Solucionario fisica serway

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© 2000 by Harcourt College Publishers. All rights reserved. Chapter 1 Even Answers 2. 623 kg/m 3 4. 4πρ (r 3 2 r 3 1 ) 3 6. 7.69 cm 8. 8.72 × 10 11 atoms/s 10. (a) 72.6 kg (b) 7.82 × 10 26 atoms 12. equation is dimensionally consistent 16. The units of G are: m 3 /kg s 2 18. 9.19 nm/s 20. (a) 3.39 × 10 5 ft 3 (b) 2.54 × 10 4 lb 22. 8.32 × 10 –4 m/s 24. 9.82 cm 26. (a) 6.31 × 10 4 AU (b) 1.33 × 10 11 AU 28. (a) 1.609 km/h (b) 88.5 km/h (c) 16.1 km/h 30. (a) 3.16 × 10 7 s/yr (b) 6.05 × 10 10 yr 32. 2.57 × 10 6 m 3 34. 1.32 × 10 21 kg 36. (a) 2.07 mm (b) 8.62 × 10 13 times as large 38. (a) 13.4 (b) 49.1 40. r Al = r Fe 3 (ρ Fe /ρ Al ) 42. ~ 10 6 km 44. ~ 10 9 drops 46. time required 50 years or more; advise against accepting the offer 48. ~ 10 5 tons 50. (a) 2 (b) 4 (c) 3 (d) 2 52. (a) 797 (b) 1.1 (c) 17.66 54. (a) 3 (b) 4 (c) 3 (d) 2 56. 5.2 m 3 , 2.7% 58. 1.79 × 10 –9 m 60. 24.6° 62. (b) A cylinder = πR 2 , A rectangular solid = lw 64. 0.141 nm 66. 289 µm 68. (a) 1000 kg (b) 5.2 × 10 –16 kg 0.27 kg (d) 1.3 × 10 –5 kg 70. Aluminum: 2.75 g cm 3 (table value is 2% smaller) Copper: 9.36 g cm 3 (table value is 5% smaller) Brass: 8.91 g cm 3 Tin: 7.68 g cm 3 Iron: 7.88 g cm 3 (table value is 0.3% smaller)

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  • 2000 by Harcourt College Publishers. All rights reserved. Chapter 1 Even Answers 2. 623 kg/m3 4. 4 (r 3 2 r 3 1) 3 6. 7.69 cm 8. 8.72 1011 atoms/s 10. (a) 72.6 kg (b) 7.82 1026 atoms 12. equation is dimensionally consistent 16. The units of G are: m3/kg s2 18. 9.19 nm/s 20. (a) 3.39 105 ft3 (b) 2.54 104 lb 22. 8.32 104 m/s 24. 9.82 cm 26. (a) 6.31 104 AU (b) 1.33 1011 AU 28. (a) 1.609 km/h (b) 88.5 km/h (c) 16.1 km/h 30. (a) 3.16 107 s/yr (b) 6.05 1010 yr 32. 2.57 106 m3 34. 1.32 1021 kg 36. (a) 2.07 mm (b) 8.62 1013 times as large 38. (a) 13.4 (b) 49.1 40. rAl = rFe 3 (Fe/Al) 42. ~ 106 km 44. ~ 109 drops 46. time required 50 years or more; advise against accepting the offer 48. ~ 105 tons 50. (a) 2 (b) 4 (c) 3 (d) 2 52. (a) 797 (b) 1.1 (c) 17.66 54. (a) 3 (b) 4 (c) 3 (d) 2 56. 5.2 m3, 2.7% 58. 1.79 109 m 60. 24.6 62. (b) Acylinder = R2, Arectangular solid = lw 64. 0.141 nm 66. 289 m 68. (a) 1000 kg (b) 5.2 1016 kg 0.27 kg (d) 1.3 105 kg 70. Aluminum: 2.75 g cm3 (table value is 2% smaller) Copper: 9.36 g cm3 (table value is 5% smaller) Brass: 8.91 g cm3 Tin: 7.68 g cm3 Iron: 7.88 g cm3 (table value is 0.3% smaller)
  • 2 Chapter 1 Even Answers 2000 by Harcourt College Publishers. All rights reserved.
  • 2000 by Harcourt College Publishers. All rights reserved. Chapter 1 Solutions *1.1 With V = (base area) (height) V = r 2 h and = m V , we have = m r 2 h = 1 kg (19.5 mm) 2 39.0 mm 10 9 mm 3 1 m 3 = 2.15 10 4 kg/m 3 1.2 = M V = M 4 3 R 3 = 3(5.64 10 26 kg) 4 (6.00 10 7 m) 3 = 623 kg/m 3 1.3 VCu = V0 Vi = 4 3 (r 3 o r 3 i ) VCu = 4 3 [ ](5.75 cm)3 (5.70 cm)3 = 20.6 cm3 = m V m = V = (8.92 g/cm3)(20.6 cm3) = 184 g 1.4 V = Vo Vi = 4 3 (r 3 2 r 3 1 ) = m V , so m = V = 4 3 (r 3 2 r 3 1) = 4 (r 3 2 r 3 1) 3 *1.5 (a) The number of moles is n = m/M, and the density is = m/V. Noting that we have 1 mole, V1 mol = mFe Fe = nFe MFe Fe = (1 mol)(55.8 g/mol) 7.86 g/cm3 = 7.10 cm3 5.7cm5.7cm 0.05 cm
  • 2 Chapter 1 Solutions 2000 by Harcourt College Publishers. All rights reserved. (b) In 1 mole of iron are NA atoms: V1 atom = V1 mol NA = 7.10 cm3 6.02 1023 atoms/mol = 1.18 1023 cm3 = 1.18 10 -29 m 3 (c) datom = 3 1.18 1029 m3 = 2.28 1010 m = 0.228 nm (d) V1 mol U = (1 mol)(238 g/mol) 18.7 g/cm3 = 12.7 cm3 V1 atom U = V1 mol U NA = 12.7 cm3 6.02 1023 atoms/mol = 2.11 1023 cm3 = 2.11 10 -29 m 3 datom U = 3 V1 atom U = 3 2.11 1029 m3 = 2.77 1010 m = 0.277 nm *1.6 r2 = r1 3 5 = (4.50 cm)(1.71) = 7.69 cm 1.7 Use m = molar mass/NA and 1 u = 1.66 10 -24 g (a) For He, m = 4.00 g/mol 6.02 10 23 mol -1 = 6.64 10 -24 g = 4.00 u (b) For Fe, m = 55.9 g/mol 6.02 10 23 mol -1 = 9.29 10 -23 g = 55.9 u (c) For Pb, m = 207 g/mol 6.02 10 23 mol -1 = 3.44 10 -22 g = 207 u
  • Chapter 1 Solutions 3 2000 by Harcourt College Publishers. All rights reserved. Goal Solution Calculate the mass of an atom of (a) helium, (b) iron, and (c) lead. Give your answers in atomic mass units and in grams. The molar masses are 4.00, 55.9, and 207 g/mol, respectively, for the atoms given. Gather information: The mass of an atom of any element is essentially the mass of the protons and neutrons that make up its nucleus since the mass of the electrons is negligible (less than a 0.05% contribution). Since most atoms have about the same number of neutrons as protons, the atomic mass is approximately double the atomic number (the number of protons). We should also expect that the mass of a single atom is a very small fraction of a gram (~1023 g) since one mole (6.02 1023) of atoms has a mass on the order of several grams. Organize: An atomic mass unit is defined as 1/12 of the mass of a carbon-12 atom (which has a molar mass of 12.0 g/mol), so the mass of any atom in atomic mass units is simply the numerical value of the molar mass. The mass in grams can be found by multiplying the molar mass by the mass of one atomic mass unit (u): 1 u = 1.66 1024 g. Analyze: For He, m = 4.00 u = (4.00 u)(1.66 1024 g/u) = 6.64 1024 g For Fe, m = 55.9 u = (55.9 u)(1.66 1024g/u) = 9.28 1023 g For Pb, m = 207 u = (207 u)(1.66 1024 g/u) = 3.44 1022 g Learn: As expected, the mass of the atoms is larger for bigger atomic numbers. If we did not know the conversion factor for atomic mass units, we could use the mass of a proton as a close approximation: 1u mp = 1.67 1024 g. *1.8 n = m M = 3.80 g 3.35 g 197 g/mol = 0.00228 mol N = (n)NA = (0.00228 mol)(6.02 10 23 atoms/mol) = 1.38 10 21 atoms t = (50.0 yr)(365 d/yr)(24.0 hr/d)(3600 s/hr) = 1.58 10 9 s N t = 1.38 10 21 atoms 1.58 10 9 s = 8.72 10 11 atoms/s 1.9 (a) m = L 3 = (7.86 g/cm 3 )(5.00 10 -6 cm) 3 = 9.83 10 -16 g (b) N = m NA Molar mass = (9.83 10 -16 g)(6.02 10 23 atoms/mol) 55.9 g/mol = 1.06 10 7 atoms
  • 4 Chapter 1 Solutions 2000 by Harcourt College Publishers. All rights reserved. 1.10 (a) The cross-sectional area is A = 2(0.150 m)(0.010 m) + (0.340 m)(0.010 m) = 6.40 10 -3 m 2 The volume of the beam is V = AL = (6.40 10 -3 m 2 )(1.50 m) = 9.60 10 -3 m 3 Thus, its mass is m = V = (7.56 10 3 kg/m 3 )(9.60 10 -3 m 3 ) = 72.6 kg (b) Presuming that most of the atoms are of iron, we estimate the molar mass as M = 55.9 g/mol = 55.9 10 -3 kg/mol. The number of moles is then n = m M = 72.6 kg 55.9 10 -3 kg/mol = 1.30 10 3 mol The number of atoms is N = nNA = (1.30 10 3 mol)(6.02 10 23 atoms/mol) = 7.82 10 26 atoms *1.11 (a) n = m M = 1.20 10 3 g 18.0 g/mol = 66.7 mol, and Npail = nNA = (66.7 mol)(6.02 10 23 molecules/mol) = 4.01 10 25 molecules (b) Suppose that enough time has elapsed for thorough mixing of the hydrosphere. Nboth = Npail mpail Mtotal = (4.01 10 25 molecules) 1.20 kg 1.32 10 21 kg , or Nboth = 3.65 10 4 molecules 1.12 r, a, b, c and s all have units of L. (s a)(s b)(s c) s = L L L L = L 2 = L Thus, the equation is dimensionally consistent. 15.0 cm 36.0 cm36.0 cm 1.00 cm 1.00 cm 15.0 cm
  • Chapter 1 Solutions 5 2000 by Harcourt College Publishers. All rights reserved. 1.13 The term s has dimensions of L, a has dimensions of LT -2 , and t has dimensions of T. Therefore, the equation, s = ka m t n has dimensions of L = (LT -2 ) m (T) n or L 1 T 0 = L m T n-2m The powers of L and T must be the same on each side of the equation. Therefore, L 1 = L m and m = 1 Likewise, equating terms in T, we see that n 2m must equal 0. Thus, n = 2m = 2 The value of k, a dimensionless constant, cannot be obtained by dimensional analysis . 1.14 2 l g = L L/T 2 = T 2 = T 1.15 (a) This is incorrect since the units of [ax] are m 2 /s 2 , while the units of [v] are m/s. (b) This is correct since the units of [y] are m, and cos(kx) is dimensionless if [k] is in m -1 . 1.16 Inserting the proper units for everything except G, kg m s2 = G[kg]2 [m]2 Multiply both sides by [m] 2 and divide by [kg] 2 ; the units of G are m 3 kg s 2 1.17 One month is 1 mo = (30 day)(24 hr/day)(3600 s/hr) = 2.592 10 6 s Applying units to the equation, V = (1.50 Mft 3 /mo)t + (0.00800 Mft 3 /mo 2 )t 2 Since 1 Mft 3 = 10 6 ft 3 , V = (1.50 10 6 ft 3 /mo)t + (0.00800 10 6 ft 3 /mo 2 )t 2
  • 6 Chapter 1 Solutions 2000 by Harcourt College Publishers. All rights reserved. Converting months to seconds, V = 1.50 10 6 ft 3 /mo 2.592 10 6 s/mo t + 0.00800 10 6 ft 3 /mo 2 (2.592 10 6 s/mo) 2 t 2 Thus, V[ft 3 ] = (0.579 ft3 /s)t + (1.19 10-9 ft3 /s2 )t2 *1.18 Apply the following conversion factors: 1 in = 2.54 cm, 1 d = 86400 s, 100 cm = 1 m, and 10 9 nm = 1 m 1 32 in/day (2.54 cm/in)(10 -2 m/cm)(10 9 nm/m) 86400 s/day = 9.19 nm/s This means the proteins are assembled at a rate of many layers of atoms each second! 1.19 Area A = (100 ft)(150 ft) = 1.50 10 4 ft 2 , so A = (1.50 10 4 ft 2 )(9.29 10 -2 m 2 /ft 2 ) = 1.39 10 3 m 2 Goal Solution A rectangular building lot is 100 ft by 150 ft. Determine the area of this lot in m2. G: We must calculate the area and convert units. Since a meter is about 3 feet, we should expect the area to be about A (30 m)(50 m) = 1 500 m2. O: Area = Length Width. Use the conversion: 1 m = 3.281 ft. A: A = L W = (100 ft) 1 m 3.281 ft (150 ft ) 1 m 3.281 ft = 1 390 m2 L: Our calculated result agrees reasonably well with our initial estimate and has the proper units of m2. Unit conversion is a common technique that is applied to many problems. 1.20 (a) V = (40.0 m)(20.0 m)(12.0 m) = 9.60 10 3 m 3 V = 9.60 10 3 m 3 (3.28 ft/1 m) 3 = 3.39 10 5 ft 3
  • Chapter 1 Solutions 7 2000 by Harcourt College Publishers. All rights reserved. (b) The mass of the air is m = airV = (1.20 kg/m 3 )(9.60 10 3 m 3 ) = 1.15 10 4 kg The student must look up weight in the index to find Fg = mg = (1.15 10 4 kg)(9.80 m/s 2 ) = 1.13 10 5 N Converting to pounds, Fg = (1.13 10 5 N)(1 lb/4.45 N) = 2.54 10 4 lb *1.21 (a) Seven minutes is 420 seconds, so the rate is r = 30.0 gal 420 s = 7.14 10 -2 gal/s (b) Converting gallons first to liters, then to m 3 , r = 7.14 10 -2 gal s 3.786 L 1 gal 10 -3 m 3 1 L r = 2.70 10 -4 m 3 /s (c) At that rate, to fill a 1-m 3 tank would take t = 1 m 3 2.70 10 -4 m 3 /s 1 hr 3600 s = 1.03 hr 1.22 v = 5.00 furlongs fortnight 220 yd 1 furlong 0.9144 m 1 yd 1 fortnight 14 days 1 day 24 hrs 1 hr 3600 s = 8.32 10 -4 m/s This speed is almost 1 mm/s; so we might guess the creature was a snail, or perhaps a sloth. 1.23 It is often useful to remember that the 1600-m race at track and field events is approximately 1 mile in length. To be precise, there are 1609 meters in a mile. Thus, 1 acre is equal in area to (1 acre) 1 mi 2 640 acres 1609 m mi 2 = 4.05 10 3 m 2
  • 8 Chapter 1 Solutions 2000 by Harcourt College Publishers. All rights reserved. 1.24 Volume of cube = L 3 = 1 quart (Where L = length of one side of the cube.) Thus, L 3 = (1 quart) 1 gallon 4 quarts 3.786 liters 1 gallon 1000 cm3 1 liter = 946 cm 3 , and L = 9.82 cm 1.25 The mass and volume, in SI units, are m = (23.94 g) 1 kg 1000 g = 0.02394 kg V = (2.10 cm 3 )(10 -2 m/cm) 3 = 2.10 10 -6 m 3 Thus, the density is = m V = 0.02394 kg 2.10 10 -6 m 3 = 1.14 10 4 kg/m 3 Goal Solution A solid piece of lead has a mass of 23.94 g and a volume of 2.10 cm3. From these data, calculate the density of lead in SI units (kg/m3). G: From Table 1.5, the density of lead is 1.13 104 kg/m3, so we should expect our calculated value to be close to this number. This density value tells us that lead is about 11 times denser than water, which agrees with our experience that lead sinks. O: Density is defined as mass per volume, in = m V . We must convert to SI units in the calculation. A: = 23.94 g 2.10 cm3 1 kg 1000 g 100 cm 1 m 3 = 1.14 104 kg/m3 L: At one step in the calculation, we note that one million cubic centimeters make one cubic meter. Our result is indeed close to the expected value. Since the last reported significant digit is not certain, the difference in the two values is probably due to measurement uncertainty and should not be a concern. One important common-sense check on density values is that objects which sink in water must have a density greater than 1 g/cm3, and objects that float must be less dense than water.
