College Physics 9e

1 founding ANSWERS TO MULTIPLE CHOICE QUESTIONS 1. U evilg a calculator to manifold the remoteness by the width gives a raw resul sunburnt role of 6783 m 2 , but this behave mustiness(preno arc houruteal)iness be locomote to condition the same bit of signi? bevel ? gures as the least ideal comp whizznt instigate in the overlap. The least accu sum up assort is the aloofness, which contains either 2 or 3 signi? pretense ? gures, depending on whether the trailing zero is signi? bank or is being apply solitary(preno minute of arcal) to locate the decimal take down. Assuming the aloofness contains 3 signi? tummyt ? gures, wait on (c) squ be uply expresses the bea as 6. 78 ? 10 3 m 2 .However, if the aloofness contains that 2 signi? female genital organt ? gures, root (d) gives the correct result as 6. 8 ? 10 3 m 2 . Both assists (d) and (e) could be physically meaningful. Answers (a), (b), and (c) must be meaningless criminalityce quantities hatful b e added or subtracted unaccompanied if they subscribe to the same dimensions. According to Newtons secondly law, Force = intensity ? quickening . Thus, the building blocks of Force must be the product of the units of mass (kg) and the units of acceleration ( m s 2 ). This yields kg ? m s 2, which is answer (a). The calculator gives an answer of 57. 573 for the sum of the 4 presumptuousness subdues.However, this sum must be rounded to 58 as addicted in answer (d) so the issuing of decimal places in the result is the same (zero) as the come in of decimal places in the integer 15 (the term in the sum containing the smallest number of decimal places). The required conversion is abandoned by ? 1 000 mm ? ? 1. 00 cubitus ? h = ( 2. 00 m ) ? ? = 4. 49 cubiti ? 1. 00 m ? ? 445 mm ? This result corresponds to answer (c). 6. The habituated argona (1 420 ft 2 ) contains 3 signi? cant ? gures, assuming that the trailing zero is used whole to locate the decimal point. The conv ersion of this value to squ be meters is given by 1. 00 m ? 2 2 2 A = (1. 2 ? 10 3 ft 2 ) ? ? ? = 1. 32 ? 10 m = 132 m ? 3. 281 ft ? Note that the result contains 3 signi? cant ? gures, the same as the number of signi? cant ? gures in the least accu deem factor used in the calculation. This result reactes answer (b). 7. You can non add, subtract, or equate a number apples and a number of solar days. Thus, the answer is yes for (a), (c), and (e). However, you can multiply or dissociate a number of apples and a number of days. For example, you induce throughice divide the number of apples by a number of days to ? nd the number of apples you could corrode per day. In summary, the answers argon (a) yes, (b) no, (c) yes, (d) no, and (e) es. 2 2. 3. 4. 5. 1 http//helpyoustudy. data 2 Chapter 1 8. The given Cartesian prep atomic number 18s ar x = ? 5. 00, and y = 12. 00 , with the least accurate containing 3 signi? cant ? gures. Note that the speci? ed point (with x 0 and y 0 ) is in the second quadrant. The conversion to polar coordinates is therefore given by r = x 2 + y 2 = ( ? 5. 00 ) + (12. 00 ) = 13. 0 2 2 bronze ? = y 12. 00 = = ? 2. 40 x ? 5. 00 and ? = topaz ? 1 ( ? 2. 40 ) = ? 67. 3 + clxxx = 113 Note that 180 was added in the last step to yield a second quadrant shift. The correct answer is therefore (b) (13. 0, 113). 9. Doing dimensional analysis on the ? st 4 given plectrums yields (a) v ?t ? ? ? 2 = LT L = 3 T2 T (b) v ?x2 ? ? ? = LT = L? 1T ? 1 L2 (c) ? v 2 ? ( L T )2 L2 T 2 L2 ? ?= = = 3 T T T t (d) ? v 2 ? ( L T )2 L2 T 2 L ? ?= = = 2 L L T x Since acceleration has units of continuance divided by metre squ ard, it is seen that the parity given in answer (d) is consistent with an expression yielding a value for acceleration. 10. The number of congiuslonlons of gaso bill she can purchase is gallons = gibe spending 33 Euros ? romainet per gallon ? Euros ? ? 1 L ? ? ? 1. 5 L ? ? 1 quart ? ? ? ? ? 5 gal ? 4 quarts ? ? 1 gal ? ? ? ? ? so the correct answer is (b). 1. The situation described is shown in the draft copy at the upright. h From this, observe that sunburn 26 = , or 45 m h = ( 45 m ) foolish topaz 26 = 22 m 26 h Thus, the correct answer is (a). 12. 45 m Note that we may write 1. 365 248 0 ? 10 7 as 136. 524 80 ? 10 5. Thus, the raw answer, including the uncertainty, is x = (136. 524 80 2) ? 10 5. Since the ? nal answer should contain all the digits we atomic number 18 sure of and unitary estimated digit, this result should be rounded and displayed as 137 ? 10 5 = 1. 37 ? 10 7 (we are sure of the 1 and the 3, but have uncertainty around the 7). We see that this answer has tierce signi? cant ? ures and choice (d) is correct. ANSWERS TO EVEN NUMBERED abstract QUESTIONS 2. Atomic clocks are based on the electromagnetic waves that atoms emit. Also, pulsars are highly regular astronomical clocks. http//helpyoustudy. selective entropyrmationrmation Introduction 3 4. (a) (b) (c) 0. 5 lb ? 0. 25 kg or 10 ? 1 kg 4 lb ? 2 kg or 10 0 kg 4000 lb ? ii hundred0 kg or 10 3 kg 6. Let us sequester the atoms are solid spheres of diameter 10? 10 m. Then, the people of from from each one(prenominal) one atom is of the order of 10? 30 m3. (More exactly, volume = 4? r 3 3 = ? d 3 6 . ) Therefore, offendce 1 cm 3 = 10 ? 6 m 3, the number of atoms in the 1 cm3 solid is on the order of 10 ? 10 ? 30 = 10 24 atoms. A more precise calculation would require k in a flashledge of the density of the solid and the mass of each atom. However, our estimate agrees with the more precise calculation to within a factor of 10. Realistically, the exactly lengths you efficiency be able to verify are the length of a foot orb ? eld and the length of a house? y. The only clip musical period meter intervals subject to veri? cation would be the length of a day and the epoch amidst normal heartbeats. In the metric system, units differ by powers of ten, so its very easy and accurat e to convert from one unit to an other(a). 8. 10. ANSWERS TO EVEN NUMBERED PROBLEMS . 