion: Action and Reaction, Momentum, Energy, Rotational Motion, Gravity, Projectile and Satellite Motion Atomic Nature of Matter, Solids, Liquids, Gases and Plasmas, Temperature, Heat and Expansion, Heat Transfer, Change of Phase Thermodynamics, Vibrations and Waves, Sound, Musical Sounds, Electrostatics, Electric Current, Magnetism, Electromagnetic Induction, Properties of Light, Color, Reflection and Refraction, Light Waves, Light Emission, Light Quanta, The Atom and the Quantum, Atomic Nucleus and Radioactivity, Nuclear Fission and Fusion, Special Theory of Relativity, General Theory of Relativity Appendices Market:Intended for those interested in learning the basics of conceptual physics

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Publisher: Addison-Wesley Longman, Inc.
Published: 2010
International: No

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Conceptual physics 12th edition hewitt solutions manual.1.1000 m1 min60 minConceptual Physics 12th Edition Hewitt Solutions ManualFull clear download (no error formatting) at:Physics 12th Edition Hewitt Test BankFull clear download (no error formatting) at:(a) Distance hiked = b + c km.(b) Displacement is a vector representing Paul’s change inposition. Drawing a diagram of Paul’s trip we can see thathis displacement is b + (–c) km east = (b –c) km east.b kmdisplacement –c km(c) Distance = 5 km + 2 km = 7 km; Displacement = (5 km – 2 km) east = 3 km east.3-2.

(a) From vdt vx.t(b) vx.We want the answer in m/s so we’ll need to convert 30 km to meters and 8 mintto seconds:30.0 km 1000 m30, 000 m; 8.0 min 60 s480 s. Then vx 30,000 m63 m.1 km 1 mint 480 s sAlternatively, we can do the conversions within the equation:vx 30.0 km 1 km63 m.tIn mi/h:8.0 min 60 s s30.0 km 1 mi18.6 mi; 8.0 min 1 h0.133 h. Les sims 3 animaux et cie cracked. Then vx 18.6 mi140 mi.1.61 km 60 min t 0.133 h h1 miOr, vx 30.0 km 60 min 1 mi140 mi. Or, vx 30.0 km 1.61 km140 mi.t 8.0 min 1 h 1.61 km h t 8.0 min 1 h hThere is usually more than one way to approach a problem and arrive at the correct answer!3-3.

(a) From vdt.vL.t(b) vL 24.0 m40 m.t 0.60s s3-4. (a) From vdt vx.t(b) vx 0.30 m30 m.t 0.010 s s3-5. (a) vd 2r.t t(b) v2r 2(400m)63 m.t 40s s.sssmm3-6. From vdt th.v(b) th 508 m34 s.v 15 m(c) Yes. At the beginning of the ride the elevator has to speed up from rest, and at the end ofthe ride the elevator has to slow down. These slower portions of the ride produce anaverage speed lower than the peak speed.3-7. Begin by getting consistent units.

Convert 100.0 yards to meters using theconversion factor on the inside cover of your textbook: 0.3048 m = 1.00 ft.1yard1ftThen 100.0 yards3ft 0.3048 m91.4 m. From vdtd 91.4 m.(b) td 91.4 m15s.t v vv 6.0 m3-8. Fromvdtd L.t v c(b) tL 1.00 m3.33 10-9s 3.33 ns. (This is 31billionths of a second!)v 3.00 108 m 33-9. From vdt d vt.(b) First, we need a consistent set of units. Since speed is in m/s let’s convert minutes to seconds:5.0 min60s300 s. Then d vt 7.5 m300s 2300 m.1min s3-10.

(a) vv0 vf v.2 2(b) d? From vdt d vtvt.2(c) dvt 2.0 s(1.5s)1.5 m.2 23-11. From vdd vtv0 vft0 vtvt.t(b) dvt 12 s(8.0s)2 2 248 m.2 23-12. From vdd vtv0 vft0 vtvt.t 2 2 2.m(b) First get consistent units: 100.0 km/h should be expressed in m/s (since the time is inseconds).100.0 km 1 h 1000 m27.8 m. Then, dvt 27.8 s(8.0 s)110m.h 3600 s 1 km s 2 2.s m2 223-13. (a) a v v2 v1.t t(b) v 40 km15 km25 km. Since our time is in seconds we need to convert kmto m:h h h h sm25 km 1hr 1000 m6.94 m.