  • Chapter 1 Solutions 9 2000 by Harcourt College Publishers. All rights reserved. 1.26 (a) We take information from Table 1.1: 1 LY = (9.46 10 15 m) 1 AU 1.50 10 11 m = 6.31 10 4 AU (b) The distance to the Andromeda galaxy is 2 10 22 m = (2 10 22 m) 1 AU 1.50 10 11 m = 1.33 10 11 AU 1.27 Natoms = mSun matom = 1.99 10 30 kg 1.67 10 -27 kg = 1.19 10 57 atoms 1.28 1 mi = 1609 m = 1.609 km; thus, to go from mph to km/h, multiply by 1.609. (a) 1 mi/h = 1.609 km/h (b) 55 mi/h = 88.5 km/h (c) 65 mi/h = 104.6 km/h. Thus, v = 16.1 km/h 1.29 (a) 6 1012 $ 1000 $/s 1 hr 3600 s 1 day 24 hr 1 yr 365 days = 190 years (b) The circumference of the Earth at the equator is 2 (6378 10 3 m) = 4.01 10 7 m. The length of one dollar bill is 0.155 m so that the length of 6 trillion bills is 9.30 10 11 m. Thus, the 6 trillion dollars would encircle the Earth 9.30 10 11 m 4.01 10 7 m = 2.32 10 4 times Goal Solution At the time of this books printing, the U.S. national debt is about $6 trillion. (a) If payments were made at the rate of $1 000 per second, how many years would it take to pay off a $6-trillion debt, assuming no interest were charged? (b) A dollar bill is about 15.5 cm long. If six trillion dollar bills were laid end to end around the Earths equator, how many times would they encircle the Earth? Take the radius of the Earth at the equator to be 6 378 km. (Note: Before doing any of these calculations, try to guess at the answers. You may be very surprised.) (a) G: $6 trillion is certainly a large amount of money, so even at a rate of $1000/second, we might guess that it will take a lifetime (~ 100 years) to pay off the debt. O: Time to repay the debt will be calculated by dividing the total debt by the rate at which it is repaid.
  • 10 Chapter 1 Solutions 2000 by Harcourt College Publishers. All rights reserved. A: T = $6 trillion $1000/s = $6 1012 ($1000/s)(3.16 107 s/yr) = 190 yr L: OK, so our estimate was a bit low. $6 trillion really is a lot of money! (b) G: We might guess that 6 trillion bills would encircle the Earth at least a few hundred times, maybe more since our first estimate was low. O: The number of bills can be found from the total length of the bills placed end to end divided by the circumference of the Earth. A: N = L C = (6 1012)(15.5 cm)(1 m/100 cm) 2 6.37 106 m = 2.32 104 times L: OK, so again our estimate was low. Knowing that the bills could encircle the earth more than 20 000 times, it might be reasonable to think that 6 trillion bills could cover the entire surface of the earth, but the calculated result is a surprisingly small fraction of the earths surface area! 1.30 (a) (3600 s/hr)(24 hr/day)(365.25 days/yr) = 3.16 10 7 s/yr (b) Vmm = 4 3 r 3 = 4 3 (5.00 10 -7 m) 3 = 5.24 10 -19 m 3 Vcube Vmm = 1 m 3 5.24 10 -19 m 3 = 1.91 10 18 micrometeorites This would take 1.91 10 18 micrometeorites 3.16 10 7 micrometeorites/yr = 6.05 10 10 yr 1.31 V = At, so t = V A = 3.78 10 -3 m 3 25.0 m 2 = 1.51 10 -4 m (or 151 m) 1.32 V = 1 3 Bh = [(13.0 acres)(43560 ft 2 /acre)] 3 (481 ft) = 9.08 10 7 ft 3 , or V = (9.08 10 7 ft 3 ) 2.83 10 -2 m 3 1 ft 3 = 2.57 10 6 m 3 h BBB hh
  • Chapter 1 Solutions 11 2000 by Harcourt College Publishers. All rights reserved. 1.33 Fg = (2.50 tons/block)(2.00 10 6 blocks)(2000 lb/ton) = 1.00 10 10 lbs 1.34 The area covered by water is Aw = 0.700 AEarth = (0.700)(4 REarth 2 ) = (0.700)(4)(6.37 10 6 m) 2 = 3.57 10 14 m 2 The average depth of the water is d = (2.30 miles)(1609 m/l mile) = 3.70 10 3 m The volume of the water is V = Awd = (3.57 10 14 m 2 )(3.70 10 3 m) = 1.32 10 18 m 3 and the mass is m = V = (1000 kg/m 3 )(1.32 10 18 m 3 ) = 1.32 10 21 kg *1.35 SI units of volume are in m 3 : V = (25.0 acre-ft) 43560 ft 2 1 acre 0.3048 m 1 ft 3 = 3.08 10 4 m 3 *1.36 (a) dnucleus, scale = dnucleus, real datom, scale datom, real = (2.40 10 -15 m) 300 ft 1.06 10 -10 m = 6.79 10 -3 ft, or dnucleus, scale = (6.79 10 -3 ft)(304.8 mm/1 ft) = 2.07 mm (b) Vatom Vnucleus = 4 r 3 atom/3 4 r 3 nucleus/3 = ratom rnucleus 3 = datom dnucleus 3 = 1.06 10 -10 m 2.40 10 -15 m 3 = 8.62 10 13 times as large 1.37 The scale factor used in the "dinner plate" model is S = 0.25 m 1.0 10 5 lightyears = 2.5 10 -6 m/lightyears The distance to Andromeda in the scale model will be Dscale = DactualS = (2.0 10 6 lightyears)(2.5 10 -6 m/lightyears) = 5.0 m
  • 12 Chapter 1 Solutions 2000 by Harcourt College Publishers. All rights reserved. 1.38 (a) AEarth AMoon = 4 rEarth 2 4rMoon 2 = rEarth rMoon 2 = (6.37 10 6 m)(100 cm/m) 1.74 10 8 cm 2 = 13.4 (b) VEarth VMoon = 4rEarth 3 /3 4rMoon 3 /3 = rEarth rMoon 3 = (6.37 10 6 m)(100 cm/m) 1.74 10 8 cm 3 = 49.1 1.39 To balance, mFe = mAl or FeVFe = AlVAl Fe 4 3 r 3 Fe = Al 4 3 r 3 Al rAl = rFe Fe Al 1/3 rAl = (2.00 cm) 7.86 2.70 1/3 = 2.86 cm 1.40 The mass of each sphere is mA1 = AlVAl = 4AlrAl 3 3 and mFe = FeVFe = 4FerFe 3 3 Setting these masses equal, 4Fer 3 Fe 3 = 4Fer 3 Fe 3 and rAl = rFe 3 Fe/Al 1.41 The volume of the room is 4 4 3 = 48 m 3 , while the volume of one ball is 4 3 0.038 m 2 3 = 2.87 10-5 m 3 . Therefore, one can fit about 48 2.87 10 -5 10 6 ping-pong balls in the room. As an aside, the actual number is smaller than this because there will be a lot of space in the room that cannot be covered by balls. In fact, even in the best arrangement, the so-called "best packing fraction" is 2 6 = 0.74 so that at least 26% of the space will be empty. Therefore, the above estimate reduces to 1.67 10 6 0.740 10 6 .