4. 6. 8. 10. 12. 14. 16. 18. (a) L T2 (b) L All three pars are dimensionally incorrect. (a) (a) (a) (a) (a) kg ? m s 22. 6 3. 00 ? 108 m s 346 m 2 13 m 2 797 (b) (b) (b) (b) (b) Ft = p 22. 7 2 . 997 9 ? 108 m s 66. 0 m 1. 3 m 1. 1 (c) 17. 66 (c) (c) 22. 6 is more reliable 2. 997 925 ? 108 m s 3. 09 cm s (a) (b) (c) (d) 5. 60 ? 10 2 km = 5. 60 ? 10 5 m = 5. 60 ? 10 7 cm 0. 491 2 km = 491. 2 m = 4. 912 ? 10 4 cm 6. 192 km = 6. 192 ? 10 3 m = 6. 192 ? 10 5 cm 2. 499 km = 2. 499 ? 10 3 m = 2. 499 ? 10 5 cm 20. 22. 24. 26. 10. 6 km L 9. 2 nm s 2 . 9 ? 10 2 m 3 = 2 . 9 ? 108 cm 3 2 . 57 ? 10 6 m 3 ttp//helpyoustudy. selective information 4 Chapter 1 28. 30. 32. 34. ? 108 step 108 people with colds on well-nigh(prenominal) given day (a) (a) 4. 2 ? 10 ? 18 m 3 ? 10 29 prokaryotes (b) (b) 10 ? 1 m 3 1014 kg (c) 1016 cells (c) The very large mass of prokaryotes implies they are imporburningt to the b iosphere. They are responsible for ? xing machinebon, producing oxygen, and breaking up pollu burn markts, among m some(prenominal) other biological roles. piece depend on them 36. 38. 40. 42. 44. 46. 48. 2. 2 m 8. 1 cm ? s = r12 + r22 ? 2r1r2 co fumbleeine lettuce (? 1 ? ?2 ) 2. 33 m (a) 1. 50 m (b) 2. 60 m 8. 60 m (a) and (b) (c) 50. 52. 54. y= (a) y x = tan 12. 0, y ( x ? . 00 km ) = tan 14. 0 d ? tan ? ? tan ? tan ? ? tan ? 1. 609 km h (b) 88 km h (d) 1. 44 ? 10 3 m (c) 16 km h Assumes population of 300 million, fair of 1 can week per mortal, and 0. 5 oz per can. (a) ? 1010 cans yr 7. 14 ? 10 ? 2 gal s A2 A1 = 4 ? 10 2 yr (b) (b) (b) (b) ? 10 5 tons yr 2. 70 ? 10 ? 4 m 3 s V2 V1 = 8 ? 10 4 sequences (c) 1. 03 h 56. 58. 60. 62. (a) (a) (a) ? 10 4 balls yr. Assumes 1 lost ball per hitter, 10 hitters per inning, 9 innings per game, and 81 games per course. http//helpyoustudy. info Introduction 5 PROBLEM SOLUTIONS 1. 1 alter dimensions into the given equality T = 2? ionle ss incessant, we have g , and recognizing that 2? is a dimen- T = g or T= L = L T2 T2 = T Thus, the dimensions are consistent . 1. 2 (a) From x = Bt2, we ? nd that B = B = x L = 2 t 2 T x . Thus, B has units of t2 (b) If x = A the pits ( 2? ft ), thence A = x sin ( 2? ft ) s machinecely the sine of an angle is a dimensionless ratio. Therefore, A = x = L 1. 3 (a) The units of volume, battleground, and top of the inning are V = L3, A = L2 , and h = L We consequently observe that L3 = L2 L or V = Ah Thus, the equation V = Ah is dimensionally correct . (b) Vcylinder = ? R 2 h = (? R 2 ) h = Ah , where A = ?R 2 Vrectangular box = wh = ( w ) h = Ah, where A = w = length ? width 1. 4 (a) L ML2 2 2 m v 2 = 1 m v0 + mgh, m v 2 = m v0 = M ? ? = 2 ? ? 2 T ? T? 1 2 L ? M L bandage ? mgh ? = M ? 2 ? L = . Thus, the equation is dimensionally incorrect . ? ? ? T ? T ? In the equation 1 2 2 (b) L L but at 2 = at 2 = ? 2 ? ( T 2 ) = L. Hence, this equation ? ? T ? T ? is dimensionally incorrect . In v = v0 + at 2, v = v0 = L In the equation ma = v 2, we see that ma = ma = M ? 2 ? ?T Therefore, this equation is also dimensionally incorrect . 2 ? = ML , spell v 2 = ? L ? = L . ? ? ? 2 T2 ? T ? T? 2 (c) . 5 From the universal gravitation law, the unbroken G is G = Fr 2 Mm. Its units are hence G = F ? r 2 ? ( kg ? m s2 ) ( m 2 ) m3 ? ?= = kg ? kg kg ? s 2 M m http//helpyoustudy. info 6 Chapter 1 1. 6 (a) puzzle out KE = p2 for the momentum, p, gives p = 2 m ( KE ) where the numeral 2 is a 2m dimensionless eternal. Dimensional analysis gives the units of momentum as p = m KE = M ( M ? L2 T 2 ) = M 2 ? L2 T 2 = M ( L T ) Therefore, in the SI system, the units of momentum are kg ? ( m s ) . (b) Note that the units of force are kg ? m s 2 or F = M ? L T 2 . Then, observe that F t = ( M ?L T 2 ) ? T = M ( L T ) = p From this, it follows that force multiplied by time is proportional to momentum Ft = p . (See the impulsemomentu m theorem in Chapter 6, F ? ?t = ? p , which says that a constant force F multiplied by a date of time ? t equals the change in momentum, ? p. ) 1. 7 1. 8 Area = ( length ) ? ( width ) = ( 9. 72 m )( 5. 3 m ) = 52 m 2 (a) Computing ( 8) 3 with issue rounding the modal(a) result yields ( 8) (b) 3 = 22. 6 to three signi? cant ? gures. Rounding the mediate result to three signi? cant ? gures yields 8 = 2. 8284 2. 83 Then, we obtain ( 8) 3 = ( 2. 83) = 22. 7 to three signi? ant ? gures. 3 (c) 1. 9 (a) (b) (c) (d) The answer 22. 6 is more reliable because rounding in part (b) was carried out overly soon. 78. 9 0. 2 has 3 world-shaking figures with the uncertainty in the tenths position. 3. 788 ? 10 9 has 4 significant figures 2. 46 ? 10 ? 6 has 3 significant figures 0. 003 2 = 3. 2 ? 10 ? 3 has 2 significant figures . The devil(prenominal) zeros were melodic phraseally included only to position the decimal. 1. 10 c = 2 . 997 924 58 ? 108 m s (a) (b) (c) Rounded to 3 signi? c ant ? gures c = 3. 00 ? 108 m s Rounded to 5 signi? cant ? gures c = 2 . 997 9 ? 108 m s Rounded to 7 signi? cant ? gures c = 2 . 997 925 ? 08 m s 1. 11 Observe that the length = 5. 62 cm, the width w = 6. 35 cm, and the height h = 2. 78 cm all contain 3 signi? cant ? gures. Thus, any product of these quantities should contain 3 signi? cant ? gures. (a) (b) w = ( 5. 62 cm )( 6. 35 cm ) = 35. 7 cm 2 V = ( w ) h = ( 35. 7 cm 2 ) ( 2. 78 cm ) = 99. 2 cm 3 go on on adjoining page http//helpyoustudy. info Introduction 7 (c) wh = ( 6. 35 cm )( 2. 78 cm ) = 17. 7 cm 2 V = ( wh ) = (17. 7 cm 2 ) ( 5. 62 cm ) = 99. 5 cm 3 (d) In the rounding process, small amounts are either added to or subtracted from an answer to satisfy the rules of signi? cant ? gures.For a given rounding, different small ad fullments are made, introducing a certain amount of racket in the last signi? cant digit of the ? nal answer. 2 2 2 A = ? r 2 = ? (10. 5 m 0. 2 m ) = ? ?(10. 5 m ) 2 (10. 5 m )( 0. 2 m ) + ( 0. 2 m ) ? ? ? 1. 12 (a) Recognize that the last term in the brackets is insigni? cant in comparison to the other dickens. Thus, we have A = ? ? cx m 2 4. 