Then av 6.94 s0.35 m.h 3600 s 1 km s t 20s s2Alternatively, we can express the speeds in m/s first and then do the calculation:11.1 m 4.17 m15 km 1hr 1000 m4.17 mand 40 km 1hr 1000 m11.1 m. Then a s s0.35 m.hr 3600 s 1 km s hr 3600 s 1 km s 20s s23-14. (a) a v v2 v1.t t(b) To make the speed units consistent with the time unit we’ll need v in m/s:v2 v14.17 mv v2 v1 20.0 km5.0 km15.0 km 1hr 1000 m4.17 m.

Then a  0.417.h h h 3600 s 1 km s t 10.0 s s2An alternative is to convert the speeds to m/s first:v1 5.0 km 1hr 1000 m1.4 m; v2 20.0 km 1hr 1000 m5.56 m.h 3600 s 1 km s h 3600 s 1 km sThen av2 v15.56 s1.4 s0.42 mm m.t 10.0 s s2m m(c) d vtv1 v2t1.4 s5.56 s10.0 s 35 m. Or,2 s 2s2d v1t 1 at21.4 m (10.0 s) 10.42 m(10.0 s)235 m.3-15. (a) a v vf v0t tv 26 m0 v v.t t(b) a  s1.3 m.t 20s s2m m(c) d? From vdd vtv0 vft26 s0 s20 s 260 m.t2 s2 2 s2Or, d v0t 1 at226 m (20 s) 11.3 m(20 s)2260 m.s(d) d =? Lonnie travels at a constant speed of 26 m/s before applying the brakes, sod vt 26 m (1.5 s) 39 m.3-16. (a) a v vf v0t tv 72 m0 v v.t t(b) a  s6.0 m.t 12 s s2.ss24 m m(c) d? Crack codes for wild games.

From vdd vtv0 vft72 s0 s(12 s) 430 m.t 22 s 22 s2Or, d v0t 1at272 m(12 s) 16.0 m(12 s)2430 m.3-17. From vdtd L 2L.2t v(b) t2L 2(1.4 m)0.19 s.vf v0 vv 15.0 m3-18. (a) vv0 vf v.2 2350 m(b) v s175 m.Note that the length of the barrel isn’t needed—yet!2(c) From vdstd L 0.40 m0.0023 s 2.3 ms.t v v 175 m3-19. (a) From vdd vtv0 vft =v0 vt.t 2 2m m(b) dv0 vt =25 s11 s(7.8 s) 140 m.2 2 3-20. There’s a time t between frames of 1s, so v=d x24 1x. (That’s 24x per24second.)(b) v 24 1x 24 1(0.15 m) 3.6 m.t 1s ss s s3-21.

Since time is not a part of the problem we can use the formula vf2v022ad andv2solve for accelerationa. Then,withv0= 0 and d = x, a.2x(b) av 1.8 10 s1.6 1015 m2 7 m2.2x 2(0.10 m) s2vf v0.1.8 10 s0 s(c) t? From vf v0 at ta 7 m m1.6 1015 ms21.1 10-8s 11 ns.Or, from vdtd L 2L 2(0.10 m)1.1 10-8s.21.8 10 st v vf v0 (v 0) 7 m.md v0 vf 2d3-22. From vt  2t with v0 0 vft.2d(b) af =? From d v0t 1at2with v0 0 d 1at2a.22d 2(402 m) m2t22d 2(402 m) m(c) vft 181; a 40.6.4.45 s st2(4.45 s)2 s23-23. From vdd vtv0 vft =v Vt.t 2 2m m(b) dv Vt =110 s250 s2 2 (3.5 s) 630 m.3-24.

Let's choose upward to be the positive direction.vf v0 0 v vFrom vf v0 at with vf 0 and a g t (b) tv 32 s3.3 s.a g gg 9.8 ms22 m 2(c) d? From vdd vtv0 vftv 0 v v 32 s  52 m.t 2 2 g 2g 2 9.8 ms22 s 2s2We get the same result with d v0t 1at232 m(3.3 s) 19.8 m(3.3 s)252m.3-25. When the potato hits the ground y = 0. Fromd v0t 1at2y v0t 1gt20 t v01gt v01gt.22 2 2s2 s2s 1000 m21.61 km 1 h h(b) v01gt 19.8 m(12 s) 59 m.In mi/h, 59 m 1 km 1 mi 3600 s130 mi.3-26. Choose downward to be the positive direction. FromFrom d v0t 1at2with v0 0, a g and d h h 1 gt2t  2h.2 2 g(b) t2h 2(25m)2.26 s 2.3s.sg 9.8 ms2s2 s(c) vf vo at 0 gt 9.8 m(2.26 s) 22 m.s22 2 m mOr, from 2ad vf v0 with a g, d h, and v0 0 vf 2gh 2 9.8(25 m) 22.sm0 000003-27. Let’s call upward the positive direction.