  • Chapter 1 Solutions 13 2000 by Harcourt College Publishers. All rights reserved. Goal Solution Estimate the number of Ping-Pong balls that would fit into an average-size room (without being crushed). In your solution state the quantities you measure or estimate and the values you take for them. G: Since the volume of a typical room is much larger than a Ping-Pong ball, we should expect that a very large number of balls (maybe a million) could fit in a room. O: Since we are only asked to find an estimate, we do not need to be too concerned about how the balls are arranged. Therefore, to find the number of balls we can simply divide the volume of an average-size room by the volume of an individual Ping-Pong ball. A: A typical room (like a living room) might have dimensions 15 ft 20 ft 8 ft. Using the approximate conversion 1 ft = 30 cm, we find Vroom 15 ft 20 ft 8 ft = 2400 ft3 30 cm 1 ft 3 = 7 107 cm3 A Ping-Pong ball has a diameter of about 3 cm, so we can estimate its volume as a cube: Vball (3 3 3) cm3 = 30 cm3 The number of Ping-Pong balls that can fill the room is N Vroom Vball = 7 107 cm3 30 cm3 = 2 106 balls ~ 106 balls L: So a typical room can hold about a million Ping-Pong balls. This problem gives us a sense of how big a million really is. *1.42 It might be reasonable to guess that, on average, McDonalds sells a 3 cm 8 cm 10 cm = 240 cm 3 medium-sized box of fries, and that it is packed 3/4 full with fries that have a cross section of 1/2 cm 1/2 cm. Thus, the typical box of fries would contain fries that stretched a total of L = 3 4 V A = 3 4 240 cm 3 (0.5 cm) 2 = 720 cm = 7.2 m 250 million boxes would stretch a total distance of (250 10 6 box)(7.2 m/box) = 1.8 10 9 m. But we require an order of magnitude, so our answer is 10 9 m = 1 million kilometers . *1.43 A reasonable guess for the diameter of a tire might be 2.5 ft, with a circumference of about 8 ft. Thus, the tire would make (50 000 mi)(5280 ft/mi)(1 rev/8 ft) = 3 10 7 rev 10 7 rev
  • 14 Chapter 1 Solutions 2000 by Harcourt College Publishers. All rights reserved. 1.44 A typical raindrop is spherical and might have a radius of about 0.1 inch. Its volume is then approximately 4 10 -3 in 3 . Since 1 acre = 43,560 ft 2 , the volume of water required to cover it to a depth of 1 inch is (1 acre)(1 inch) = (1 acre in) 43,560 ft2 1 acre 144 in 2 1 ft 2 6.3 10 6 in 3 . The number of raindrops required is n = volume of water required volume of a single drop 6.3 10 6 in 3 4 10 -3 in 3 = 1.6 10 9 10 9 *1.45 In order to reasonably carry on photosynthesis, we might expect a blade of grass to require at least 1/16 in 2 = 43 10 -5 ft 2 . Since 1 acre = 43,560 ft 2 , the number of blades of grass to be expected on a quarter-acre plot of land is about n= total area area per blade = (0.25 acre)(43,560 ft 2 /acre) 43 10 -5 ft 2 /blade = 2.5 10 7 blades 10 7 blades 1.46 Since you have only 16 hours (57,600 s) available per day, you can count only $57,600 per day. Thus, the time required to count $1 billion dollars is t = 10 9 dollars 5.76 10 4 dollars/day 1 year 365 days = 47.6 years Since you are at least 18 years old, you would be beyond age 65 before you finished counting the money. It would provide a nice retirement, but a very boring life until then. We would not advise it. 1.47 Assume the tub measure 1.3 m by 0.5 m by 0.3 m. One-half of its volume is then V = (0.5)(1.3 m)(0.5 m)(0.3 m) = 0.10 m 3 The mass of this volume of water is mwater = waterV= (1000 kg/m 3 )(0.10 m 3 ) = 100 kg ~10 2 kg Pennies are now mostly zinc, but consider copper pennies filling 50% of the volume of the tub. The mass of copper required is mcopper = copperV = (8930 kg/m 3 )(0.10 m 3 ) = 893 kg ~10 3 kg
  • Chapter 1 Solutions 15 2000 by Harcourt College Publishers. All rights reserved. *1.48 The typical person probably drinks 2 to 3 soft drinks daily. Perhaps half of these were in aluminum cans. Thus, we will estimate 1 aluminum can disposal per person per day. In the U.S. there are 250 million people, and 365 days in a year, so (250 10 6 cans/day)(365 days/year) 10 10 cans are thrown away or recycled each year. Guessing that each can weighs around 1/10 of an ounce, we estimate this represents (10 10 cans)(0.1 oz/can)(1 lb/16 oz)(1 ton/2000 lb) 3.1 10 5 tons/year. 10 5 tons 1.49 Assume: Total population = 10 7 ; one out of every 100 people has a piano; one tuner can serve about 1,000 pianos (about 4 per day for 250 weekdays, assuming each piano is tuned once per year). Therefore, # tuners ~ 1 tuner 1000 pianos 1 piano 100 people (10 7 people) = 100 1.50 (a) 2 (b) 4 (c) 3 (d) 2 1.51 (a) r 2 = (10.5 m 0.2 m) 2 = [ ](10.5 m) 2 2(10.5 m)(0.2 m) + (0.2 m) 2 = 346 m 2 13 m 2 (b) 2r = 2 (10.5 m 0.2 m) = 66.0 m 1.3 m 1.52 (a) 756.?? 37.2? 0.83 + 2.5? 796./5/3 = 797 (b) 0.0032 (2 s.f.) 356.3 (4 s.f.) = 1.14016 = (2 s.f.) 1.1 (c) 5.620 (4 s.f.) (> 4 s.f.) = 17.656 = (4 s.f.) 17.66
  • 16 Chapter 1 Solutions 2000 by Harcourt College Publishers. All rights reserved. 1.53 r = (6.50 0.20) cm = (6.50 0.20) 10 -2 m m = (1.85 0.02) kg = m 4 3 r 3 also, = m m + 3r r In other words, the percentages of uncertainty are cumulative. Therefore, = 0.02 1.85 + 3(0.20) 6.50 = 0.103 = 1.85 4 3 (6.5 10 -2 m) 3 = 1.61 10 3 kg/m 3 and = (1.61 0.17) 10 3 kg/m 3 1.54 (a) 3 (b) 4 (c) 3 (d) 2 1.55 The distance around is 38.44 m + 19.5 m + 38.44 m + 19.5 m = 115.88 m, but this answer must be rounded to 115.9 m because the distance 19.5 m carries information to only one place past the decimal. 115.9 m 1.56 V = 2V1 + 2V2 = 2(V1 + V2) V1 = (17.0 m + 1.0 m + 1.0 m)(1.0 m)(0.09 m) = 1.70 m 3 V2 = (10.0 m)(1.0 m)(0.090 m) = 0.900 m 3 V = 2(1.70 m 3 ) + 2(0.900 m 3 ) = 5.2 m 3 l1 l1 = 0.12 m 19.0 m = 0.0063 w1 w1 = 0.01 m 1.0 m = 0.010 t1 t1 = 0.1 cm 9.0 cm = 0.011 V V = 0.006 + 0.010 + 0.011 = 0.027 = 2.7% 19.0 m 36.0 cm10.0 m 19.0 m
  • Chapter 1 Solutions 17 2000 by Harcourt College Publishers. All rights reserved. *1.57 It is desired to find the distance x such that x 100 m = 1000 m x (i.e., such that x is the same multiple of 100 m as the multiple that 1000 m is of x) . Thus, it is seen that x 2 = (100 m)(1000 m) = 1.00 10 5 m 2 , and therefore x = 1.00 10 5 m 2 = 316 m . 1.58 The volume of oil equals V = 9.00 10 -7 kg 918 kg/m 3 = 9.80 10 10 m 3 . If the diameter of a molecule is d, then that same volume must equal d(r 2 ) = (thickness of slick)(area of oil slick) where r = 0.418 m. Thus, d = 9.80 10 -10 m 3 (0.418 m) 2 = 1.79 10 -9 m 1.59 Atotal = (N)(Adrop) = Vtotal Vdrop (Adrop) = Vtotal 4r 3 /3 (4r 2 ) = 3Vtotal r = 3 30.0 10 -6 m 3 2.00 10 -5 m = 4.50 m 2 1.60 ' (deg) (rad) tan() sin() difference 15.0 0.262 0.268 0.259 3.47% 20.0 0.349 0.364 0.342 6.43% 25.0 0.436 0.466 0.423 10.2% 24.0 0.419 0.445 0.407 9.34% 24.4 0.426 0.454 0.413 9.81% 24.5 0.428 0.456 0.415 9.87% 24.6 0.429 0.458 0.416 9.98% 24.6 24.7 0.431 0.460 0.418 10.1% 1.61 2r = 15.0 m r = 2.39 m h r = tan55.0 h = (2.39 m)tan(55.0) = 3.41 m
  • 18 Chapter 1 Solutions 2000 by Harcourt College Publishers. All rights reserved. 5555 h rr h
  • Chapter 1 Solutions 19 2000 by Harcourt College Publishers. All rights reserved. *1.62 (a) [V] = L 3 , [A] = L 2 , [h] = L [V] = [A][h] L 3 = L 3 L = L 3 . Thus, the equation is dimensionally correct. (b) Vcylinder = R 2 h = (R 2 )h = Ah, where A = R 2 Vrectangular object = lwh = ( lw)h = Ah, where A = lw 1.63 The actual number of seconds in a year is (86,400 s/day)(365.25 day/yr) = 31,557,600 s/yr The percentage error in the approximation is thus ( 107 s/yr) (31,557,600 s/yr) 31,557,600 s/yr 100% = 0.449% *1.64 From the figure, we may see that the spacing between diagonal planes is half the distance between diagonally adjacent atoms on a flat plane. This diagonal distance may be obtained from the Pythagorean theorem, Ldiag = L 2 + L 2 . Thus, since the atoms are separated by a distance L = 0.200 nm, the diagonal planes are separated 1 2 L 2 + L 2 = 0.141 nm *1.65 (a) The speed of flow may be found from v = (Vol rate of flow) (Area: D 2 /4) = 16.5 cm 3 /s (6.30 cm) 2 /4 = 0.529 cm/s (b) Likewise, at a 1.35 cm diameter, v = 16.5 cm 3 /s (1.35 cm) 2 /4 = 11.5 cm/s *1.66 t = V A = V D 2 /4 = 4(12.0 cm 3 ) (23.0 cm) 2 = 0.0289 cm 1 m 100 cm 10 6 m 1 m = 289 m 1.67 V20 mpg = (10 8 cars)(10 4 mi/yr) 20 mi/gal = 5.0 10 10 gal/yr V25 mpg = (10 8 cars)(10 4 mi/yr) 25 mi/gal = 4.0 10 10 gal/yr Fuel saved = V25 mpg V20 mpg = 1.0 10 10 gal/yr
  • 20 Chapter 1 Solutions 2000 by Harcourt College Publishers. All rights reserved.
  • Chapter 1 Solutions 21 2000 by Harcourt College Publishers. All rights reserved. 1.68 (a) 1 cubic meter of water has a mass m = V = (1.00 10 -3 kg/cm 3 )(1.00 m 3 )(10 2 cm/m) 3 = 1000 kg (b) As a rough calculation, we treat each item as if it were 100% water. cell: m = V = Error! R 3 ) = Error! D 3 ) = ( )1000 kg/m 3 1 6 (1.0 10 -6 m) 3 = 5.2 10 -16 kg kidney: m = V = Error! R 3 ) = (1.00 10 -3 kg/cm 3 )Error! 3 = Error! fly: m = 4 D 2 h = (1 10 -3 kg/cm 3 ) 4 (2.0 mm) 2 (4.0 mm)(10 -1 cm/mm) 3 = 1.3 10 -5 kg 1.69 The volume of the galaxy is r 2 t = (10 21 m) 2 10 19 m ~ 10 61 m 3 If the distance between stars is 4 10 16 m, then there is one star in a volume on the order of (4 10 16 m) 3 ~ 10 50 m 3 . The number of stars is about 10 61 m 3 10 50 m 3 /star ~ 10 11 stars
  • 22 Chapter 1 Solutions 2000 by Harcourt College Publishers. All rights reserved. 1.70 The density of each material is = m V = m r 2 h = 4m D 2 h Al: = 4(51.5 g) (2.52 cm) 2 (3.75 cm) = 2.75 g cm 3 The tabulated value 2.70 g cm 3 is 2% smaller. Cu: = 4(56.3 g) (1.23 cm) 2 (5.06 cm) = 9.36 g cm 3 The tabulated value 8.92 g cm 3 is 5% smaller. Brass: = 4(94.4 g) (1.54 cm) 2 (5.69 cm) = 8.91 g cm 3 Sn: = 4(69.1 g) (1.75 cm) 2 (3.74 cm) = 7.68 g cm 3 Fe: = 4(216.1 g) (1.89 cm) 2 (9.77 cm) = 7.88 g cm 3 The tabulated value 7.86 g cm 3 is 0.3% smaller.
  • 2000 by Harcourt College Publishers. All rights reserved. Chapter 2 Even Answers 2. (a) 180 km (b) 63.4 km/h 4. (a) 50.0 m/s (b) 41.0 m/s 6. (a) 2v1v2/(v1 + v2) (b) 0 8. (a) 27.0 m (b) xf = 27.0 m + (18.0 m/s)t + (3.00 m/s2)(t)2 (c) 18.0 m/s 10. (b) vt = 5.0 s = 23 m/s, vt = 4.0 s = 18 m/s, vt = 3.0 s = 14 m/s, vt = 2.0 s = 9.0 m/s (c) 4.6 m/s2 (d) 0 12. 4.00 m/s2, sign indicates that acceleration is in negative x direction 14. (a) 20.0 m/s, 5.00 m/s (b) 262 m 16. (c) 4 m/s2 (d) 34 m (e) 28 m 18. (a) 13.0 m/s (b) 10.0 m/s, 16.0 m/s (c) 6.00 m/s2 (d) 6.00 m/s2 20. (f) The spacing of the successive positions would change with less regularity. 22. (a) 5.25 m/s2 (b) 168 m (c) 52.5 m/s 24. 160 ft 26. (a) 1.87 km (b) 1.46 km (c) a1 = 3.3 m/s2 (0 t 15 s), a2 = 0 (15 s t 40 s), a3 = 5.0 m/s2 (40 s t 50 s) (d) (i) x1 = (1.67 m/s2)t2, (ii) x2 = (50 m/s)t 375 m, (iii) x3 = (250 m/s)t (2.5 m/s2)t2 4375 m (e) 37.5 m/s 28. (a) 12.7 m/s (b) -2.30 m/s 30. (a) x = (30.0t t2) m, v = (30.0 2.00t) m/s (b) 225 m 32. 3.10 m/s 34. (a) 4.90 105 m/s2 (b) 3.57 104 s (c) 18.0 cm 36. 200 m 38. (a) 4.98 109 s (b) 1.20 1015 m/s2 40. 11.4 s, 212 m 42. $99.4/h 44. 1.79 s 46. gh 48. (a) 96.0 ft/s downward (b) 3.07 103 ft/s2 upward (c) 3.13 102 s 50. (a) 98.0 m/s (b) 490 m 52. 7.96 s 54. (a) a = (10.0 107 m/s3)t + 3.00 105 m/s2; x = (1.67 107 m/s3)t3 + (1.50 105 m/s2)t2 (b) 3.00 103 s (c) 450 m/s (d) 0.900 m 56. (a) 0.111 s (b) 5.53 m/s 58. 48.0 mm 60. (a) 15.0 s (b) 30.0 m/s (c) 225 m 62. 155 s, 129 s 64. ~ 103 m/s2 66. (a) 26.4 m (b) 6.82% 68. 1.38 103 m 70. (c) v 2 boy h , 0 (d) vboy, 0 72. (b) a = 1.63 m/s2 downward
  • 2 Chapter 2 Even Answers 2000 by Harcourt College Publishers. All rights reserved.