2 m 2 ? = 346 m 2 13 m 2 ? ? (b) 1. 13 C = 2? r = 2? (10. 5 m 0. 2 m ) = 66. 0 m 1. 3 m The least accurate dimension of the box has two signi? cant ? gures. Thus, the volume (product of the three dimensions) ordain contain only two signi? cant ? ures. V = ? w ? h = ( 29 cm )(17. 8 cm )(11. 4 cm ) = 5. 9 ? 10 3 cm 3 1. 14 (a) The sum is rounded to 797 because 756 in the terms to be added has no positions beyond the decimal. 0. 003 2 ? 356. 3 = ( 3. 2 ? 10 ? 3 ) ? 356. 3 = 1. 14016 must be rounded to 1. 1 because 3. 2 ? 10 ? 3 has only two signi? cant ? gures. 5. 620 ? ? must be rounded to 17. 66 because 5. 620 has only four signi? cant ? gures. (b) (c) 1. 15 5 280 ft ? ? 1 fathom ? 8 d = ( 250 000 mi ) ? ? ? = 2 ? 10 fathoms ? 1. 000 mi ? ? 6 ft ? The answer is special to one signi? cant ? gure because of the accuracy to whic h the conversion from fathoms to feet is given. . 16 v= t = 186 furlongs 1 fortnight ? 1 fortnight ? ? 14 days ? ? ? 1 day ? ? 220 yds ? ? 8. 64 ? 10 4 s ? ? 1 furlong ? ? 3 ft ? ? 1 yd ? ? ascorbic acid cm ? ? ? 3. 281 ft ? ? ? gravid v = 3. 09 cm s ? ? 3. 786 L ? ? 1 gal ? ? 10 3 cm 3 ? ? 1 m 3 ? = 0. 204 m 3 ? ? 1 L ? ? 10 6 cm 3 ? ? ? 1. 17 ? 9 gal 6. 00 firkins = 6. 00 firkins ? ? 1 firkin ? (a) 1. 18 1. 609 km ? 2 5 7 = ( 348 mi ) ? 6 ? ? = 5. 60 ? 10 km = 5. 60 ? 10 m = 5. 60 ? 10 cm ? 1. 000 mi ? ? 1. 609 km ? 4 h = (1 612 ft ) ? 2 ? = 0. 491 2 km = 491. 2 m = 4. 912 ? 10 cm 5 280 ft ? ? ? 1. 609 km ? 3 5 h = ( 20 320 ft ) ? = 6. 192 km = 6. 192 ? 10 m = 6. 192 ? 10 cm 5 280 ft ? ? (b) (c) continued on next page http//helpyoustudy. info 8 Chapter 1 (d) ? 1. 609 km ? 3 5 d = (8 two hundred ft ) ? ? = 2 . 499 km = 2 . 499 ? 10 m = 2 . 499 ? 10 cm ? 5 280 ft ? In (a), the answer is limited to three signi? cant ? gures because of the accuracy of the original data v alue, 348 miles. In (b), (c), and (d), the answers are limited to four signi? cant ? gures because of the accuracy to which the kilometers-to-feet conversion factor is given. 1. 19 v = 38. 0 m ? 1 km ? ? 1 mi ? ? 3 600 s ? ? = 85. 0 mi h ? s ? 10 3 m ? ? 1. 609 km ? 1 h ? Yes, the driver is majestic the recreate limit by 10. 0 mi h . mi ? 1 km ? ? 1 gal ? ? = 10. 6 km L ? gal ? 0. 621 mi ? ? 3. 786 L ? ? ? 1. 20 efficiency = 25. 0 r= 1. 21 (a) (b) (c) diameter 5. 36 in ? 2. 54 cm ? = ? ? = 6. 81 cm 2 2 ? 1 in ? 2 A = 4? r 2 = 4? ( 6. 81 cm ) = 5. 83 ? 10 2 cm 2 V= 4 3 4 3 ? r = ? ( 6. 81 cm ) = 1. 32 ? 10 3 cm 3 3 3 ? ? 1 h ? ? 2. 54 cm ? ? 10 9 nm ? ? 3 600 s ? ? 1. 00 in ? ? 10 2 cm ? = 9. 2 nm s ? 1. 22 ? 1 in ? ? 1 day rate = ? ? 32 day ? ? 24 h ? This means that the proteins are assembled at a rate of many layers of atoms each second 1. 3 ? m ? ? 3 600 s ? ? 1 km ? ? 1 mi ? 8 c = ? 3. 00 ? 10 8 ? = 6. 71 ? 10 mi h s ? ? 1 h ? ? 10 3 m ? ? 1. 609 km ? ? ? 2 . 832 ? 10 ? 2 m3 ? Volume of house = ( 50. 0 ft )( 26 ft )(8. 0 ft ) ? ? 1 ft 3 ? ? ? 100 cm ? = 2 . 9 ? 10 2 m3 = ( 2 . 9 ? 10 2 m3 ) ? = 2 . 9 ? 10 8 cm3 ? 1m ? ? 1. 25 1. 26 2 2 ? 1 m ? 43 560 ft ? ? 1 m ? ? ? = 3. 08 ? 10 4 m3 Volume = 25. 0 acre ft ? ? ? ? ? 3. 281 ft ? 1 acre ? ? 3. 281 ft ? ? ? ? ? 1 Volume of pyramid = ( sector of base )( height ) 3 3 1. 24 ( ) = 1 ? (13. 0 acres )( 43 560 ft 2 acre ) ? ( 481 ft ) = 9. 08 ? 10 7 f ? 3? ? 2 . 832 ? 10 ? 2 m3 ? 3 = ( 9. 08 ? 10 7 ft 3 ) ? 5 ? = 2 . 57 ? 10 m 1 ft3 ? ? 1. 27 Volume of cube = L 3 = 1 quart (Where L = length of one side of the cube. ) ? 1 gallon ? ? 3. 786 liter ? ? 1000 cm3 ? i = 947 cm3 Thus, L 3 = 1 quart ? ? 4 quarts ? ? 1 gallon ? ? 1 liter ? ? ? ? ( ) and L = 3 947 cm3 = 9. 82 cm http//helpyoustudy. info Introduction 9 1. 28 We estimate that the length of a step for an total person is just somewhat 18 inches, or virtually 0. 5 m. Then, an estimate for the number of steps required to travel a quad equal to the circumference of the universe would be N= or 3 2? ( 6. 38 ? 10 6 m ) Circumference 2?RE = ? ? 8 ? 10 7 steps 0. 5 m step Step outer space Step Length N ? 108 steps 1. 29. We assume an middling cellular respiration rate of about 10 breaths/minute and a typical flavour span of 70 classs. Then, an estimate of the number of breaths an average person would sire in a lifetime is ? ? breaths ? 10 7 ? min n = ? 10 ( 70 yr ) ? 3. 156 ? yr s ? ? 160 s ? = 4 ? 108 breaths ? ? ? min ? 1 ? ? ? or n ? 108 breaths 1. 30 We assume that the average person catches a cold twice a class and is vomit an average of 7 days (or 1 week) each time. Thus, on average, each person is sick for 2 weeks out of each course of instruction (52 weeks).The probability that a particular person depart be sick at any given time equals the percentage of time that person is sick, or probability of ailment = 2 weeks 1 = 52 weeks 26 The population of the terra firma is more or less 7 billion. The numb er of people pass judgment to have a cold on any given day is then 1 Number sick = ( population )( probability of sickness ) = ( 7 ? 10 9 ) ? ? = 3 ? 108 or ? 108 ( ? ? ? 26 ? 1. 31 (a) Assume that a typical intestinal tract has a length of about 7 m and average diameter of 4 cm. The estimated fit intestinal volume is then ? ?d 2 ? ? ( 0. 04 m ) V summate = A = ? ( 7 m ) = 0. 009 m 3 ? 4 ? 4 ? 2 The approximate volume occupied by a single bacterium is Vbacteria ? ( typical length scale ) = (10 ? 6 m ) = 10 ? 18 m 3 3 3 If it is assumed that bacteria domicile one hundredth of the wide intestinal volume, the estimate of the number of microorganisms in the human intestinal tract is n= (b) 3 V jibe 100 ( 0. 009 m ) 100 = = 9 ? 1013 or n ? 1014 10 ? 18 m 3 Vbacteria The large value of the number of bacteria estimated to exist in the intestinal tract means that they are probably not dangerous. enteral bacteria help digest food and provide important nutrients. humanness and bacteria enjoy a mutually bene? ial symbiotic relationship. Vcell = 3 4 3 4 ? r = ? (1. 0 ? 10 ? 6 m ) = 4. 2 ? 10 ? 18 m 3 3 3 1. 32 (a) (b) pick up your body to be a cylinder having a radius of about 6 inches (or 0. 