Since the trajectory is symmetric, vf = –v0.gtThen from vf v0 at, with a g v0 v0 gt 2v0 gt v02.2gt 9.8 m(4.0s)m(b) v02 20.2 s(c) d? From vdd vtv0 vftv0t20 st 2 2(2.0 s) 20 m.2We use t = 2.0 s because we are only considering the time to the highest pointrather than the whole trip up and down.3-28. Let’s call upward the positive direction. Since no time is given, usevf2v022ad with a = –g, vf = 0 at the top, and d = (y – 2 m).v22(g)(y 2m) v 2g(y 2m).s2 s s(b) v0 2g(y 2 m) 2 9.8 m(20 m 2 m) 18.8 m 19 m.3-29. (a) Taking upward to be the positive direction, from2 2 22ad vf v0 with a g and d h vf v0 2gh. So on the way upvf v22gh.(b) From above, on the way down vf v22gh, same magnitude but opposite direction as(a).22gh v v0 v22gh(c) From avf v0ttvf v0 v0 0.a g gm m2 m 2 ms2m vf v0s a9.5 s 16 s9.8 m(d) vf v0 2gh  16 s2 9.8 (8.5 m) 9.5. T  2.6 s.s23-30.

Taking upward to be the positive direction, from2 2 22ad vf v0 with a g and d h vf v0 2gh. The displacement d isnegative because upward direction was taken to be positive, and the water balloon endsup below the initial position. The final velocity is negative because the water balloon isheading downward (in the negative direction) when it lands.22gh v v0 v22gh(b) t =? From avf v0ttvf v0 v0 0.a g g.(c) vf =? Still taking upward to be the positive direction, from2 2 2 2 22ad vf v0 with initial velocity = –v0,a g and d h vf v0 2gh vf v0 2gh.We take the negative square root because the balloon is going downward. Note that thefinal velocity is the same whether the balloon is thrown straight up or straight down withinitial speed v0.v0 v024 (h)22m22 m 2 m ms2(d) vf v0 2gh  5.0 s2 9.8 (11.8 m) 16 sfor the balloon whether it istossed upward or downward.

For the balloon tossed upward,m mtvf v0 16 s 5 s2.1 s.a 9.8 ms23-31. (a) Call downward the positive direction, origin at the top.From d v0t 1at2with a g, d ² y h h v0t 1gt2 1gt2v0t h 0.2From the general form of the quadratic formula x 2b b24ac2a2we identifygag, b v, and c h, which gives t2  v0 v022gh.2 0g gTo get a positive value for the time we take the positive root, and getv0 + v02+ 2ght.g(b) From2 2 2 2 22ad vf v0 with initial velocity v0,a g and d h vf v0 2gh vf v0 2gh.v0 v022ghOr you could start with vf v0 at v0 gv0g + 2gh.m m 29.8 m (3.5 m)(c) t v0 v022gh g 3.2 s 3.2 s s29.8 ms20.58 s.;2 m 2 m ms2vf v0 2gh  3.2 s2 9.8 (3.5 m) 8.9 s3-32. (a) From d v0t 1at2a 2(d v0t).t2(b) a2(d v0t) 2120 m 13 s5.0 s4.4 m.t2(5.0 s)2 s2s s2s(c) vf v0 at 13 m4.4 m(5s) 35 m.(d) 35 m 1 km 1 mi 3600 s78 mi. This is probably not a safe speed for driving ins 1000 m 1.61 km 1 h han environment that would have a traffic light!3-33.