  • 2000 by Harcourt College Publishers. All rights reserved. Chapter 2 Solutions *2.1 (a) v = 2.30 m/s (b) v = x t = 57.5 m 9.20 m 3.00 s = 16.1 m/s (c) v = x t = 57.5 m 0 m 5.00 s = 11.5 m/s 2.2 (a) Displacement = (8.50 104 m/h) 35.0 60.0 h + 130 103 m x = (49.6 + 130) 103 m = 180 km (b) Average velocity = displacement time = 180 km (35.0 + 15.0) 60.0 + 2.00 h = 63.4 km/h 2.3 (a) vav = x t = 10 m 2 s = 5 m/s (b) vav = 5 m 4 s = 1.2 m/s (c) vav = x2 x1 t2 t1 = 5 m 10 m 4 s 2 s = 2.5 m/s (d) vav = x2 x1 t2 t1 = 5 m 5 m 7 s 4 s = 3.3 m/s (e) vav = x2 x1 t2 t1 = 0 0 8 0 = 0 m/s 2.4 x = 10t2 For t(s) = 2.0 2.1 3.0 x(m) = 40 44.1 90 (a) v = x t = 50m 1.0 s = 50.0 m/s (b) v = x t = 4.1 m 0.1 s = 41.0 m/s
  • 2 Chapter 2 Solutions 2000 by Harcourt College Publishers. All rights reserved. 2.5 (a) Let d represent the distance between A and B. Let t1 be the time for which the walker has the higher speed in 5.00 m/s = d t1 . Let t2 represent the longer time for the return trip in 3.00 m/s = d t2 . Then the times are t1 = d (5.00 m/s) and t2 = d (3.00 m/s) . The average speed is: v = Total distance Total time = d + d d (5.00 m/s) + d (3.00 m/s) = 2d (8.00 m/s)d (15.0 m2/s2) v = 2(15.0 m2/s2) 8.00 m/s = 3.75 m/s (b) She starts and finishes at the same point A. With total displacement = 0, average velocity = 0 2.6 (a) v = Total distance Total time Let d be the distance from A to B. Then the time required is d v1 + d v2 . And the average speed is v = 2d d v1 + d v2 = 2v1v2 v1 + v2 (b) With total displacement zero, her average velocity is 0 . 2.7 (a) 5 0 5 2 4 6 x (m) t (s) (b) v = slope = 5.00 m (3.00 m) (6.00 s 1.00 s) = 8.00 m 5.00 s = 1.60 m/s
  • Chapter 2 Solutions 3 2000 by Harcourt College Publishers. All rights reserved. 2.8 (a) At any time, t, the displacement is given by x = (3.00 m/s2)t2. Thus, at ti = 3.00 s: xi = (3.00 m/s2)(3.00 s)2 = 27.0 m (b) At tf = 3.00 s + t :xf = (3.00 m/s2)(3.00 s + t)2, or xf = 27.0 m + (18.0 m/s)t + (3.00 m/s2)(t)2 (c) The instantaneous velocity at t = 3.00 s is: v = lim t o xf xi t = lim t 0 [(18.0 m/s) + (3.00 m/s2)t], or v = 18.0 m/s 2.9 (a) at ti = 1.5 s, xi = 8.0 m (Point A) at tf = 4.0 s, xf = 2.0 m (Point B) v = xf xi tf ti = (2.0 8.0) m (4 1.5) s = 6.0 m 2.5 s = 2.4 m/s (b) The slope of the tangent line is found from points C and D. (tC = 1.0 s, xC = 9.5 m) and (tD = 3.5 s, xD = 0), v 3.8 m/s (c) The velocity is zero when x is a minimum. This is at t 4 s . 2.10 (b) At t = 5.0 s, the slope is v 58 m 2.5 s 23 m/s At t = 4.0 s, the slope is v 54 m 3 s 18 m/s At t = 3.0 s, the slope is v 49 m 3.4 s 14 m/s At t = 2.0 s, the slope is v 36 m 4.0 s 9.0 m/s (c) a = v t 23 m/s 5.0 s 4.6 m/s2 (d) Initial velocity of the car was zero . 4 2 0 8 6 3210 4 5 6 t (s) 10 12 x (m) A C B D A C B D 60 40 20 0 0 2 4 x (m) t (s) 20 0 0 2 4 v (m/s) t (s)
  • 4 Chapter 2 Solutions 2000 by Harcourt College Publishers. All rights reserved. 2.11 (a) v = (5 0) m (1 0) s = 5 m/s (b) v = (5 10) m (4 2) s = 2.5 m/s (c) v = (5 m 5 m) (5 s 4 s) = 0 (d) v = 0 (5 m) (8 s 7 s) = +5 m/s 2.12 a = vf vi tf ti = 0 60.0 m/s 15.0 s 0 = 4.00 m/s2 The negative sign in the result shows that the acceleration is in the negative x direction. *2.13 Choose the positive direction to be the outward perpendicular to the wall. v = vi + at a = v t = 22.0 m/s (25.0 m/s) 3.50 103 s = 1.34 104 m/s2 2.14 (a) Acceleration is constant over the first ten seconds, so at the end v = vi + at = 0 + (2.00 m/s2)(10.0 s) = 20.0 m/s Then a = 0 so v is constant from t = 10.0 s to t = 15.0 s. And over the last five seconds the velocity changes to v = vi + at = 20.0 m/s (3.00 m/s2)(5.00 s) = 5.00 m/s (b) In the first ten seconds x = xi + vit + 1 2 at2 = 0 + 0 + 1 2 (2.00 m/s2)(10.0 s) 2 = 100 m Over the next five seconds the position changes to x = xi + vit + 1 2 at2 = 100 m + 20.0 m/s (5.00 s) + 0 = 200 m And at t = 20.0 s x = xi + vit + 1 2 at2 = 200 m + 20.0 m/s (5.00 s) + 1 2 (3.00 m/s2)(5.00 s) 2 = 262 m 4 2 0 2 4 8 6 321 4 5 6321 4 5 6 t (s) 7 87 8 10 x (m) 6
  • Chapter 2 Solutions 5 2000 by Harcourt College Publishers. All rights reserved. *2.15 (a) Acceleration is the slope of the graph of v vs t. For 0 < t < 5.00 s, a = 0 For 15.0 s < t < 20.0 s, a = 0 For 5.0 s < t < 15.0 s, a = vf vi tf ti a = 8.00 (8.00) 15.0 5.00 = 1.60 m/s2 We can plot a(t) as shown. (b) a = vf vi tf ti (i) For 5.00 s < t < 15.0 s, ti = 5.00 s, vi = 8.00 m/s tf = 15.0 s, vf = 8.00 m/s; a = vf vi tf ti = 8.00 (8.00) 15.0 5.00 = 1.60 m/s2 (ii) ti = 0, vi = 8.00 m/s, tf = 20.0 s, vf = 8.00 m/s a = vf vi tf ti = 8.00 (8.00) 20.0 0 = 0.800 m/s2 0.0 1.0 1050 15 20 t (s) 1.6 2.0 a (m/s2 )
  • 6 Chapter 2 Solutions 2000 by Harcourt College Publishers. All rights reserved. 2.16 (a) See the Graphs at the right. Choose x = 0 at t = 0 At t = 3 s, x = 1 2 (8 m/s)(3 s) = 12 m At t = 5 s, x = 12 m + (8 m/s)(2 s) = 28 m At t = 7 s, x = 28 m + 1 2 (8 m/s)(2 s) = 36 m (b) For 0 < t < 3 s, a = (8 m/s)/3 s = 2.67 m/s2 For 3 < t < 5 s, a = 0 (c) For 5 s < t < 9 s, a = (16 m/s)/4 s = 4 m/s2 (d) At t = 6 s, x = 28 m + (6 m/s)(1 s) = 34 m (e) At t = 9 s, x = 36 m + 1 2 ( 8 m/s) 2 s = 28 m 2.17 x = 2.00 + 3.00t t2, v = dx d t = 3.00 2.00t, a = dv d t = 2.00 At t = 3.00 s: (a) x = (2.00 + 9.00 9.00) m = 2.00 m (b) v = (3.00 6.00) m/s = 3.00 m/s (c) a = 2.00 m/s2 2.18 (a) At t = 2.00 s, x = [3.00(2.00)2 2.00(2.00) + 3.00] m = 11.0 m At t = 3.00 s, x = [3.00(9.00)2 2.00(3.00) + 3.00] m = 24.0 m so v = x t = 24.0 m 11.0 m 3.00 s 2.00 s = 13.0 m/s 20 0 40 50 5 t (s) 1010 x (m) 0 10 10 55 t (s) 1010 v (m/s) 0 5 5 t (s) 1010 a (m/s2 ) 5
  • Chapter 2 Solutions 7 2000 by Harcourt College Publishers. All rights reserved. (b) At all times the instantaneous velocity is v = d d t (3.00t2 2.00t + 3.00) = (6.00t 2.00) m/s At t = 2.00 s, v = [6.00(2.00) 2.00] m/s = 10.0 m/s At t = 3.00 s, v = [6.00(3.00) 2.00] m/s = 16.0 m/s (c) a = v t = 16.0 m/s 10.0 m/s 3.00 s 2.00 s = 6.00 m/s2 (d) At all times a = d d t (6.00 2.00) = 6.00 m/s2 (This includes both t= 2.00 s and t= 3.00 s). 2.19 (a) a = v t = 8.00 m/s 6.00 s = 4 3 m/s2 (b) Maximum positive acceleration is at t = 3 s, and is approximately 2 m/s2 (c) a = 0, at t = 6 s , and also for t > 10 s (d) Maximum negative acceleration is at t = 8 s, and is approximately 1.5 m/s2
  • 8 Chapter 2 Solutions 2000 by Harcourt College Publishers. All rights reserved. *2.20 a = reading order = velocity = acceleration b c d e f One way of phrasing the answer: The spacing of the successive positions would change with less regularity. Another way: The object would move with some combination of the kinds of motion shown in (a) through (e). Within one drawing, the acceleration vectors would vary in magnitude and direction. *2.21 From v 2 f = vi 2 + 2ax, we have (10.97 103 m/s)2 = 0 + 2a(220 m), so that a = 2.74 105 m/s2 which is 2.79 104 times g 2.22 (a) Assuming a constant acceleration: a = vf vi t = 42.0 m/s 8.00 s = 5.25 m/s2 (b) Taking the origin at the original position of the car, x = 1 2 (vi + vf) t = 1 2 (42.0 m/s)(8.00 s) = 168 m (c) From vf = vi + at, the velocity 10.0 s after the car starts from rest is: vf = 0 + (5.25 m/s2)(10.0 s) = 52.5 m/s
  • Chapter 2 Solutions 9 2000 by Harcourt College Publishers. All rights reserved. *2.23 (a) x xi = 1 2 (vi + v) t becomes 40 m = 1 2 (vi + 2.80 m/s)(8.50 s) which yields vi = 6.61 m/s (b) a = v vi t = 2.80 m/s 6.61 m/s 8.50 s = 0.448 m/s2 2.24 Suppose the unknown acceleration is constant as a car moving at vi = 35.0 mi/h comes to a v = 0 stop in x xi = 40.0 ft. We find its acceleration from v2 = v 2 i + 2a(x xi ) a = (v2 v 2 i ) 2(x xi ) = 0 (35.0 mi/h)2 2(40.0 ft) 5280 ft 1 mi 2 1 h 3600 s 2 = 32.9 ft/s2 Now consider a car moving at vi = 70.0 mi/h and stopping to v = 0 with a = 32.9 ft/s2. From the same equation its stopping distance is x xi = v2 vi 2 2a = 0 (70.0 mi/h)2 2(32.9 ft/s2) 5280 ft 1 mi 2 1 h 3600 s 2 = 160 ft 2.25 Given vi = 12.0 cm/s when xi = 3.00 cm (t = 0), and at t = 2.00 s, x = 5.00 cm x = vit + 1 2 at2; x xi = vit + 1 2 at2; 5.00 3.00 = 12.0(2.00) + 1 2 a (2.00)2 ; 8.00 = 24.0 + 2a a = 32.0 2 = 16.0 cm/s2
  • 10 Chapter 2 Solutions 2000 by Harcourt College Publishers. All rights reserved. Goal Solution A body moving with uniform acceleration has a velocity of 12.0 cm/s in the positive x direction when its x coordinate is 3.00 cm. If its x coordinate 2.00s later is 5.00 cm, what is the magnitude of its acceleration? G: Since the object must slow down as it moves to the right and then speeds up to the left, the acceleration must be negative and should have units of cm/s2. O: First we should sketch the problem to see what is happening: x (cm) 5 0 5 5 0 5 initial final Here we can see that the object travels along the x-axis, first to the right, slowing down, and then speeding up as it travels to the left in the negative x direction. We can show the position as a function of time with the notation: x(t) x(0) = 3.00 cm, x(2.00) = 5.00 cm, and v(0) = 12.0 cm/s A: Use the kinematic equation x xi = vit + 1 2 at2, and solve for a. a = 2(x xi vit) t2 a = 2[5.00 cm 3.00 cm (12.0 cm/s)(2.00 s)] (2.00 s)2 a = 16.0 cm/s2 L: The acceleration is negative as expected and it has the correct units of cm/s2. It also makes sense that the magnitude of the acceleration must be greater than 12 cm/s2 since this is the acceleration that would cause the object to stop after 1 second and just return the object to its starting point after 2 seconds.