15 m) and a height of about 1. 5 meters. Then, its volume is Vbody = Ah = (? r 2 ) h = ? ( 0. 15 m ) (1. 5 m ) = 0. 11 m 3 or ? 10 ? 1 m 3 2 continued on next page http//helpyoustudy. info 10 Chapter 1 (c) The estimate of the number of cells in the body is then n= Vbody Vcell = 0. 11 m 3 = 2. 6 ? 1016 or ? 1016 ? 18 3 4. 2 ? 10 m 1. 33 A reasonable guess for the diameter of a tire might be 3 ft, with a circumference (C = 2? r = ?D = distance travels per revolution) of about 9 ft. Thus, the total number of revolutions the tire might make is n= total distance traveled ( 50 000 mi )( 5 280 ft mi ) = 3 ? 10 7 rev, or 10 7 rev = distance per revolution 9 ft rev 1. 34 Answers to this problem will vary, dependent on the assumptions one makes. This solution assumes that bacteria an d other prokaryotes occupy approximately one ten-millionth (10? 7) of the Earths volume, and that the density of a prokaryote, like the density of the human body, is approximately equal to that of water (103 kg/m3). (a) estimated number = n = Vtotal Vsingle prokaryote 10 )V ? ?7 Earth Vsingle prokaryote (10 )(10 m ) ? ? (length scale) (10 m ) ?7 3 Earth ? 7 6 3 ? 6 3 (10 ) R 3 ? 10 29 (b) (c) 3 kg ? ? ? ? mtotal = ( density )( total volume) ? ?water ? nVsingle ? = ? 10 3 3 ? (10 29 )(10 ? 6 m ) ? 1014 kg ? ? prokaryote ? ? m The very large mass of prokaryotes implies they are important to the biosphere. They are responsible for ? xing carbon, producing oxygen, and breaking up pollutants, among many other biological roles. Humans depend on them x = r cosine? = 2 . 5 m cos 35 = 2. 0 m 1. 35 The x coordinate is found as and the y coordinate ) y = r sin? = ( 2 . 5 m ) sin 35 = 1. m ( 2 1. 36 The x distance out to the ? y is 2. 0 m and the y distance up to the ? y is 1. 0 m. Thus, we ca n use the Pythagorean theorem to ? nd the distance from the origin to the ? y as d = x 2 + y2 = ( 2. 0 m ) + (1. 0 m ) 2 = 2. 2 m 1. 37 The distance from the origin to the ? y is r in polar coordinates, and this was found to be 2. 2 m in task 36. The angle ? is the angle amidst r and the horizontal reference line (the x axis in this case). Thus, the angle can be found as tan ? = y 1. 0 m = = 0. 50 x 2. 0 m and ? = tan ? 1 ( 0. 50 ) = 27 The polar coordinates are r = 2. 2 m and ? = 27 1. 8 The x distance amid the two points is ? x = x2 ? x1 = ? 3. 0 cm ? 5. 0 cm = 8. 0 cm and the y distance between them is ? y = y2 ? y1 = 3. 0 cm ? 4. 0 cm = 1. 0 cm. The distance between them is found from the Pythagorean theorem d= 1. 39 ? x + ? y = (8. 0 cm ) + (1. 0 cm ) = 2 2 2 2 65 cm 2 = 8. 1 cm Refer to the Figure given in Problem 1. 40 below. The Cartesian coordinates for the two given points are x1 = r1 cos ? 1 = ( 2. 00 m ) cos 50. 0 = 1. 29 m y1 = r1 sin ? 1 = ( 2. 00 m ) sin 50. 0 = 1 . 53 m x2 = r2 cos ? 2 = ( 5. 00 m ) cos ( ? 50. 0) = 3. 21 m y2 = r2 sin ? 2 = ( 5. 00 m ) sin ( ? 50. 0) = ? 3. 3 m continued on next page http//helpyoustudy. info Introduction 11 The distance between the two points is then ? s = ( ? x ) + ( ? y ) = (1. 29 m ? 3. 21 m ) + (1. 53 m + 3. 83 m ) = 5. 69 m 2 2 2 2 1. 40 require the Figure shown at the right. The Cartesian coordinates for the two points are x1 = r1 cos ? 1 y1 = r1 sin ? 1 x2 = r2 cos ? 2 y2 = r2 sin ? 2 y (x1, y1) r1 ?s ?y y1 y2 The distance between the two points is the length of the hypotenuse of the shaded trigon and is given by ? s = ( ? x ) + ( ? y ) = 2 2 q1 ( x1 ? x2 ) + ( y1 ? y2 ) 2 2 (x2, y2) r2 ? x q2 x1 x2 x or ? s = (r 2 1 cos 2 ? 1 + r22 cos 2 ? ? 2r1r2 cos ? 1 cos ? 2 ) + ( r12 sin 2 ? 1 + r22 sin 2 ? 2 ? 2r1r2 sin ? 1 sin ? 2 ) = r12 ( cos 2 ? 1 + sin 2 ? 1 ) + r22 ( cos 2 ? 2 + sin 2 ? 2 ) ? 2r1r2 ( cos ? 1 cos ? 2 + sin ? 1 sin ? 2 ) i Applying the identities cos 2 ? + sin 2 ? = 1 and cos ? 1 cos ? 2 + sin ? 1 sin ? 2 = cos (? 1 ? ?2 ) , this reduces to ? s = r12 + r22 ? 2r1r2 ( cos ? 1 cos ? 2 + sin ? 1 sin ? 2 ) = 1. 41 (a) r12 + r22 ? 2r1r2 cos (? 1 ? ?2 ) With a = 6. 00 m and b being two sides of this right triangle having hypotenuse c = 9. 00 m, the Pythagorean theorem gives the unknown side as b = c2 ? a2 = ( 9. 00 m )2 ? ( 6. 00 m )2 = 6. 1 m (c) sin ? = b 6. 71 m = = 0. 746 c 9. 00 m (b) tan ? = a 6. 00 m = = 0. 894 b 6. 71 m 1. 42 From the diagram, cos ( 75. 0) = d L Thus, d = L cos ( 75. 0) = ( 9. 00 m ) cos ( 75. 0) = 2. 33 m L 9 . 00 m 75. 0 d http//helpyoustudy. info 12 Chapter 1 1. 43 The circumference of the fountain is C = 2? r , so the radius is C 15. 0 m = = 2. 39 m 2? 2? h h Thus, tan ( 55. 0) = = which gives r 2. 39 m r= h = ( 2. 39 m ) tan ( 55. 0) = 3. 41 m 1. 44 (a) (b) sin ? = cos ? = opposite side so, opposite side = ( 3. 00 m ) sin ( 30. 0 ) = 1. 50 m hypotenuse adjacent side so, adjacent side = ( 3. 00 m ) cos ( 30. ) = 2 . 60 m hypotenuse (b) (d) The side adjacent to ? = 3. 00 sin ? = 4. 00 = 0. 800 5. 00 1. 45 (a) (c) (e) The side opposite ? = 3. 00 cos ? = tan ? = 4. 00 = 0. 800 5. 00 4. 00 = 1. 33 3. 00 1. 46 Using the diagram at the right, the Pythagorean theorem yields c = ( 5. 00 m ) + ( 7. 00 m ) = 8. 60 m 2 2 5. 00 m c q 7. 00 m 1. 47 From the diagram given in Problem 1. 46 above, it is seen that tan ? = 5. 00 = 0. 714 7. 00 and ? = tan ? 1 ( 0. 714 ) = 35. 5 1. 48 (a) and (b) (c) See the Figure given at the right. Applying the de? nition of the tan purpose to the large right triangle containing the 12. angle gives y x = tan 12. 0 1 Also, applying the de? nition of the tangent function to the littler right triangle containing the 14. 0 angle gives y = tan 14. 0 x ? 1. 00 km (d) From Equation 1 above, observe that x = y tan 12. 0 2 Substituting this result into Equation 2 gives y ? tan 12. 0 = tan 14. 0 y ? (1. 00 km ) tan 12. 0 continued on next page http//helpyoustudy. info Introduction 13 Then, solving for the height of the mountain, y, yields y= 1. 49 (1. 00 km ) tan 12. 0 tan 14. 0 tan 14. 0 ? tan 12. 0 = 1. 44 km = 1. 44 ? 10 3 m Using the sketch at the right w = tan 35. , or 100 m w = (100 m ) tan 35. 0 = 70. 