(a) From x vtv0 vft v2xv.2ft0x x(b) avf v0 (2 t v0 )v0 2 t 2v02x v0t t t t2t. 22sv0 v024 (h)21 0 21v v a t2212x 2(95m) m m(c) vf  v0 t13 3.0.11.9s s sm m ma 2x v0295m 13 s0.84 mor avf v03.0 s13 s0.84 m.t2t (11.9s)211.9s s2t 11.9s s2 2 23-34. (a) From 2ad vf v0 with d L vf v0 2aL. This is Rita’s speed at the bottom ofthe hill. To get her time to cross the highway: From vdt(b) td 25m1.54s.tdvfd.v0 2aLv0 2aL 3.0 m 22 1.5 m(85m)s23-35. (a) Since v0 is upward, call upward the positive direction and put the origin at the ground.ThenFrom d v0t 1at2with a g, d ² y h h v0t 1gt2 1gt2v0t h 0.2From the general form of the quadratic formula x 2b b24ac2a2we identifygag, b v, and c h, which gives t2  v0 v022gh.2 0g g2 2 2 2 2(b) From 2ad vf v0 with a g and d h vf v0 2gh vf v0 2gh.m m 29.8 m (14.7m)(c) t v0 v022gh g 22 s 22 s s29.8 ms20.82s or 3.67s. SoAnthony has to have the ball leave his had either 0.82s or 3.67s before midnight. Thefirst time corresponds to the rock hitting the bell on the rock’s way up, and the secondtime is for the rock hitting the bell on the way down.2 m 2 m ms2vf v0 2gh  22 s2 9.8 (14.7m) 14 s.3-31.

The rocket starts at rest and after time t1 it has velocity v1 and has risen to a heighth1. Taking upward to be the positive direction, from vf v0 at with v0 0 v1 at1.(b) h =? From d v t 1at2with h1 d and v0 0 h1  1at2.(c) h2 =? For this stage of the problem the rocket has initial velocity v1, vf = 0, a = –g andthe distance risen d = h2.2 22 2 f 0 0 v2 2(at1)2 2 21.1 v2 1 1a 2From 2ad vf v0 d  h2 2a 2(g) 2g.2g 2g(d) tadditional =?

To get the additional rise time of the rocket: Fromavf v0tvf v0 0 v1 at1.t additionala g g(e) The maximum height of the rocket is the sum of the answers from (a) and (b) =2hmax h1 h21at2a2t12g1at21  g.2(f) tfalling =? Keeping upward as the positive direction, now v0 = 0, a = –g and d = –hmax.From d v0t 1at2hmax1(g)t222hmax22 2 at1 1gat2(g a) ttfalling  1 2g ga 1g2at1 t1a(g + a) 1g(g) ttotal t1 tadditional tfalling t1  a(g + a).g g(h) vruns out of fuel v1 at1 120 m(1.70 s) 204 m; h 1at1  120 (1.70 s) 173 m.s22a2t1 s2 1 m 2s 1 2 2 s22 120 m(1.70 s)2hadditional h22g2123 m.s2 2 9.8 mtadditional at1 g120 m(1.70 s)s29.8 ms220.8 s.hmax 173 m + 2123 m 2296 m 2300 m.tfalling 2hmax g2(2300 m)9.8 ms221.7 s.ttotal t1 tadditional tfalling 1.7 s 20.8 s 21.7 s 44.2 s.3-32. Vtotal distance x x 2x1.14 x.total time t 0.75t 1.75t t(b) v 1.14 x 1.14140 km80 km.t 2 hr hr3-33.

(a) vtotal distance. From vdd vt.total timedwalk djogSo v tvwalktwalk vjogtjog v(30 min) 2v(30 min) 3v(30 min)1.5 v.twalk tjog twalk tjog 30 min 30 min 2(30 min)(b) v 1.5v 1.51.0 m1.5 m.s s(c) dto cabin vttotal v(twalk tjog ) 1.5 m(30 min +30 min) 60 s5400 m = 5.4 km.s 1 min3-34. (a) vtotal distance.

From vdd vt.total time tSo vdslow dfast vslowtslow vfasttfast v(1 h)4v(1 h) 5v(1 h)2.5 v.tslow tfast tslow tfast 1h 1h 2 h(b) v 2.5v 2.525 km63 km.h h.1 2 1 2 1 2sm3-35. (a) vtotal distance. From vd td x.total time t v vSo vd1 d2 2x 2x 2v vxv vv vt1 t2 x x 1 1 v2 v12v1v2 v(1.5v)  1.5v22 2 1.2v.Note that the average velocity is biased towardv2 v1  1.5v v 2.5vthe lower speed since you spend more time driving at the lower speed than the higher speed.(b) v 1.2v 1.228 km34 km.h h3-36. From VAtti dAttid Vt.The time that Atti runs = the time that Judyt Attiwalks, which is tx. So d Vx Vx.v Attiv vm(b) XVx4.5 s(150m) 450 m.v 1.5 m3-37. Vd 3 m2 m.t 1.5 s s3-38. Call upward the positive direction.From vf2v022ad with d h, vf 0 and a g2v 2 2 2 m 2hvf 0 v0 v014.7s11 m.s22a 2(g) 2g 2 9.8 m3-39.