  • Chapter 2 Solutions 11 2000 by Harcourt College Publishers. All rights reserved. 2.26 (a) Total displacement = area under the (v, t) curve from t = 0 to 50 s. x = 1 2 (50 m/s)(15 s) + (50 m/s)(40 15)s + 1 2 (50 m/s)(10 s) = 1875 m (b) From t = 10 s to t = 40 s, displacement (area under the curve) is x = 1 2 (50 m/s + 33 m/s)(5 s) + (50 m/s)(25 s) = 1457 m (c) 0 t 15 s: a1 = v t = (50 0) m/s 15 s 0 = 3.3 m/s2 15 s < t < 40 s: a2 = 0 40 s t 50 s: a3 = v t = (0 50) m/s 50 s 40 s = 5.0 m/s2 (d) (i) x1 = 0 + 1 2 a1t2 = 1 2 (3.3 m/s2) t2, or x1 = (1.67 m/s2)t2 (ii) x2 = 1 2 (15 s) [50 m/s 0] + (50 m/s)(t 15 s), or x2 = (50 m/s)t 375 m (iii) For 40 s t 50 s, x3 = area under v vs t from t = 0 to 40 s + 1 2 a3(t 40 s)2 + (50 m/s)(t 40 s) or x3 = 375 m + 1250 m + 1 2 (5.0 m/s2)(t 40 s) 2 + (50 m/s)(t 40 s) which reduces to x3 = (250 m/s)t (2.5 m/s2)t2 4375 m (e) v = total displacement total elapsed time = 1875 m 50 s = 37.5 m/s *2.27 (a) Compare the position equation x = 2.00 + 3.00t 4.00t2 to the general form x = xi + vit + 1 2 at2 to recognize that: xi = 2.00 m, vi = 3.00 m/s, and a = 8.00 m/s2 The velocity equation, v = vi + at, is then v = 3.00 m/s (8.00 m/s2)t. The particle changes direction when v = 0, which occurs at t = 3 8 s. The position at this time is: x = 2.00 m + (3.00 m/s)Error! s) (4.00 m/s2)Error! s) 2 = Error! 5 0 5 10 20 30 40 50 a (m/s2) t (s)
  • 12 Chapter 2 Solutions 2000 by Harcourt College Publishers. All rights reserved. (b) From x = xi + vit + 1 2 at2, observe that when x = xi, the time is given by t = 2vi a . Thus, when the particle returns to its initial position, the time is t = 2(3.00 m/s) 8.00 m/s2 = 3 4 s and the velocity is v = 3.00 m/s (8.00 m/s2) 3 4 s = 3.00 m/s 2.28 vi = 5.20 m/s (a) v(t = 2.50 s) = vi + at = 5.20 m/s + (3.00 m/s2)(2.50 s) = 12.7 m/s (b) v(t = 2.50 s) = vi + at = 5.20 m/s + (3.00 m/s2)(2.50 s) = 2.30 m/s 2.29 (a) x = 1 2 at2 (Eq 2.11) 400 m = 1 2 (10.0 m/s2) t2 t = 8.94 s (b) v = at (Eq 2.8) v = (10.0 m/s2)(8.94 s) = 89.4 m/s 2.30 (a) Take ti = 0 at the bottom of the hill where xi = 0, vi = 30.0 m/s, and a = 2.00 m/s2. Use these values in the general equation x = xi + vi t + 1 2 at2 to find x = 0 + 30.0t m/s + 1 2 (2.00 m/s2) t2 when t is in seconds x = (30.0t t2)m To find an equation for the velocity, use v = vi + at = 30.0 m/s + (2.00 m/s2)t v = (30.0 2.00t) m/s
  • Chapter 2 Solutions 13 2000 by Harcourt College Publishers. All rights reserved. (b) The distance of travel x becomes a maximum, xmax, when v = 0 (turning point in the motion). Use the expressions found in part (a) for v to find the value of t when x has its maximum value: From v = (30.0 2.00t) m/s , v = 0 when t = 15.0 s Then xmax = (30.0t t2) m = (30.0)(15.0) (15.0)2 = 225 m 2.31 (a) vi = 100 m/s, a = 5.00 m/s2 v2 = v 2 i + 2ax 0 = (100)2 2(5.00)x x = 1000 m and t = 20.0 s (b) No, at this acceleration the plane would overshoot the runway. *2.32 In the simultaneous equations vx = vxi + axt x xi = 1 2 (vxi + vx)t we have vx = vxi (5.60 m/s2)(4.20 s) 62.4 m = 1 2 (vxi+ vx)4.20 s So substituting for vxi gives 62.4 m = 1 2 [vx + (5.60 m/s2)(4.20 s) + vx]4.20 s 14.9 m/s = vx + 1 2 (5.60 m/s2)(4.20 s) vx = 3.10 m/s
  • 14 Chapter 2 Solutions 2000 by Harcourt College Publishers. All rights reserved. *2.33 Take any two of the standard four equations, such as vx = vxi + axt x xi = 1 2 (vxi + vx)t solve one for vxi, and substitute into the other: vxi = vx axt x xi = 1 2 (vx axt + vx) t Thus x xi = vxt 1 2 axt2 Back in problem 32, 62.4 m = vx(4.20 s) 1 2 (5.60 m/s2)(4.20 s) 2 vx = 62.4 m 49.4 m 4.20 s = 3.10 m/s 2.34 We assume the bullet is a cylinder. It slows down just as its front end pushes apart wood fibers. (a) a = v2 v 2 i 2x = (280 m/s)2 (420 m/s)2 2(0.100 m) = 4.90 105 m/s2 (b) t = 0.100 350 + 0.020 280 = 3.57 10 -4 s (c) vi = 420 m/s, v = 0; a = 4.90 105 m/s2; v2 = vi 2 + 2ax x = v2 vi 2 2a = vi 2 2a = (420 m/s)2 (2 4.90 105 m/s2) x = 0.180 m *2.35 (a) The time it takes the truck to reach 20.0 m/s is found from v = vi + at, solving for t yields t = v vi a = 20.0 m/s 0 m/s 2.00 m/s2 = 10.0 s The total time is thus 10.0 s + 20.0 + 5.00 s = 35.0 s
  • Chapter 2 Solutions 15 2000 by Harcourt College Publishers. All rights reserved. (b) The average velocity is the total distance traveled divided by the total time taken. The distance traveled during the first 10.0 s is x1 = v t = 0 + 20.0 2 (10.0) = 100 m The distance traveled during the next 20.0 s is x2 = vit + 1 2 at2 = (20.0)(20.0) + 0 = 400 m, a being 0 for this interval. The distance traveled in the last 5.00 s is x3 = v t = 20.0 + 0 2 (5.00) = 50.0 m The total distance x = x1 + x2 + x3 = 100 + 400 + 50.0 = 550 m, and the average velocity is given by v = x t = 550 35.0 = 15.7 m/s *2.36 Using the equation x = vit + 1 2 at2 yields x = 20.0(40.0) 1.00(40.0)2/2 = 0, which is obviously wrong. The error occurs because the equation used is for uniformly accelerated motion, which this is not. The acceleration is 1.00 m/s2 for the first 20.0 s and 0 for the last 20.0 s. The distance traveled in the first 20.0 s is: x = vit + 1 2 at2 = (20.0)(20.0) 1.00(20.02)/2 = 200 m During the last 20.0 s, the train is at rest. Thus, the total distance traveled in the 40.0 s interval is 200 m . 2.37 (a) a = v vi t = 632(5280/3600) 1.40 = 662 ft/s2 = 202 m/s2 (b) x = vit + 1 2 at2 = (632)(5280/3600)(1.40) 1 2 662(1.40)2 = 649 ft = 198 m 2.38 We have vi = 2.00 104 m/s, v = 6.00 106 m/s, x xi = 1.50 102 m (a) x xi = 1 2 (vi + v) t t = 2(x xi) vi + v = 2(1.50 102 m) 2.00 104 m/s + 6.00 106 m/s = 4.98 109 s
  • 16 Chapter 2 Solutions 2000 by Harcourt College Publishers. All rights reserved. (b) v2 = v 2 i + 2a(x xi ) a = v2 v 2 i 2(x xi) = (6.00 106 m/s)2 (2.00 104 m/s)2 2(1.50 102 m) = 1.20 1015 m/s2 2.39 (a) Take initial and final points at top and bottom of the incline. If the ball starts from rest, vi = 0, a = 0.500 m/s2, x xi = 9.00 m Then v2 = v 2 i + 2a(x xi) = 02 + 2(0.5 00 m/s2) 9.00 m v = 3.00 m/s (b) x xi = vit + 1 2 at2 9.00 m = 0 + 1 2 (0.500 m/s2) t2 t = 6.00 s (c) Take initial and final points at the bottom of the planes and the top of the second plane, respectively. vi = 3.00 m/s v = 0 x xi = 15.00 m v2 = v 2 i + 2a(x xi) gives a = (v2 v 2 i ) 2(x xi ) = [0 (3.00 m/s)2] 2(15.0 m) = 0.300 m/s2 (d) Take initial point at the bottom of the planes and final point 8.00 m along the second: vi = 3.00 m/s x xi = 8.00 m a = 0.300 m/s2 v2 = v 2 i + 2a(x xi) = (3.00 m/s)2 + 2( 0.300 m/s2)(8.00 m) = 4.20 m2/s2 v = 2.05 m/s
  • Chapter 2 Solutions 17 2000 by Harcourt College Publishers. All rights reserved. 2.40 Take the original point to be when Sue notices the van. Choose the origin of the x-axis at Sue's car. For her we have xis = 0 vis = 30.0 m/s as = 2.00 m/s2 so her position is given by xs(t) = xis + vis t + 1 2 ast2 = (30.0 m/s)t + 1 2 (2.00 m/s2) t2 For the van, xiv = 155 m viv = 5.00 m/s av = 0 and xv(t) = xiv + vivt + 1 2 avt2 = 155 m + (5.00 m/s)t + 0 To test for a collision, we look for an instant tc when both are at the same place: 30.0tc t 2 c = 155 + 5.00tc 0 = t 2 c 25.0tc + 155 From the quadratic formula tc = 25.0 (25.0)2 4(155) 2 = 13.6 s or 11.4 s The smaller value is the collision time. (The larger value tells when the van would pull ahead again if the vehicles could move through each other). The wreck happens at position 155 m + (5.00 m/s)(11.4 s) = 212 m . 2.41 Choose the origin (y = 0, t = 0) at the starting point of the ball and take upward as positive. Then, yi = 0, vi = 0, and a = g = 9.80 m/s2. The position and the velocity at time t become: y yi = vit + 1 2 at2 y = 1 2 gt2 = 1 2 (9.80 m/s2) t2 and v = vi + at v = gt = (9.80 m/s2)t (a) at t = 1.00 s: y = 1 2 (9.80 m/s2)(1.00 s) 2 = 4.90 m at t = 2.00 s: y = 1 2 (9.80 m/s2)(2.00 s) 2 = 19.6 m at t = 3.00 s: y = 1 2 (9.80 m/s2)(3.00 s) 2 = 44.1 m
  • 18 Chapter 2 Solutions 2000 by Harcourt College Publishers. All rights reserved. (b) at t = 1.00 s: v = (9.80 m/s2)(1.00 s) = 9.80 m/s at t = 2.00 s: v = (9.80 m/s2)(2.00 s) = 19.6 m/s at t = 3.00 s: v = (9.80 m/s2)(3.00 s) = 29.4 m/s *2.42 Assume that air resistance may be neglected. Then, the acceleration at all times during the flight is that due to gravity, a = g = 9.80 m/s2. During the flight, Goff went 1 mile (1609 m) up and then 1 mile back down. Determine his speed just after launch by considering his upward flight: v2 = v 2 i + 2a(y yi) 0 = v 2 i 2(9.80 m/s2)(1609 m) vi = 178 m/s His time in the air may be found by considering his motion from just after launch to just before impact: y yi = vit + 1 2 at2 0 = (178 m/s)t 1 2 (9.80 m/s2) t2 The root t = 0 describes launch; the other root, t = 36.2 s, describes his flight time. His rate of pay may then be found from pay rate = $1.00 36.2 s = 0.0276 $ s 3600 s 1 h = $99.4/h 2.43 (a) y = vit + 1 2 at2 4.00 = (1.50)vi (4.90)(1.50)2 and vi = 10.0 m/s upward (b) v = vi + at = 10.0 (9.80)(1.50) = 4.68 m/s v = 4.68 m/s downward 2.44 We have y = 1 2 gt2 + vit + yi 0 = (4.90 m/s2)t2 (8.00 m/s)t + 30.0 m Solving for t, t = 8.00 64.0 + 588 9.80 Using only the positive value for t, we find t = 1.79 s
  • Chapter 2 Solutions 19 2000 by Harcourt College Publishers. All rights reserved. *2.45 The bill starts from rest vi = 0 and falls with a downward acceleration of 9.80 m/s2 (due to gravity). Thus, in 0.20 s it will fall a distance of y = vit 1 2 gt2 = 0 (4.90 m/s2)(0.20 s)2 = 0.20 m This distance is about twice the distance between the center of the bill and its top edge ( 8 cm). Thus, David will be unsuccessful . Goal Solution Emily challenges her friend David to catch a dollar bill as follows. She holds the bill vertically, as in Figure P2.45, with the center of the bill between Davids index finger and thumb. David must catch the bill after Emily releases it without moving his hand downward. If his reaction time is 0.2 s, will he succeed? Explain your reasoning. G: David will be successful if his reaction time is short enough that he can catch the bill before it falls half of its length (about 8 cm). Anyone who has tried this challenge knows that this is a difficult task unless the catcher cheats by anticipating the moment the bill is released. Since Davids reaction time of 0.2 s is typical of most people, we should suspect that he will not succeed in meeting Emilys challenge. O: Since the bill is released from rest and experiences free fall, we can use the equation y = 1 2 gt2 to find the distance y the bill falls in t = 0.2 s A: y = 1 2 (9.80 m/s2)(0.2 s) 2 = 0.196 m > 0.08 m Since the bill falls below Davids fingers before he reacts, he will not catch it. L: It appears that even if David held his fingers at the bottom of the bill (about 16 cm below the top edge), he still would not catch the bill unless he reduced his reaction time by tensing his arm muscles or anticipating the drop. *2.46 At any time t, the position of the ball released from rest is given by y1 = h 1 2 gt2. At time t, the position of the ball thrown vertically upward is described by y2 = vit 1 2 gt2. The time at which the first ball has a position of y1 = h/2 is found from the first equation as h/2 = h 1 2 gt2, which yields t = h/g . To require that the second ball have a position of y2 = h/2 at this time, use the second equation to obtain h/2 = vi h/g 1 2 g(h/g). This gives the required initial upward velocity of the second ball as vi = g h .
  • 20 Chapter 2 Solutions 2000 by Harcourt College Publishers. All rights reserved. 2.47 (a) v = vi gt (Eq. 2.8) v = 0 when t = 3.00 s, g = 9.80 m/s2, vi = gt = (9.80 m/s2)(3.00 s) = 29.4 m/s (b) y = 1 2 (v + vi) t = 1 2 (29.4 m/s)(3.00 s) = 44.1 m Goal Solution A baseball is hit such that it travels straight upward after being struck by the bat. A fan observes that it requires 3.00 s for the ball to reach its maximum height. Find (a) its initial velocity and (b) the maximum height it reaches. G: We can expect the initial speed of the ball to be somewhat greater than the speed of the pitch, which might be about 60 mph (~30 m/s), so an initial upward velocity off the bat of somewhere between 20 and 100 m/s would be reasonable. We also know that the length of a ball field is about 300 ft. (~100m), and a pop-fly usually does not go higher than this distance, so a maximum height of 10 to 100 m would be reasonable for the situation described in this problem. O: Since the balls motion is entirely vertical, we can use the equation for free fall to find the initial velocity and maximum height from the elapsed time. A: Choose the initial point when the ball has just left contact with the bat. Choose the final point at the top of its flight. In between, the ball is in free fall for t = 3.00 s and has constant acceleration, a = -g = -9.80 m/s2. Solve the equation vyf = vyi gt for vyi when vyf = 0 (when the ball reaches its maximum height). (a) vyi = vyf + gt = 0 + (9.80 m/s2)(3.00 s) = 29.4 m/s (upward) (b) The maximum height in the vertical direction is yf = vyi t + 1 2 at2 = (29.4 m/s)(3.00 s) + 1 2 (9.80 m/s2)(3.00 s) 2 = 44.1 m L: The calculated answers seem reasonable since they lie within our expected ranges, and they have the correct units and direction. We must remember that it is possible to solve a problem like this correctly, yet the answers may not seem reasonable simply because the conditions stated in the problem may not be physically possible (e.g. a time of 10 seconds for a pop fly would not be realistic).