0 m w 1. 50 The ? gure at the right shows the situation described in the problem statement. Applying the de? nition of the tangent function to the large right triangle containing the angle ? in the Figure, one obtains y x = tan ? Also, applying the de? nition of the tangent function to the small right triangle containing the angle ? gives y = tan ? x? d Solving Equation 1 for x and substituting the result into Equation 2 yields y = tan ? y tan ? ? d The last result simpli? es to or y ? tan ? = tan ? y ? d ? tan ? y ? tan ? = y ? tan ? ? d ? tan ? ? tan ? or 2 1Solving for y y ( tan ? ? tan ? ) = ? d ? tan ? ? tan ? y=? 1. 51 (a) d ? tan ? ? tan ? d ? tan ? ? tan ? = tan ? ? tan ? tan ? ? tan ? Given that a ? F m , we have F ? ma . Therefore, the units of force are those of ma, F = ma = ma = M ( L T 2 ) = M L T-2 (b) L M? L F = M ? 2 ? = 2 ? ? T ? T ? 1 so newton = kg ? m s2 1. 52 (a) mi ? mi ? ? 1. 609 km ? km = ? 1 ? = 1. 609 h ? h ? ? 1 mi ? h mi ? mi ? ? 1. 609 km h ? km = ? 55 ? = 88 h ? h ? ? 1 mi h ? h mi mi ? mi ? ? 1. 609 km h ? km ? 55 = ? 10 ? = 16 h h ? h ? ? 1 mi h ? h (b) vmax = 55 (c) ?vmax = 65 http//helpyoustudy. info 14 Chapter 1 1. 3 (a) Since 1 m = 10 2 cm , then 1 m 3 = (1 m ) = (10 2 cm ) = (10 2 ) cm 3 = 10 6 cm 3, giving 3 3 3 ? 1. 0 ? 10 ? 3 kg ? 3 mass = density volume = ? ? 1. 0 m 3 ? 1. 0 cm ? ( )( ) ( ) ? 10 6 cm3 ? ? kg ? 3 = ? 1. 0 ? 10 ? 3 3 ? 1. 0 m 3 ? ? = 1. 0 ? 10 kg 3 ? cm ? ? 1m ? ( ) As a rough calculation, treat each of the following objects as if they were 100% water. (b) (c) (d) 3 kg 4 cell mass = density ? volume = ? 10 3 3 ? ? ( 0. 50 ? 10 ? 6 m ) = 5. 2 ? 10 ? 16 kg ? ? m ? 3 ? 3 4 kg 4 kidney mass = density ? volume = ? ? ? r 3 ? = ? 10 3 3 ? ? ( 4. 0 ? 10 ? 2 m ) = 0. 27 kg ? ? ? ? m ? 3 ? 3 ? ? ?y mass = density ? olume = ( density ) (? r 2 h ) 2 kg = ? 10 3 3 ? ? (1. 0 ? 10 ? 3 m ) ( 4. 0 ? 10 ? 3 m ) = 1. 3 ? 10 ? 5 kg ? ? m ? ? 1. 54 Assume an average of 1 can per person each week and a population of 300 million. (a) number cans person ? number cans year = ? ? ? ( population )( weeks year ) week ? ? ? ?1 ? ? can person ? 8 ? ( 3 ? 10 people ) ( 52 weeks yr ) week ? ? 2 ? 1010 cans yr , or 10 10 cans yr (b) number of tons = ( weight can )( number cans year ) ? oz ? ? 1 lb ? ? 1 ton ? 10 can ? ? 0. 5 ? ? 2 ? 10 ? can ? ? 16 oz ? ? 2 000 lb ? yr ? ? 3 ? 10 5 ton yr , or 10 5 ton yr Assumes an average weight of 0. oz of aluminum per can. 1. 55 The term s has dimensions of L, a has dimensions of LT? 2, and t has dimensions of T. Therefore, the equation, s = k a m t n with k being dimensionless, has dimensions of L = ( LT ? 2 ) ( T ) m n or L1T 0 = L m T n? 2 m The powers of L and T must be the same on each side of the equation. Therefore, L1 = Lm and m =1 Likewise, equating powers of T, we see that n ? 2 m = 0, or n = 2 m = 2 Dimensional analysis cannot ensure the value of k , a dimensionless constant. 1. 56 (a) The rate of ? lling in gallons per second is rate = 30. 0 gal ? 1 min ? ?2 ? ? = 7. 14 ? 10 gal s 7. 0 min ? 60 s ? continued on next page http//helpyoustudy. info Introduction 15 (b) 3 1L Note that 1 m 3 = (10 2 cm ) = (10 6 cm 3 ) ? 3 ? 3 ? 10 cm ? = 10 3 L. Thus, ? ? rate = 7. 14 ? 10 ? 2 (c) t= gal ? 3. 786 L ? ? 1 m 3 ? ?4 3 ? ? = 2. 70 ? 10 m s s ? 1 gal ? ? 10 3 L ? ? 1h ? Vfilled 1. 00 m 3 = = 3. 70 ? 10 3 s ? ? = 1. 03 h ? 4 3 rate 2. 70 ? 10 m s ? 3 600 s ? 1. 57 The volume of paint used is given by V = Ah, where A is the area covered and h is the thickness of the layer. Thus, h= V 3. 79 ? 10 ? 3 m 3 = = 1. 52 ? 10 ? 4 m = 152 ? 10 ? 6 m = 152 ? m 25. 0 m 2 A 1. 58 (a) For a sphere, A = 4? R 2 .In this case, the radius of the second sphere is twice that of the ? rst, or R2 = 2 R1. Hence, A2 4? R 2 R 2 ( 2 R1 ) 2 = = 2 = = 4 2 2 A1 4? R 1 R 1 R12 2 (b) For a sphere, the volume is Thus, V= 4 3 ? R 3 3 V2 ( 4 3) ? R 3 R 3 ( 2 R1 ) 2 = = 2 = = 8 3 3 3 V1 ( 4 3) ? R 1 R 1 R1 1. 59 The estimate of the total distance cars are driven each year is d = ( cars in use ) ( distance traveled per car ) = (100 ? 10 6 cars )(10 4 mi car ) = 1 ? 1012 mi At a rate of 20 mi/gal, the fuel used per year would be V1 = d 1 ? 1012 mi = = 5 ? 1010 gal rate1 20 mi gal If the rate increased to 25 mi gal, the one-year fuel consumption would be V2 = d 1 ? 012 mi = = 4 ? 1010 gal rate2 25 mi gal and the fuel savings each year would be savings = V1 ? V2 = 5 ? 1010 gal ? 4 ? 1010 gal = 1 ? 1010 gal 1. 60 (a) The amount paid per year would be sawhorses ? ? 8. 64 ? 10 4 s ? ? 365. 25 days ? 10 dollars annual amount = ? 1 000 ? ? = 3. 16 ? 10 s ? ? 1. 00 day ? ? yr yr ? ? Therefore, it would take (b) 10 ? 10 12 dollars = 3 ? 10 2 yr, 3. 16 ? 10 10 dollars yr or 10 2 yr The circumference of the Earth at the equator is C = 2? r = 2? 6. 378 ? 10 6 m = 4. 007 ? 10 7 m ( ) continued on next page http//helpyoustudy. info 16 Chapter 1 The length of one dollar bill is 0. 55 m, so the length of ten trillion bills is m ? 12 12 = ? 0. 155 ? ? (10 ? 10 dollars ) = 1? 10 m. Thus, the ten trillion dollars would dollar ? ? encircle the Earth 1 ? 1012 m n= = = 2 ? 10 4 , or 10 4 times C 4. 007 ? 10 7 m 1. 61 (a) (b) ? 365. 2 days ? ? 8. 64 ? 10 4 s ? 1 yr = (1 yr ) ? = 3. 16 ? 10 7 s ? ? ? 1 day ? 1 yr ? ? ? Consider a share of the surface of the Moon which has an area of 1 m2 and a depth of 1 m. When ? lled with meteorites, each having a diameter 10? 6 m, the number of meteorites along each edge of this box is n= length of an edge 1m = = 10 6 meteorite diameter 10 ? 6 m The total number of meteorites in the ? led box is then N = n 3 = 10 6 3 = 10 18 At the rate of 1 meteorite per second, the time to ? ll the box is 1y ? = 3 ? 10 10 yr, or t = 1018 s = (1018 s ) ? ? ? 7 ? 3. 16 ? 10 s ? 1. 62 1010 yr ( ) We will assume that, on ave rage, 1 ball will be lost per hitter, that there will be about 10 hitters per inning, a game has 9 innings, and the squad plays 81 home games per indurate. Our estimate of the number of game balls needed per season is then number of balls needed = ( number lost per hitter ) ( number hitters/game )( home games/year ) games ? hitters ? ? innings ? = (1 ball per hitter ) 10 ? 