From vdd vtv0 vft0 27.5 s(8.0 s) 110 m.t 2 2 3-40. Let's take down as the positive direction. From d v0t 1at2with v0 0 and a g d 1gt22 2t2d 2(16 m)1.8s.g 9.8 ms2m m3-41. A v tvf v0t12 s 0 s3 s4 m.s2.m m3-42.

A v tvf v0t75 s 0 s2.5 s230 m.s22 2s23-43. With v0 0, d v0t 1at2becomes d 1at2 1 2.0 m(8.0 s)264 m.mms2d3-44. With v0 0, d v0t 1at2becomes d 1at2a 2(5.0 m)2.5 m.2 22 2 2t2s2(2.0 s)2 s23-45. With v0 0, d v0t 1at2becomes d 1at2 13.5 m(5.5 s)253 m.3-46. Here we’ll take upwards to be the positive direction, with a =g and vf = 0.s2 sFrom vf2v022ad v02vf22(g)d v0 2gd 2 9.8 m(3.0 m) 7.7 m.3-47. We can calculate the time for the ball to reach its maximum height (where the velocitywill be zero) and multiply by two to get its total time in the air. Here we’ll take upward tobe the positive direction, with a=g.From avf v0tvf v0 v0 v018 s1.84 s.This is the time to reacht a g g 9.8 ms2the maximum height.

The total trip will take 2 1.84 s = 3.7 s, which is less than 4 s.Alternatively, this can be done in one step with by recognizing that since the trajectory issymmetric vf = –v0.Then from vf v0 at, with a g v0 v0 gt 2v0 gtt2v0218 s 3.7 s.g 9.8 ms23-48. Since she throws and catches the ball at the same height, vf v0. Calling upwardthe positive direction, a = –g.2gt 9.8 m(3.0 s)mFrom vf v0 at v0 v0 (g)t 2v0 gt v02 15.2 s3-49.

For a ball dropped with v0 = 0 and a = +g (taking downward to be the positive direction),fallen, 1stsecond 2 2s2d v0t 1at2 1 9.8 m(1 s)24.9 m. At the beginning of the 2ndsecondwe have v0 = 9.8 m/s sofallen, 2ndsecond 2 s 2s2dd ndv0t 1at29.8 m(1 s) 1 9.8 m(1 s)214.7 m.

The.22ratiofallen, 2 second 14.7 m3. More generally, the distance fallen from rest in a timedfallen, 1stsecond4.9 mt is d 1gt2. In the next time interval t the distance fallen isdfrom time t to 2t v0t 1at2(gt)t 1gt2 3gt2. The ratios of these two distances is2 2 23 2dfrom time t to 2t 2 gt3.dfrom rest in time t1gt2.ssd d2v03-50. Call upward the positive direction. From vf2v022ad with d h, vf 0 and a g2v 2 2 2 m 2hvf 0 v0 v01,000s51,000 m 50 km.s22a 2(g) 2g 2 9.8 m3-51. With d h, v0 22 m, a –g and t 3.5 s, d v0t 1at2becomesss 2 2s2h (22 m)(3.5 s) 19.8 m(3.5 s)217 m3-52.

From vdtd 65 m5.0 s.t v 13 mm m3-53. From a v tvf v0tt vf v0a28 s0 s7.0 ms24.0 s.3-54. From vdtd d 2d.2t v vf v0 v(b) a =? With v0 0 and vf v, vf2– v022ad becomes a.2d2 m 2(c) t2d 2(140 m)10 s; av  28 s  2.8 m.v 28 m 2d 2(140 m) s23-55.

From vdd vtv0 vft0 s25 smt 2 2m(5.0 s) 63 m.2462 mi 1 km3-56. From v t 0.621 mi8.5 min.t v 28,000 km 1 hh 60 min3-57. With vf 0, vf2– v022ad becomesmi 1 km 1000 m 1 h 2v 2a 2d220 h 0.621 mi 1 km 3600 s 2(800 m)–6.05 ms2–6 m.s23-58.