  • Chapter 2 Solutions 21 2000 by Harcourt College Publishers. All rights reserved. 2.48 Take downward as the positive y direction. (a) While the woman was in free fall, y = 144 ft, vi = 0, and a = g = 32.0 ft/s2 Thus, y = vit + 1 2 at2 144 ft = 0 + (16.0 ft/s2)t2 giving tfall = 3.00 s. Her velocity just before impact is: v = vi + gt = 0 + (32.0 ft/s2)(3.00 s) = 96.0 ft/s . (b) While crushing the box, vi = 96.0 ft/s, v = 0, and y = 18.0 in = 1.50 ft. Therefore, a = v2 v 2 i 2(y) = 0 (96.0 ft/s)2 2(1.50 ft) = 3.07 103 ft/s2, or a = 3.07 103 ft/s2 upward (c) Time to crush box: t = y v = y (v + vi)/2 = 2(1.50 ft) 0 + 96.0 ft/s or t = 3.13 102 s 2.49 Time to fall 3.00 m is found from Eq. 2.11 with vi = 0, 3.00 m = 1 2 (9.80 m/s2) t2; t = 0.782 s (a) With the horse galloping at 10.0 m/s, the horizontal distance is vt = 7.82 m (b) t = 0.782 s
  • 22 Chapter 2 Solutions 2000 by Harcourt College Publishers. All rights reserved. 2.50 Time to top = 10.0 s. v = vi gt (a) At the top, v = 0. Then, t = vi g = 10.0 s vi = 98.0 m/s (b) h = vit 1 2 gt2 At t = 10.0 s, h = (98.0)(10.0) 1 2 (9.80)(10.0) 2 = 490 m 2.51 vi = 15.0 m/s (a) v = vi gt = 0 t = vi g = 15.0 m/s 9.80 m/s2 = 1.53 s (b) h = vit 1 2 gt2 = v 2 i 2g = 225 19.6 m= 11.5 m (c) At t = 2.00 s v = vi gt = 15.0 19.6 = 4.60 m/s a = g = 9.80 m/s2 2.52 y = 3.00t3 At t = 2.00 s, y = 3.00(2.00)3 = 24.0 m, and vy = dy d t = 9.00t2 = 36.0 m/s If the helicopter releases a small mailbag at this time, the equation of motion of the mailbag is yb = ybi + vit 1 2 gt2 = 24.0 + 36.0t 1 2 (9.80) t2 Setting yb = 0, 0 = 24.0 + 36.0t 4.90t2 Solving for t, (only positive values of t count), t = 7.96 s
  • Chapter 2 Solutions 23 2000 by Harcourt College Publishers. All rights reserved. 2.53 (a) J = da d t = constant da = Jdt a = J dt = Jt + c2 but a = ai when t = 0 so c1 = ai, Therefore, a = Jt + ai a = dv d t ; dv = adt v = adt = (Jt + ai)dt = 1 2 Jt2 + ait + c2 but v = vi when t = 0, so c2 = vi and v = 1 2 Jt2 + ait + vi v = dx d t ; dx = vdt x = vdt = 1 2 Jt2 + ait + vi dt x = 1 6 Jt3 + 1 2 ait2 + vit + c3 x = xi when t = 0, so c3 = xi Therefore, x = 1 6 Jt3 + 1 2 ait2 + vit + xi (b) a2 = (Jt + ai)2 = J2t2 + a 2 i + 2Jait a2 = a 2 i + (J2t2 + 2Jait) a2 = a 2 i + 2J 1 2 Jt2 + ait Recall the expression for v: v = 1 2 Jt2 + ait + vi So (v vi) = 1 2 Jt2 + ait Therefore, a2 = a 2 i + 2J(v vi)
  • 24 Chapter 2 Solutions 2000 by Harcourt College Publishers. All rights reserved. 2.54 (a) a = dv d t = d d t [ 5.00 107 t2 + 3.00 105 t] Error! t + 3.00 105 m/s2 ) Take xi = 0 at t = 0. Then v = dx d t x 0 = 0 t vdt = 0 t (5.00 107 t2 + 3.00 105t) dt x = 5.00 107 t3 3 + 3.00 105 t2 2 x = (1.67 107 m/s3)t3 + (1.50 105 m/s2)t2 (b) The bullet escapes when a = 0, at (10.0 107 m/s3)t + 3.00 105 m/s2 = 0 t = 3.00 105 s 10.0 107 = 3.00 10-3 s (c) New v = ( 5.00 107)(3.00 10 -3)2 + (3.00 105)(3.00 10 -3) v = 450 m/s + 900 m/s = 450 m/s (d) x = (1.67 107)(3.00 10 -3)3 + (1.50 105)(3.00 10 -3)2 x = 0.450 m + 1.35 m = 0.900 m 2.55 a = dv d t = 3.00v2, vi = 1.50 m/s Solving for v, dv d t = 3.00v2 v = vi v v2dv = 3.00 t = 0 0 dt 1 v + 1 vi = 3.00t or 3.00t = 1 v 1 vi When v = vi 2 , t = 1 3.00 vi = 0.222 s
  • Chapter 2 Solutions 25 2000 by Harcourt College Publishers. All rights reserved. 2.56 (a) The minimum distance required for the motorist to stop, from an initial speed of 18.0 m/s, is x = v2 v 2 i 2a = 0 (18.0 m/s)2 2(4.50 m/s2) = 36.0 m Thus, the motorist can travel at most (38.0 m 36.0 m) = 2.00 m before putting on the brakes if he is to avoid hitting the deer. The maximum acceptable reaction time is then tmax = 2.00 m vi = 2.00 m 18.0 m/s = 0.111 s (b) In 0.300 s, the distance traveled at 18.0 m/s is x = vit1 = (18.0 m/s)(0.300) = 5.40 m The displacement for an acceleration 4.50 m/s2 is 38.0 5.40 = 32.6 m. v2 = v 2 i + 2ax = (18.0 m/s)2 2(4.50 m/s2)(32.6 m) = 30.6 m2/s2 v = 30.6 = 5.53 m/s 2.57 The total time to reach the ground is given by y yi = vit + 1 2 at2 0 25.0 m = 0 + 1 2 (9.80 m/s2) t2 t = 2(25.0 m) 9.80 m/s2 = 2.26 s The time to fall the first fifteen meters is found similarly: 15.0 m = 0 1 2 (9.80 m/s2) t 2 1 t1 = 1.75 s So t t1 = 2.26 s 1.75 s = 0.509 s suffices for the last ten meters.
  • 26 Chapter 2 Solutions 2000 by Harcourt College Publishers. All rights reserved. *2.58 The rate of hair growth is a velocity and the rate of its increase is an acceleration. Then vi = 1.04 mm/d and a = 0.132 mm/d w . The increase in the length of the hair (i.e., displacement) during a time of t = 5.00 w = 35.0 d is x = vit + 1 2 at2 x = (1.04 mm/d)(35.0 d) + 1 2 (0.132 mm/d w)(35.0 d)(5.00 w) or x = 48.0 mm 2.59 Let path (#1) correspond to the motion of the rocket accelerating under its own power. Path (#2) is the motion of the rocket under the influence of gravity with the rocket still rising. Path (#3) is the motion of the rocket under the influence of gravity, but with the rocket falling. The data in the table is found for each phase of the rocket's motion. (#1): v2 (80.0)2 = 2(4.00)(1000); therefore v = 120 m/s 120 = 80.0 + (4.00)t giving t = 10.0 s (#2): 0 (120)2 = 2(9.80)x giving x = 735 m 0 120 = 9.80t giving t = 12.2 s This is the time of maximum height of the rocket. (#3): v2 0 = 2(9.80)(1735) v = 184 = (9.80)t giving t = 18.8 s (a) ttotal = 10 + 12.2 + 18.8 = 41.0 s (b) xtotal = 1.73 km (c) vfinal = 184 m/s t x v a 0 Launch 0 0 80 +4.00 #1 End Thrust 10.0 1000 120 +4.00 #2 Rise Upwards 22.2 1735 0 9.80 #3 Fall to Earth 41.0 0 184 9.80 0 3 1 2
  • Chapter 2 Solutions 27 2000 by Harcourt College Publishers. All rights reserved. 2.60 Distance traveled by motorist = (15.0 m/s)t Distance traveled by policeman = 1 2 (2.00 m/s2) t2 (a) intercept occurs when 15.0t =t2 t = 15.0 s (b) v (officer) = (2.00 m/s2)t = 30.0 m/s (c) x (officer) = 1 2 (2.00 m/s2) t2 = 225 m *2.61 Area A1 is a rectangle. Thus, A1 = hw = vit. Area A2 is triangular. Therefore A2 = 1 2 bh = 1 2 t(v vi). The total area under the curve is A = A1 + A2 = vit + (v vi)t/2 and since v vi = at A = vit + 1 2 at2 The displacement given by the equation is: x = vit + 1 2 at2, the same result as above for the total area. 2.62 a1 = 0.100 m/s2, a2 = 0.500 m/s2 x = 1000 m = 1 2 a1t 2 1 + v1t2 + 1 2 a2t 2 2 t = t1 + t2 and v1 = a1t1 = a2t2 1000 = 1 2 a1t 2 1 + a1t1 a1t1 a2 + 1 2 a2 a1t1 a2 2 0 vi t t v v A2 A1
  • 28 Chapter 2 Solutions 2000 by Harcourt College Publishers. All rights reserved. 1000 = 1 2 a1 1 a1 a2 t 2 1 t1 = 20,000 1.20 = 129 s t2 = a1t1 a2 = 12.9 0.500 26 s Total time = t = 155 s 2.63 (a) Let x be the distance traveled at acceleration a until maximum speed v is reached. If this achieved in time t1 we can use the following three equations: x = (v + vi) 2 t1, 100 x = v(10.2 t1) and v = vi + at1 The first two give 100 = 10.2 1 2 t1 v = 10.2 1 2 t1 at1 a = 200 (20.4 t1)t1 . For Maggie a = 200 (18.4)(2.00) = 5.43 m/s2 For Judy a = 200 (17.4)(3.00) = 3.83 m/s2 (b) v = at1 Maggie: v = (5.43)(2.00) = 10.9 m/s Judy: v = (3.83)(3.00) = 11.5 m/s (c) At the six-second mark x = 1 2 at 2 1 + v(6.00 t1) Maggie: x = 1 2 (5.43)(2.00) 2 + (10.9)(4.00) = 54.3 m Judy: x = 1 2 (3.83)(3.00) 2 + (11.5)(3.00) = 51.7 m Maggie is ahead by 2.62 m .