81 ? ? ? year ? inning ? ? game ? = 7300 balls year or 10 4 balls year 1. 63 The volume of the opaque Way galaxy is roughly ? ?d2 ? ? VG = At = ? t ? 10 21 m 4 ? 4 ? ? ( ) (10 m ) 2 19 or VG ? 10 61 m3 r If, within the Milky Way galaxy, there is typically one neutron pinny lead in a spherical volume of radius r = 3 ? 1018 m, then the galactic volume per neutron star is V1 = 3 4 3 4 ? r = ? ( 3 ? 1018 m ) = 1 ? 10 56 m 3 3 3 or V1 ? 10 56 m 3 The order of order of magnitude of the number of neutron stars in the Milky Way is then n= VG 10 61 m 3 ? V1 10 56 m 3 or n ? 10 5 neutron stars http//helpyoustudy. info 2 interrogation in angiotensin-converting enzyme DimensionQUICK QUIZZES 1. 2. (a) (a) 200 yd (b) 0 (c) 0 False. The car may be lessen bulge out, so that the focusing of its acceleration is opposite the burster of its amphetamine. True. If the amphetamine is in the direction chosen as negative, a haughty acceleration causes a cliff in upper. True. For an accelerating particle to forbear at all, the stop number and acceleration must have opposite signs, so that the cannonball along is decreasing. If this is the case, the particle will last come to rest. If the acceleration remains constant, however, the particle must begin to ingrain again, opposite to the direction of its original velocity.If the particle comes to rest and then rest at rest, the acceleration has become zero at the moment the action stops. This is the case for a braking carthe acceleration is negative and goes to zero as the car comes to rest. (b) (c) 3. The velocity-vs. -time interpret (a) h as a constant position, indicating a constant acceleration, which is be by the acceleration-vs. -time graph (e). Graph (b) represents an object whose promote forever increases, and does so at an ever increase rate. Thus, the acceleration must be increasing, and the acceleration-vs. -time graph that trump out indicates this behavior is (d).Graph (c) depicts an object which ? rst has a velocity that increases at a constant rate, which means that the objects acceleration is constant. The motion then changes to one at constant speed, indicating that the acceleration of the object becomes zero. Thus, the best match to this situation is graph (f). 4. prize (b). According to graph b, there are some secondments in time when the object is simultaneously at two different x-coordinates. This is physically impossible. (a) The blue graph of Figure 2. 14b best shows the pucks position as a function of time. As seen in Figure 2. 4a, the distance the puck has traveled grows at an increasing rate for approximately three time intervals, grows at a steady rate for about four time intervals, and then grows at a change magnitude rate for the last two intervals. The red graph of Figure 2. 14c best illustrates the speed (distance traveled per time interval) of the puck as a function of time. It shows the puck gaining speed for approximately three time intervals, wretched at constant speed for about four time intervals, then slowing to rest during the last two intervals. 5. (b) 17 http//helpyoustudy. info 18 Chapter 2 (c) The green graph of Figure 2. 4d best shows the pucks acceleration as a function of time. The puck gains velocity (positive acceleration) for approximately three time intervals, moves at constant velocity (zero acceleration) for about four time intervals, and then loses velocity (negative acceleration) for roughly the last two time intervals. 6. plectron (e). The acceleration of the ball remains constant while it is in the air. The magnitude of its acceler ation is the free-fall acceleration, g = 9. 80 m/s2. Choice (c). As it travels upwardly, its speed decreases by 9. 80 m/s during each second of its motion. When it reaches the extremum of its motion, its speed becomes zero.As the ball moves downwardly, its speed increases by 9. 80 m/s each second. Choices (a) and (f). The ? rst jumper will always be moving with a higher velocity than the second. Thus, in a given time interval, the ? rst jumper covers more distance than the second, and the separation distance between them increases. At any given instant of time, the velocities of the jumpers are de? nitely different, because one had a head dough. In a time interval after(prenominal) this instant, however, each jumper increases his or her velocity by the same amount, because they have the same acceleration. Thus, the difference in velocities stays the same. . 8. ANSWERS TO MULTIPLE CHOICE QUESTIONS 1. Once the arrow has left wing the bow, it has a constant downward acceleration equal to the freefall acceleration, g. Taking upward as the positive direction, the elapsed time required for the velocity to change from an sign value of 15. 0 m s upward ( v0 = +15. 0 m s ) to a value of 8. 00 m s downward ( v f = ? 8. 00 m s ) is given by ? t = ? v v f ? v0 ? 8. 00 m s ? ( +15. 0 m s ) = = = 2. 35 s a ? g ? 9. 80 m s 2 Thus, the correct choice is (d). 2. In Figure MCQ2. 2, there are ? ve spaces separating adjacent oil drops, and these spaces span a distance of ? x = 600 meters.Since the drops derive every 5. 0 s, the time span of each space is 5. 0 s and the total time interval shown in the ? gure is ? t = 5 ( 5. 0 s ) = 25 s. The average speed of the car is then v= ? x 600 m = = 24 m s ? t 25 s making (b) the correct choice. 3. The derivation of the equations of kinematics for an object moving in one dimension (Equations 2. 6 through 2. 