From d v0t 1at2 1at2v0t d 0. From the general form of the quadratic formula2b b24ac x 2a.22we identify a 1a, b v0, and c h, which givesv0 v0241 a(d) 2t2 v0 v0 2ad. To get a positive answer for t we takea a45 m  45 m223.2 m (440m)sthe positive root, which gives us t s s23.2 ms27.7s.g23-59. The candy bar just clears the top of the balcony with height 4.2m + 1.1m = 5.3 m.With vf 0, vf2– v022ad with v0 and ² y positive and a g v02vf22(g)hs2 s sv0 2gh 2 9.8 m(5.3m) 10.19 m 10.2 m. The total time is the time for theway to the top of the balcony rail plus the time to fall 1.1 m to the floor of the balcony.2d 2(5.3 m)tup? From d vft 1 at2with vf 0 and a g d 1(g)t2tup1.04 s.2 2 g 9.8 ms22h 2(1.1 m)tdown? From d v0t 1 at2with v0 0, a g and d ² y h h 1(g)t2tdown 02 2g 9.8 ms2So ttotal tup tdown 1.040 s 0.47 s 1.51s.

An alternative route is: Since v0 isupward, call upward the positive direction and put the origin at the ground. ThenFrom d v0t 1at2with a g, d ² y 4.2m d v0t 1gt2 1gt2v0t d 0.2From the general form of the quadratic formula x 2b b24ac2a2we identifyv0 v024 (d) 22gdag, b v, and c d, which gives t2 v0 v0210.19 m 0g g10.19 m229.8 m (4.2m)s s s29.8 ms20.57s or 1.51s. The first answercorresponds to the candy reaching 4.2 m but not having gone over the top balcony railyet. The second answer is the one we want, where the candy has topped the rail andarrives 4.2 m above the ground.3-58. Consider the subway trip as having three parts—a speeding up part, a constant speed part,and a slowing down part. Dtotal dspeeding up dconstant speed dslowing down.s2 0 2 2 s2For dspeeding up, v0 0, a 1.5 mand t 12 s, so d v t 1at2 11.5 m(12 s)2108 m.For dconstant speed vt. From the speeding up part we had v0 0, a 1.5 ms and t 12 ss2 s sso v v0 at 1.5 m (12 s) 18 m and so d 18 m (38 s) 684 mFor dslowing down, vf 0, a –1.5 mand t 12 s, so d v t 1at2 1-1.5 m(12 s)2108 m.s2 f 2 2 s2.So dtotal dspeeding up dconstant speed dslowing down 108 m 684 m 108 m 900 m.3-59.

One way to approach this is to use Phil’s average speed to find how far he has run duringthe time it takes for Mala to finish the race.From vdd v t100.0 m(12.8 s) 94.1 m. Since Phil has onlyt Phil Phil Mala13.6 straveled 94.1 m when Mala crosses the finish line, he is behind by100 m 94.1 m 5.9 m 6 m.600h3-60. The time for Terrence to land from his maximum height is the same as the time ittakes for him to rise to his maximum height. Let’s consider the time for him to land froma height of 0.6 m. Taking down as the positive direction:From d v0t 1at2with v0 0 and a g d 1gt22 2t2d 2(0.6 m)0.35s.g 9.8 ms2His total time in the air would be twice this amount, 0.7 s.3-61. Vd 1 mi80 mi.t 45 s 31 hs3-62.

Vtotal distance. If we call the distance she drives d, then from vdtd.total time t vSo vdthere dback 2d 2d 22vtherevbacktthere tback dvd dv1v1v vback therevthere back there backvback vtherethere backv v40 km60 km  2400km 22 h h2 h48 km. Note that the average velocity is biased60 km 40 km  100 km hh h h toward the lower speed since Norma spends more time driving at the lower speed than atthe higher speed.Conceptual Physics 12th Edition Hewitt Solutions ManualFull clear download (no error formatting) at:Physics 12th Edition Hewitt Test BankFull clear download (no error formatting) at:physics 12th edition answersconceptual physics 12th edition pdfconceptual physics 12th edition paul g. Hewitt pdfconceptual physics 12th edition access code.conceptual physics 12th edition ebookhewitt conceptual physics 12th edition pdfconceptual physics 12th edition onlineconceptual physics hewitt 12th edition pdf free download12th Edition, Conceptual Physics, Hewitt, Solutions Manual.