  • Chapter 2 Solutions 29 2000 by Harcourt College Publishers. All rights reserved. *2.64 Let the ball fall 1.50 m. It strikes at speed given by: v 2 x = v 2 xi + 2a(x xi ) v 2 x = 0 + 2(9.80 m/s2)(1.50 m) vx = 5.42 m/s and its stopping is described by v 2 x = v 2 xi + 2ax(x xi) 0 = (5.42 m/s)2 + 2ax(102 m) ax = 29.4 m2/s2 2.00 102 m = +1.47 103 m/s2 Its maximum acceleration will be larger than the average acceleration we estimate by imagining constant acceleration, but will still be of order of magnitude ~ 103 m/s2 . 2.65 Acceleration a = 3.00 m/s2 Deceleration a' = 4.50 m/s2 (a) Keeping track of speed and time for each phase of motion, v0 = 0, v1 = 12.0 m/s t01 = 4.00 s v1 = 12.0 m/s t1 = 5.00 s v1 = 12.0 m/s, v2 = 0 t12 = 2.67 s v2 = 0 m/s, v3 = 18.0 m/s t23 = 6.00 s v3 = 18.0 m/s t3 = 20 .0 s v3 = 18.0 m/s, v4 = 6.00 m/s t34 = 2.67 s v4 = 6.00 m/s t4 = 4 .00 s v4 = 6.00 m/s, v5 = 0 t45 = 1.33 s t = 45.7 s (b) x = vi ti = 6.00(4.00) + 12.0(5.00) + 6.00(2.67) + 9.00(6.00) + 18.0(20.0) + 12.0(2.67) + 6.00(4.00) + 3.00(1.33) = 574 m (c) v = x t = 574 m 45.7 s = 12.6 m/s
  • 30 Chapter 2 Solutions 2000 by Harcourt College Publishers. All rights reserved. (d) tWALK = 2x vWALK = 2(574 m) (1.50 m/s) = 765 s (about 13 minutes, and better exercise!) 2.66 (a) d = 1 2 (9.80) t 2 1 d = 336t2 t1 + t2 = 2.40 336t2 = 4.90(2.40 t2)2 4.90t 2 2 359.5t2 + 28.22 = 0 t2 = 359.5 (359.5)2 4(4.90)(28.22) 9.80 t2 = 359.5 358.75 9.80 = 0.0765 s d = 336t2 = 26.4 m (b) Ignoring the sound travel time, d = 1 2 (9.80)(2.40) 2 = 28.2 m, an error of 6.82% . 2.67 (a) y = vi1t + 1 2 at2 = 50.0 = 2.00t + 1 2 (9.80) t2 t = 2.99 s after the first stone is thrown. (b) y = vi2t + 1 2 at2 and t = 2.99 1.00 = 1.99 s substitute 50.0 = vi2(1.99) + 1 2 (9.80)(1.99) 2 vi2 = 15.4 m/s downward (c) v1 = vi1 + at = 2.00 + (9.80)(2.99) = 31.3 m/s v2 = vi2 + at = 15.3 + (9.80)(1.99) = 34.9 m/s 2.68 The time required for the car to come to rest and the time required to regain its original speed of 25.0 m/s are both given by t = |v| |a| = 25.0 m/s 2.50 m/s2 . The total distance the car travels in these two intervals is xcar = x1 + x2 = (25.0 m/s + 0) 2 (10.0 s) + (0 + 25.0 m/s) 2 (10.0 s) = 250 m
  • Chapter 2 Solutions 31 2000 by Harcourt College Publishers. All rights reserved. The total elapsed time when the car regains its original speed is ttotal = 10.0 s + 45.0 s + 10.0 s = 65.0 s The distance the train has traveled in this time is xtrain = (25.0 m/s)(65.0 s) = 1.63 103 m Thus, the train is 1.63 103 m 250 m = 1.38 103 m ahead of the car. 2.69 (a) We require xs = xk when ts = tk + 1.00 xs = 1 2 (3.50 m/s2)(tk + 1.00) 2 = 1 2 (4.90 m/s2)(tk) 2 = xk tk + 1.00 = 1.183tk tk = 5.46 s (b) xk = 1 2 (4.90 m/s2)(5.46 s) 2 = 73.0 m (c) vk = (4.90 m/s2)(5.46 s) = 26.7 m/s vs = (3.50 m/s2)(6.46 s) = 22.6 m/s 2.70 (a) In walking a distance x, in a time t, the length of rope l is only increased by x sin . The pack lifts at a rate x t sin . v = x t sin = vboy x l = vboy x x2 + h2 (b) a = dv d t = vboy l dx d t + vboyx d d t 1 l a = vboy vboy l vboyx l2 dl d t , but dl d t = v a = v 2 boy l 1 x2 l2 = v 2 boy l h2 l2 = h2v 2 boy (x2 + h2)3/2 (c) v 2 boy h , 0 (d) vboy, 0 x h m l vboy av
  • 32 Chapter 2 Solutions 2000 by Harcourt College Publishers. All rights reserved. 2.71 h = 6.00 m, vboy = 2.00 m/s v = x t sin = vboy x l = vboyx (x2 + h2)1/2 However x = vboyt v = v 2 boyt (v 2 boyt2 + h2)1/2 = 4t (4t2 + 36)1/2 (a) t(s) v(m/s) 0 0 0.5 0.32 1 0.63 1.5 0.89 2 1.11 2.5 1.28 3 1.41 3.5 1.52 4 1.60 4.5 1.66 5 1.71 (b) From problem 2.70 above, a = h2v 2 boy (x2 + h2)3/2 = h2v 2 boy (v 2 boyt2 + h2)3/2 = 144 (4t2 + 36)3/2 t(s) a(m/s2) 0 0.67 0.5 0.64 1 0.57 1.5 0.48 2 0.38 2.5 0.30 3 0.24 3.5 0.18 4 0.14 4.5 0.11 5 0.09 0.6 0.3 0.0 1.2 0.9 3210 4 5 t (s) 1.5 1.8 v (m/s) 0.2 0.1 0.0 0.4 0.3 3210 4 5 t (s) 0.5 0.6 0.7 a (m/s2 )
  • Chapter 2 Solutions 33 2000 by Harcourt College Publishers. All rights reserved. 2.72 Time t(s) Height h(m) h (m) t (s) v (m/s) midpt time t (s) 0.00 5.00 0.75 0.25 3.00 0.13 0.25 5.75 0.65 0.25 2.60 0.38 0.50 6.40 0.54 0.25 2.16 0.63 0.75 6.94 0.44 0.25 1.76 0.88 1.00 7.38 0.34 0.25 1.36 1.13 1.25 7.72 0.24 0.25 0.96 1.38 1.50 7.96 0.14 0.25 0.56 1.63 1.75 8.10 0.03 0.25 0.12 1.88 2.00 8.13 0.06 0.25 0.24 2.13 2.25 8.07 0.17 0.25 0.68 2.38 2.50 7.90 0.28 0.25 1.12 2.63 2.75 7.62 0.37 0.25 1.48 2.88 3.00 7.25 0.48 0.25 1.92 3.13 3.25 6.77 0.57 0.25 2.28 3.38 3.50 6.20 0.68 0.25 2.72 3.63 3.75 5.52 0.79 0.25 3.16 3.88 4.00 4.73 0.88 0.25 3.52 4.13 4.25 3.85 0.99 0.25 3.96 4.38 4.50 2.86 1.09 0.25 4.36 4.63 4.75 1.77 1.19 0.25 4.76 4.88 5.00 0.58 acceleration = slope of line is constant. a = 1.63 m/s2 = 1.63 m/s2 downward 4.00 2.00 0.00 2.00 4.00 321 4 5321 4 5 t (s) 6.00 v (m/s)
  • 34 Chapter 2 Solutions 2000 by Harcourt College Publishers. All rights reserved. 2.73 The distance x and y are always related by x2 + y2 = L2. Differentiating this equation with respect to time, we have 2x dx d t + 2y dy d t = 0 Now dy d t is vB, the unknown velocity of B; and dx d t = v. From the equation resulting from differentiation, we have d y d t = x y dx d t = x y (v) But y x = tan so vB = 1 tan v When = 60.0, vB = v tan 60.0 = v 3 3 = 0.577v Goal Solution Two objects, A and B, are connected by a rigid rod that has a length L. The objects slide along perpendicular guide rails, as shown in Figure P2.73. If A slides to the left with a constant speed v, find the velocity of B when = 60.0. G: The solution to this problem may not seem obvious, but if we consider the range of motion of the two objects, we realize that B will have the same speed as A when = 45, and when = 90, then vB = 0. Therefore when = 60, we should expect vB to be between 0 and v. O: Since we know a distance relationship and we are looking for a velocity, we might try differentiating with respect to time to go from what we know to what we want. We can express the fact that the distance between A and B is always L, with the relation: x2+ y2 = L2. By differentiating this equation with respect to time, we can find vB = dy/dt in terms of dx/dt = vA = v. A: Differentiating x2 + y2 = L2 gives us 2x dx d t + 2y dy d t = 0 Substituting and solving for the speed of B: vB = dy d t = x y dx d t = x y (v) Now from the geometry of the figure, we notice that y x = tan , so vB = v tan When = 60.0, vB = v tan 60 = v 3 = 0.577v (B is moving up) L: Our answer seems reasonable since we have specified both a magnitude and direction for the velocity of B, and the speed is between 0 and v in agreement with our earlier prediction. In this and many other physics problems, we can find it helpful to examine the limiting cases that define boundaries for the answer. L y x v A B x O y
  • Chapter 2 Solutions 35 2000 by Harcourt College Publishers. All rights reserved.
  • 2000 by Harcourt College Publishers. All rights reserved. Chapter 3 Even Answers 2. (a) 8.60 m (b) 4.47 m, -63.4, 4.24 m, 135 4. (a) (2.17, 1.25) m and (1.90, 3.29) m (b) 4.55 m 6. (a) r, 180 (b) 2r, 180 + (c) 3r, 8. 14 km, 65 N of E 10. 310 km at 57 S of W 12. 9.54 N, 57.0 above the x-axis 14. 7.92 m at 4.34 N of W 16. (a) ~105 m upward (b) ~103 m upward 18. 5.24 km at 25.9 N of W 20. 86.6 m, - 50.0 m 22. 358 m at 2.00 S of E 24. |B| = 7.81, = 59.2, = 39.8, = 67.4 26. 788 miles at 48.0 NE of Dallas 28. (b) 5.00i + 4.00j, 6.40 at 38.7, 1.00i + 8.00j, 8.06 at 97.2 30. Cx = 7.30 cm, Cy = 7.20 cm 32. 6.22 blocks at 110 counterclockwise from east 34. (a) 4.47 m at = 63.4 (b) 8.49 m at = 135 36. 42.7 yards 38. 4.64 m at 78.6 N of E 40. 1.43 104 m at 32.2 above the horizontal 42. 106 44. - 220i + 57.6j, 227 paces at 165 46. (a) (3.12i + 5.02j 2.20k) km (b) 6.31 km 48. (a) (15.1i + 7.72j) cm (b) (7.72i + 15.1j) cm (c) (+7.72i + 15.1j) cm 50. (a) 74.6 N of E (b) 470 km 52. a = 5.00, b = 7.00 54. 2 tan1(1/n) 56. (3.60i + 7.00j) N, 7.87 N at 97.8counterclockwise from horizontal 58. 2.00 m/s j, it is the velocity vector 60. (a) (10.0 m, 16.0 m)
  • 2 Chapter 3 Even Answers 2000 by Harcourt College Publishers. All rights reserved.
  • 2000 by Harcourt College Publishers. All rights reserved. Chapter 3 Solutions *3.1 x = r cos = (5.50 m) cos 240 = (5.50 m)(0.5) = 2.75 m y = r sin = (5.50 m) sin 240 = (5.50 m)(0.866) = 4.76 m 3.2 (a) d = (x2 x1)2 + (y2 y1)2 = (2.00 [3.00]2) + (4.00 3.00)2 d = 25.0 + 49.0 = 8.60 m (b) r1 = (2.00)2 + (4.00)2 = 20.0 = 4.47 m 1 = tan1 4.00 2.00 = 63.4 r2 = (3.00)2 + (3.00)2 = 18.0 = 4.24 m 2 = 135 measured from + x axis. 3.3 We have 2.00 = r cos 30.0 r = 2.00 cos 30.0 = 2.31 and y = r sin 30.0 = 2.31 sin 30.0 = 1.15 3.4 (a) x = r cos and y = r sin , therefore x1 = (2.50 m) cos 30.0, y1 = (2.50 m) sin 30.0, and (x1, y1) = (2.17, 1.25) m x2 = (3.80 m) cos 120, y2 = (3.80 m) sin 120, and (x2, y2) = (1.90, 3.29) m (b) d = (x)2 + (y)2 = 16.6 + 4.16 = 4.55 m
  • 2 Chapter 3 Solutions 2000 by Harcourt College Publishers. All rights reserved. 3.5 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 = x2 + y2 = (2.00 m)2 + (1.00 m)2 = 5.00 m2 = 2.24 m (b) = Arctan 1 2 = 26.6; r = 2.24 m, 26.6 3.6 We have r = x2 + y2 and = Arctan y x (a) The radius for this new point is (x) 2 + y 2 = x 2 + y 2 = r and its angle is Arctan y (x) = 180 (b) (2x)2 + (2y)2 = 2r 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 angle 180 + . (c) (3x)2 + (3y)2 = 3r 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 angle . 3.7 (a) The distance d from A to C is d = x2 + y2 where x = (200) + (300 cos 30.0) = 460 km and y = 0 + (300 sin 30.0) = 150 km d = (460)2 + (150)2 = 484 km (b) tan = y x = 150 460 = 0.326 = tan-1(0.326) = 18.1 N of W 3.8 R 14 km = 65 N of E d 300 km C B 200 km A 3030 R 13 km 6 km
  • Chapter 3 Solutions 3 2000 by Harcourt College Publishers. All rights reserved. 3.9 tan 35.0 = x 100 m x = (100 m)(tan 35.0) = 70.0 m 35.035.0 100 m x 3.10 R = 310 km at 57 S of W base R B A 200 km100 km0 E 3.11 (a) Using graphical methods, place the tail of vector B at the head of vector A. The new vector A + B has a magnitude of 6.1 a t 112 from the x-axis. (b) The vector difference A B is found by placing the negative of vector B at the head of vector A. The resultant vector A B has magnitude 14.8 units at an angle of 22 from the + x-axis. y x A + B A A B B B O
  • 4 Chapter 3 Solutions 2000 by Harcourt College Publishers. All rights reserved. 3.12 Find the resultant F1 + F2 graphically by placing the tail of F2 at the head of F1. The resultant force vector F1 + F2 is of magnitude 9.5 N and at an angle of 57 above the x-axis . 0 1 2 3 N x y F1 F2 F1 + F2 3.13 (a) d = 10.0i = 10.0 m since the displacement is a straight line from point A to point B. (b) The actual distance walked is not equal to the straight-line displacement. The distance follows the curved path of the semi-circle (ACB). s = 1 2 (2 r) = 5 = 15.7 m (c) If the circle is complete, d begins and ends at point A. Hence, d = 0 . 3.14 Your sketch should be drawn to scale, and should look somewhat like that pictured below. The angle from the westward direction, , can be measured to be 4 N of W , and the distance R from the sketch can be converted according to the scale to be 7.9 m . E N W S 30.030.0 R 8.20 meters 8.20 meters 15.0 meters 3.50 meters 3.50 meters 15.0 meters B d A C 5.00 m5.00 m
  • Chapter 3 Solutions 5 2000 by Harcourt College Publishers. All rights reserved. 3.15 To find these vector expressions graphically, we draw each set of vectors. Measurements of the results are taken using a ruler and protractor. (a) A + B = 5.2 m at 60 (b) A B = 3.0 m at 330 B A + B 4 m2 m a A 0 B A B A 4 m2 m b 0 (c) B A = 3.0 m at 150 (d) A 2B = 5.2 m at 300 B B A A 4 m2 m c 0 2B A 2B A 4 m2 m d 0 *3.16 (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, 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 ~ 105(0.03 m) + 102(1 m) ~103 m upward .
  • 6 Chapter 3 Solutions 2000 by Harcourt College Publishers. All rights reserved. 3.17 The scale drawing for the graphical solution should be similar to the figure at the right. The magnitude and direction of the final displacement from the starting point are obtained by measuring d and on the drawing and applying the scale factor used in making the drawing. The results should be d 420 ft and 3 3.18 x y 0 km 3.00 km 1.41 1.41 4.00 0 2.12 2.12 4.71 2.29 R = |x|2 + |y|2 = 5.24 km = tan1 y x = 154 or = 25.9 N of W 3.19 Call his first direction the x direction. R = 10.0 m i + 5.00 m(j) + 7.00 m(i) = 3.00 m i 5.00 m j = (3.00)2 + (5.00)2 m at Arctan 5 3 to the right R = 5.83 m at 59.0 to the right from his original motion 3.20 Coordinates of super-hero are: x = (100 m) cos (30.0) = 86.6 m y = (100 m) sin (30.0) = 50.0 m 135 ft 200 ft d 135 ft y x 40.040.0 30.030.0 E N 3.00 km 2.00 km 4.00 km 45.045.0 45.045.0 3.00 km3.00 km 3.00 km 2.00 km 4.00 km R t
  • Chapter 3 Solutions 7 2000 by Harcourt College Publishers. All rights reserved. 100 m x y 30.0
  • 8 Chapter 3 Solutions 2000 by Harcourt College Publishers. All rights reserved. 3.21 The person would have to walk 3.10 sin(25.0) = 1.31 km north , and 3.10 cos(25.0) = 2.81 km east . 3.22 + x East, + y North x = 250 + 125 cos 30 = 358 m y = 75 + 125 sin 30 150 = 12.5 m d = (x)2 + (y)2 = (358)2 + (12.5)2 = 358 m tan = (y) (x) = 12.5 358 = 0.0349 = 2.00 d = 358 m at 2.00 S of E *3.23 Let the positive x-direction be eastward, positive y-direction be vertically upward, and the positive z-direction be southward. The total displacement is then d = (4.80 cm i + 4.80 cm j) + (3.70 cm j 3.70 cm k) or d = 4.80 cm i + 8.50 cm j 3.70 cm k (a) The magnitude is d = (4.80)2 + (8.50)2 + (3.70)2 cm = 10.4 cm (b) Its angle with the y-axis follows from cos = 8.50 10.4 , giving = 35.5 . 3.24 B = Bxi + Byj + B2k B = 4.00i + 6.00j + 3.00k |B|= (4.00)2 + (6.00)2 + (3.00)2 = 7.81 = cos1 4.00 7.81 = 59.2 = cos1 6.00 7.81 = 39.8 = cos1 3.00 7.81 = 67.4
  • Chapter 3 Solutions 9 2000 by Harcourt College Publishers. All rights reserved. 3.25 Ax = 25.0 Ay = 40.0 A = A 2 x + A 2 y = (25.0)2 + (40.0)2 = 47.2 units From the triangle, we find that = 58.0, so that = 122 Goal Solution A vector has an x component of 25.0 units and a y component of 40.0 units. Find the magnitude and direction of this vector. r x y 3025201510 5 0 5 10 40 30 20 10 G: First we should visualize the vector either in our mind or with a sketch. Since the hypotenuse of the right triangle must be greater than either the x or y components that form the legs, we can estimate the magnitude of the vector to be about 50 units. The direction of the vector appears to be about 120 from the +x axis. O: The graphical analysis and visual estimates above may be sufficient for some situations, but we can use trigonometry to obtain a more precise result. A: The magnitude can be found by the Pythagorean theorem: r = x2 + y2 r = (25.0 units)2 + (40 units)2 = 47.2 units We observe that tan = y x (if we consider x and y to both be positive) . = tan1 y x = tan1 40.0 25.0 = tan1 (1.60) = 58.0 The angle from the +x axis can be found by subtracting from 180. = 180 58 = 122 L: Our calculated results agree with our graphical estimates. We should always remember to check that our answers are reasonable and make sense, especially for problems like this where it is easy to mistakenly calculate the wrong angle by confusing coordinates or overlooking a minus sign. Quite often the direction angle of a vector can be specified in more than one way, and we must choose a notation that makes the most sense for the given problem. If compass directions were stated in this question, we could have reported the vector angle to be 32.0 west of north or a compass heading of 328. t 40.040.0 25.025.0 A x y
  • 10 Chapter 3 Solutions 2000 by Harcourt College Publishers. All rights reserved. *3.26 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: dDCeast = dDAeast + dACeast = 730 cos 5.00 560 sin 21.0 = 527 miles. dDCnorth = dDAnorth + dACnorth = 730 sin 5.00 + 560 cos 21.0 = 586 miles. By the Pythagorean theorem, d = (dDCeast)2 + (dDCnorth)2 = 788 mi Then tan = dDCnorth dDCeast = 1.11 and = 48.0. Thus, Chicago is 788 miles at 48.0 north east of Dallas . 3.27 x = d cos = (50.0 m)cos(120) = 25.0 m y = d sin = (50.0 m)sin(120) = 43.3 m d = (25.0 m)i + (43.3 m)j 3.28 (a) A + B A B B B A B A + B A B B B A B (b) C = A + B = 2.00i + 6.00j + 3.00i 2.00j = 5.00i + 4.00j C = 25.0 + 16.0 at Arctan 4 5 C = 6.40 at 38.7 D = A B = 2.00i + 6.00j 3.00i + 2.00j = 1.00i + 8.00j D = (1.00)2 + (8.00)2 at Arctan 8.00 (1.00)
  • Chapter 3 Solutions 11 2000 by Harcourt College Publishers. All rights reserved. D = 8.06 at (180 82.9) = 8.06 at 97.2
  • 12 Chapter 3 Solutions 2000 by Harcourt College Publishers. All rights reserved. 3.29 d = (x1 + x2 + x3)2 + (y1 + y2 + y3)2 = (3.00 5.00 + 6.00)2 + (2.00 + 3.00 + 1.00)2 = 52.0 = 7.21 m = tan-1 6.00 4.00 = 56.3 3.30 A = 8.70i + 15.0j B = 13.2i 6.60j A B + 3C = 0 3C = B A = 21.9i 21.6j C = 7.30i 7.20j or Cx = 7.30 cm Cy = 7.20 cm 3.31 (a) (A + B) = (3i 2j) + (i 4j) = 2i 6j (b) (A B) = (3i 2j) (i 4j) = 4i + 2j (c) A + B = 22 + 62 = 6.32 (d) A B = 42 + 22 = 4.47 (e) A + B = tan1 6 2 = 71.6 = 288 A B = tan1 2 4 = 26.6 3.32 Let i = east and j = north. R = 3.00b j + 4.00b cos 45 i + 4.00b sin 45 j 5.00b i R = 2.17b i + 5.83b j R = 2.172 + 5.832 b at Arctan 5.83 2.17 N of W = 6.22 blocks at 110 counterclockwise from east
  • Chapter 3 Solutions 13 2000 by Harcourt College Publishers. All rights reserved. 3.33 x = r cos and y = r sin , therefore: (a) x = 12.8 cos 150, y = 12.8 sin 150, and (x, y) = (11.1i + 6.40j) m (b) x = 3.30 cos 60.0, y = 3.30 sin 60.0, and (x, y) = (1.65i + 2.86j) cm (c) x = 22.0 cos 215, y = 22.0 sin 215, and (x, y) = (18.0i 12.6j) in 3.34 (a) D = A + B + C = 2i + 4j D = 22 + 42 = 4.47 m at = 63.4 (b) E = A B + C = 6i + 6j E = 62 + 62 = 8.49 m at = 135 3.35 d1 = (3.50j) m d2 = 8.20 cos 45.0i + 8.20 sin 45.0j = (5.80i + 5.80j) m d3 = (15.0i) m R = d1 + d2 + d3 = (15.0 + 5.80)i + (5.80 3.50)j = (9.20i + 2.30j) m (or 9.20 m west and 2.30 m north) The magnitude of the resultant displacement is |R|= R 2 x + R 2 y = (9.20)2 + (2.30)2 = 9.48 m The direction is = Arctan 2.30 9.20 = 166 3.36 Refer to the sketch R = A + B + C = 10.0i 15.0j + 50.0i = 40.0i 15.0j R = [(40.0)2 + (15.0)2]1/2 = 42.7 yards |A| = 10.0 |B| = 15.0 |C| = 50.0 R
  • 14 Chapter 3 Solutions 2000 by Harcourt College Publishers. All rights reserved. 3.37 (a) F = F1 + F2 F = 120 cos (60.0)i + 120 sin (60.0)j 80.0 cos (75.0)i + 80.0 sin (75.0)j F = 60.0i + 104j 20.7i + 77.3j = (39.3i + 181j) N F = (39.3)2 + (181)2 = 185 N ; = tan1 181 39.3 = 77.8 (b) F3 = F = (39.3i 181j) N Goal Solution The helicopter view in Figure P3.37 shows two people pulling on a stubborn mule. Find (a) the single force that is equivalent to the two forces shown and (b) the force that a third person would have to exert on the mule to make the resultant force equal to zero. The forces are measured in units of newtons. G: The resultant force will be larger than either of the two individual forces, and since the two people are not pulling in exactly the same direction, the magnitude of the resultant should be less than the sum of the magnitudes of the two forces. Therefore, we should expect 120 N < R < 200 N. The angle of the resultant force appears to be straight ahead and perhaps slightly to the right. If the stubborn mule remains at rest, the ground must be exerting on the animal a force equal to the resultant R but in the opposite direction. 75 60 80 N 120 N O: We can find R by adding the components of the two force vectors. A: F1 = (120 cos 60)i N + (120 sin 60)j N = 60.0i N + 103.9j N F2 = (80 cos 75)i N + (80 sin 75)j N = 20.7i N + 77.3j N R = F1 + F2 = 39.3i N + 181.2j N R = |R| = (39.3)2 + (181.2)2 = 185 N The angle can be found from the arctan of the resultant components. = tan1 y x = tan1 181.2 39.3 = tan1 (4.61) = 77.8 counterclockwise from the +x axis The opposing force that the either the ground or a third person must exert on the mule, in order for the overall resultant to be zero, is 185 N at 258 counterclockwise from +x. L: The resulting force is indeed between 120 N and 200 N as we expected, and the angle seems reasonable as well. The process applied to solve this problem can be used for other statics problems encountered in physics and engineering. If another force is added to act on a system that is already in equilibrium (sum of the forces is equal to zero), then the system may accelerate. Such a system is now a dynamic one and will be the topic of Chapter 5.
  • Chapter 3 Solutions 15 2000 by Harcourt College Publishers. All rights reserved. 3.38 East North x y 0m 4.00 m 1.41 1.41 0.500 0.866 +0.914 4.55 R = x 2 + y 2 = 4.64 m at 78.6 N of E 3.39 A = 3.00 m, A = 30.0, B = 3.00 m, B = 90.0 Ax = A cos A = 3.00 cos 30.0 = 2.60 m, Ay = A sin A = 3.00 sin 30.0 = 1.50 m so, A = Axi + Ayj = (2.60i + 1.50j) m Bx = 0, By = 3.00 m so B = 3.00j m A + B = (2.60i + 1.50j) + 3.00j = (2.60i + 4.50j) m *3.40 The y coordinate of the airplane is constant and equal to 7.60 103 m whereas the x coordinate is given by x = vit where vi is the constant speed in the horizontal direction. At t = 30.0 s we have x = 8.04 103, so vi = 268 m/s. The position vector as a function of time is P = (268 m/s)t i + (7.60 103 m)j. At t = 45.0 s, P = [1.21 104 i + 7.60 103 j] m. The magnitude is P = (1.21 104)2 + (7.60 103)2 m = 1.43 104 m and the direction is = Arctan 7.60 103 1.21 104 = 32.2 above the horizontal 3.41 We have B = R A Ax = 150 cos 120 = 75.0 cm Ay = 150 sin 120 = 130 cm Rx = 140 cos 35.0 = 115 cm Ry = 140 sin 35.0 = 80.3 cm Therefore, B = [115 (75)]i + [80.3 130]j = (190i 49.7j) cm A B x y 120.0 35.035.0 120.0 R = A + B
  • 16 Chapter 3 Solutions 2000 by Harcourt College Publishers. All rights reserved. B = [1902 + (49.7)2]1/2 = 196 cm , = tan1 49.7 190 = 14.7
  • Chapter 3 Solutions 17 2000 by Harcourt College Publishers. All rights reserved. *3.42 Since A + B = 6.00j, we have (Ax + Bx)i + (Ay + By)j = 0i + 6.00 j giving Ax + Bx = 0, or Ax = Bx (1) and Ay + By = 6.00 (2) Since both vectors have a magnitude of 5.00, we also have: A 2 x + A 2 y = B 2 x + B 2 y = (5.00)2 From Ax = Bx, it is seen that A 2 x = B 2 x . Therefore A 2 x + A 2 y = B 2 x + B 2 y gives A 2 y = B 2 y . Then Ay = By, and Equation (2) gives Ay = By = 3.00. Defining as the angle between either A or B and the y axis, it is seen that cos = Ay A = By B = 3.00 5.00 = 0.600 and = 53.1 The angle between A and B is then = 2 = 106 . 3.43 (a) A = 8.00i + 12.0j 4.00 k (b) B = A/4 = 2.00i + 3.00j 1.00k (c) C = 3A = 24.0i 36.0j + 12.0k 3.44 R = 75.0 cos 240i + 75.0 sin 240j + 125 cos 135i + 125 sin 135j + 100 cos 160i + 100 sin 160j R = 37.5i 65.0j 88.4i + 88.4j 94.0i + 34.2j R = 220i + 57.6j R = (220)2 + 57.62 at Arctan 57.6 220 above the x-axis R = 227 paces at 165 3.45 (a) C = A + B = (5.00i 1.00j 3.00k) m |C|= (5.00)2 + (1.00)2 + (3.00)2 m = 5.92 m (b) D = 2A B = (4.00i 11.0j + 15.0k) m tt A + B AB x y = 2t = 2
  • 18 Chapter 3 Solutions 2000 by Harcourt College Publishers. All rights reserved. |D|= (4.00)2 + (11.0)2 + (15.0)2 m = 19.0 m
  • Chapter 3 Solutions 19 2000 by Harcourt College Publishers. All rights reserved. *3.46 The displacement from radar station to ship is S = (17.3 sin 136i + 17.3 cos 136j) km = (12.0i 12.4j) km From station to plane, the displacement is P = (19.6 sin 153i + 19.6 cos 153j + 2.20k) km, or P = (8.90i 17.5j + 2.20k) km. (a) From plane to ship the displacement is D = S P = (3.12i + 5.02j 2.20k) km (b) The distance the plane must travel is D= |D| = (3.12)2 + (5.02)2 + (2.20)2 km = 6.31 km 3.47 The hurricane's first displacement is 41.0 km h (3.00 h) at 60.0 N of W, and its second displacement is 25.0 km h (1.50 h) due North. With i representing east and j representing north, its total displacement is: 41.0 km h cos 60.0 (3.00 h)(i) + 41.0 km h sin 60.0 (3.00 h) j + 25.0 km h (1.50 h) j = 61.5 km (i) + 144 km j with magnitude (61.5 km)2 + (144 km)2 = 157 km *3.48 (a) E = (17.0 cm) cos 27.0i + (17.0 cm) sin 27.0j E = (15.1i + 7.72j) cm (b) F = (17.0 cm) sin 27.0i + (17.0 cm) cos 27.0j F = (7.72i + 15.1j) cm (c) G = +(17.0 cm) sin 27.0i + (17.0 cm) cos 27.0j G= (+7.72i + 15.1j) cm F G E x y 27.0 27.0 27.0 27.0 27.0 27.0
  • 20 Chapter 3 Solutions 2000 by Harcourt College Publishers. All rights reserved. 3.49 Ax = 3.00, Ay = 2.00 (a) A = Axi + Ayj = 3.00i + 2.00j (b) |A|= A 2 x + A 2 y = (3.00)2 + (2.00)2 =