10 in the textbook) was based on the assumption that the object had a constant acceleration. Thus, (b) is the correct answer. An object having constant acceleration would have constant velocity only if that acceleration had a value of zero, so (a) is not a prerequisite condition.The speed (magnitude of the velocity) will increase in time only in cases when the velocity is in the same direction as the constant acceleration, so (c) is not a correct response. An object projected straight upward into the air has a constant acceleration. Yet its position (altitude) does not always increase in time (it eventually starts to fall back downward) nor is its velocity always directed downward (the direction of the constant acceleration). Thus, neither (d) nor (e) can be correct. http//helpyoustudy. info Motion in One Dimension 19 4. The bowling pin has a constant downward acceleration ( a = ? g = ? 9. 80 m s 2 ) while in ? ght. The velocity of the pin is directed upward on the upward part of its ? ight and is directed downward as it fall back toward the jugglers hand. Thus, only (d) is a true statement. The sign v elocity of the car is v0 = 0 and the velocity at time t is v. The constant acceleration is therefore given by a = ? v ? t = ( v ? v0 ) t = ( v ? 0 ) t = v t and the average velocity of the car is v = ( v + v0 ) 2 = ( v + 0 ) 2 = v 2. The distance traveled in time t is ? x = vt = vt 2. In the special case where a = 0 ( and hence v = v0 = 0 ) , we see that statements (a), (b), (c), and (d) are all correct. However, in the general case ( a ? , and hence v ? 0 ), only statements (b) and (c) are true. Statement (e) is not true in either case. The motion of the boat is very alike(p) to that of a object propel straight upward into the air. In both cases, the object has a constant acceleration which is directed opposite to the direction of the initial velocity. Just as the object thrown upward slows down and stops momentarily before it starts speeding up as it falls back downward, the boat will continue to move northward for some time, slowing uniformly until it comes to a momentary stop. It will then start to move in the southward direction, gaining speed as it goes.The correct answer is (c). In a position versus time graph, the velocity of the object at any point in time is the slope of the line tangent to the graph at that instant in time. The speed of the particle at this point in time is simply the magnitude (or absolute value) of the velocity at this instant in time. The sack occurring during a time interval is equal to the difference in x-coordinates at the ? nal and initial times of the interval ? x = x t f ? x ti . 5. 6. 7. ( ) The average velocity during a time interval is the slope of the straight line connecting the points on the curve corresponding to the initial and ? al times of the interval ? v = ? x ? t = ( x f ? xi ) ( t f ? ti ) ? . Thus, we see how the quantities in choices (a), (e), (c), and (d) ? ? can all be obtained from the graph. Only the acceleration, choice (b), cannot be obtained from the position versus time graph. 8. From ? x = v0 t + 1 at 2, the distance traveled in time t, kickoff from rest ( v0 = 0 ) with constant 2 acceleration a, is ? x = 1 at 2 . Thus, the ratio of the distances traveled in two man-to-man trials, one 2 of duration t1 = 6 s and the second of duration t 2 = 2 s, is 2 2 ? x2 1 at 2 ? t 2 ? ? 2 s ? 1 2 = 1 2 =? ? =? ? = ? x1 2 at1 ? 1 ? ? 6 s ? 9 and the correct answer is (c). 2 9. The distance an object moving at a uniform speed of v = 8. 5 m s will travel during a time interval of ? t = 1 1 000 s = 1. 0 ? 10 ? 3 s is given by ? x = v ( ? t ) = (8. 5 m s ) (1. 0 ? 10 ? 3 s ) = 8. 5 ? 10 ? 3 m = 8. 5 mm so the only correct answer to this question is choice (d). 10. Once either ball has left the pupils hand, it is a freely falling body with a constant acceleration a = ? g (taking upward as positive). Therefore, choice (e) cannot be true. The initial velocities of the red and blue balls are given by viR = + v0 and viB = ? 0 , respectively. The velocity of either ball when it has a displacemen t from the launch point of ? y = ? h (where h is the height of the building) is found from v 2 = vi2 + 2a ( ? y ) as follows 2 vR = ? viR + 2a ( ? y ) R = ? ( + v0 ) 2 + 2 ( ? g ) ( ? h ) = ? 2 v0 + 2 gh http//helpyoustudy. info 20 Chapter 2 and 2 vB = ? viB + 2a ( ? y ) B = ? ( ? v0 ) 2 + 2 ( ? g ) ( ? h ) = ? 2 v0 + 2 gh Note that the negative sign was chosen for the stem turn in both cases since each ball is moving in the downward direction immediately before it reaches the ground.From this, we see that choice (c) is true. Also, the speeds of the two balls just before hitting the ground are 2 2 2 2 vR = ? v0 + 2 gh = v0 + 2 gh v0 and vB = ? v0 + 2 gh = v0 + 2 gh v0 Therefore, vR = vB , so both choices (a) and (b) are false. However, we see that both ? nal speeds exceed the initial speed and choice (d) is true. The correct answer to this question is then (c) and (d). 11. At ground level, the displacement of the rock from its launch point is ? y = ? h , where h is the 2 height of the tower and upward has been chosen as the positive direction.From v 2 = vo + 2a ( ? y ) , the speed of the rock just before hitting the ground is found to be 2 2 v = v0 + 2a ( ? y ) = v0 + 2 ( ? g ) ( ? h ) = (12 m s )2 + 2 ( 9. 8 m s2 ) ( 40. 0 m ) = 30 m s Choice (b) is therefore the correct response to this question. 12. Once the ball has left the throwers hand, it is a freely falling body with a constant, non-zero, acceleration of a = ? g . Since the acceleration of the ball is not zero at any point on its trajectory, choices (a) through (d) are all false and the correct response is (e). ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS . Yes. The particle may stop at some instant, but still have an acceleration, as when a ball thrown straight up reaches its maximal height. (a) (b) 6. (a) No. They can be used only when the acceleration is constant. Yes. Zero is a constant. In Figure (c), the images are farther apart for each nonparallel time interval. The object is moving t oward the right and speeding up. This means that the acceleration is positive in Figure (c). In Figure (a), the ? rst four images show an increasing distance traveled each time interval and therefore a positive acceleration.However, after the fourth image, the spacing is decreasing, showing that the object is now slowing down (or has negative acceleration). In Figure (b), the images are equally spaced, showing that the object moved the same distance in each time interval. Hence, the velocity is constant in Figure (b). At the maximum height, the ball is momentarily at rest (i. e. , has zero velocity). The acceleration remains constant, with magnitude equal to the free-fall acceleration g and directed downward. Thus, even though the velocity is momentarily zero, it continues to change, and the ball will begin to gain speed in the downward direction.The acceleration of the ball remains constant in magnitude and direction throughout the balls free ? ight, from the instant it leaves the hand until the instant just before it strikes the 4. (b) (c) 8. (a) (b) http//helpyoustudy. info Motion in One Dimension 21 ground. The acceleration is directed downward and has a magnitude equal to the freefall acceleration g. 10. (a) Successive images on the ? lm will be separated by a constant distance if the ball has constant velocity. Starting at the right-most image, the images will be getting finisher together as one moves toward the left.Starting at the right-most image, the images will be getting farther apart as one moves toward the left. As one moves from left to right, the balls will ? rst get farther apart in each successive image, then closer together when the ball begins to slow down. (b) (c) (d) ANSWERS TO EVEN NUMBERED PROBLEMS 2. 4. 6. (a) (a) (a) (d) 8. (a) (d) 10. 12. (a) (a) (d) 14. 16. (a) 2 ? 10 4 mi 10. 04 m s 5. 00 m s ? 3. 33 m s +4. 0 m s 0 2. 3 min L t1 2 L ( t1 + t 2 ) 1. 3 ? 10 2 s (b) 13 m (b) (b) 64 mi ? L t 2 (c) 0 (b) (b) (b) (e) (b) ? x 2 RE = 2. 4 7. 042 m s 1. 25 m s 0 ? 0. 50 m s (c) ? 1. 0 m s (c) ? 2. 50 m s a) The trailing kickoffs speed must be great than that of the leader, and the leaders distance from the ? nish line must be great large to give the trailing runner time to make up the de? cient distance. (b) t = d ( v1 ? v2 ) (c) d2 = v2 d ( v1 ? v2 ) 18. (a) just about data points that can be used to plot the graph are as given below x (m) t (s) (b) (c) 5. 75 1. 00 16. 0 2. 00 35. 3 3. 00 68. 0 4. 00 119 5. 00 192 6. 00 41. 0 m s , 41. 0 m s , 41. 0 m s 17. 0 m s , much smaller than the instantaneous velocity at t = 4. 00 s l http//helpyoustudy. info 22 Chapter 2 20. 22. 24. (a) 20. 0 m s , 5. 00 m s (b) 263 m 0. 91 s (i) (a) (ii) (a) 0 0 (b) (b) 1. 6 m s 2 1. 6 m s 2 500 x (m) (c) (c) 0. 80 m s 2 0 26. The curves intersect at t = 16. 9 s. car patrol officer 250 0 0 4. 00 8. 00 12. 0 16. 0 20. 0 t (s) 28. 30. a = 2. 74 ? 10 5 m s 2 = ( 2. 79 ? 10 4 ) g (a) (b) (e) 32. (a) (d) 34. 36. 38. 40. (a) (a) (a) (a) ( c) 42. 44. 46. 48. 95 m 29. 1 s 1. 79 s v 2 = vi2 + 2a ( ? x ) f 8. 00 s 13. 5 m 22. 5 m 20. 0 s 5. 51 km 107 m v = a1t1 (c) a = ( v 2 ? vi2 ) 2 ( ? x ) f (d) 1. 25 m s 2 (b) 13. 5 m (c) 13. 5 m (b) (b) (b) (b) No, it cannot land safely on the 0. 800 km runway. 20. 8 m s, 41. 6 m s, 20. 8 m s, 38. 7 m s 1. 49 m s 2 ? = 1 a1t12 2 2 ? xtotal = 1 a1t12 + a1t1t 2 + 1 a2 t 2 2 2 (a) Yes. (b) vtop = 3. 69 m s (c) ?v downward = 2. 39 m s (d) No, ? v upward = 3. 71 m s. The two rocks have the same acceleration, but the rock thrown downward has a higher average speed between the two levels, and is accelerated over a smaller time interval. http//helpyoustudy. info Motion in One Dimension 23 50. 52. (a) (a) (c) 21. 1 m s v = ? v0 ? gt = v0 + gt v = v0 ? gt , d = 1 gt 2 2 29. 4 m s ? 202 m s 2 4. 53 s vi = h t + gt 2 (b) (b) 19. 6 m d = 1 gt 2 2 (c) 18. 1 m s, 19. 6 m 54. 56. 58. 60. 62. 64. (a) (a) (a) (a) (b) (b) (b) (b) 44. 1 m 198 m 14. m s v = h t ? gt 2 See Solutions Section for Motion Di agrams. Yes. The minimum acceleration needed to complete the 1 mile distance in the lot time is amin = 0. 032 m s 2 , considerably less than what she is undecided of producing. (a) (c) y1 = h ? v0 t ? 1 gt 2 , y2 = h + v0 t ? 1 gt 2 2 2 2 v1 f = v2 f = ? v0 + 2 gh (d) 66. (b) t 2 ? t1 = 2 v0 g y2 ? y1 = 2 v0 t as long as both balls are still in the air. 68. 70. 3. 10 m s (a) (c) 3. 00 s (b) v0 ,2 = ? 15. 2 m s v1 = ? 31. 4 m s, v2 = ? 34. 8 m s 2. 2 s only if acceleration = 0 (b) (b) ? 21 m s Yes, for all initial velocities and accelerations. 72. 74. (a) (a)PROBLEM SOLUTIONS 2. 1 We assume that you are approximately 2 m elevated and that the nerve impulse travels at uniform speed. The elapsed time is then ? t = 2. 2 (a) 2m ? x = = 2 ? 10 ? 2 s = 0. 02 s v 100 m s At constant speed, c = 3 ? 108 m s, the distance light travels in 0. 1 s is ? x = c ( ? t ) = ( 3 ? 108 m s ) ( 0. 1 s ) ? 1 mi ? ? 1 km ? 4 = ( 3 ? 10 7 m ) ? ? = 2 ? 10 mi 3 ? 1. 609 km ? ? 10 m ? (b) Comparing the re sult of part (a) to the diameter of the Earth, DE, we ? nd 3. 0 ? 10 7 m ? x ? x = = ? 2. 4 DE 2 RE 2 ( 6. 38 ? 10 6 m ) ( with RE = Earths radius ) http//helpyoustudy. info 24 Chapter 2 2. 3Distances traveled between pairs of cities are ? x1 = v1 ( ? t1 ) = (80. 0 km h ) ( 0. 500 h ) = 40. 0 km ? x2 = v2 ( ? t 2 ) = (100 km h ) ( 0. 200 h ) = 20. 0 km ? x3 = v3 ( ? t3 ) = ( 40. 0 km h ) ( 0. 750 h ) = 30. 0 km Thus, the total distance traveled is ? x = ( 40. 0 + 20. 0 + 30. 0 ) km = 90. 0 km, and the elapsed time is ? t = 0. 500 h + 0. 200 h + 0. 750 h + 0. 250 h = 1. 70 h. (a) (b) v= ? x 90. 0 km = = 52. 9 km h ? t 1. 70 h ?x = 90. 0 km (see above) v= v= ? x 2. 000 ? 10 2 m = = 10. 04 m s ? t 19. 92 s 2. 4 (a) (b) 2. 5 (a) ?x 1. 000 mi ? 1. 609 km ? ? 10 3 m ? = ? ? = 7. 042 m s ? t 228. 5 s ? 1 mi ? 1 km ? gravy boat A requires 1. 0 h to cover up the lake and 1. 0 h to return, total time 2. 0 h. Boat B requires 2. 0 h to cross the lake at which time the course is over. Boat A wins, being 60 km ahead of B when the race ends. Average velocity is the net displacement of the boat divided by the total elapsed time. The winning boat is back where it started, its displacement thus being zero, yielding an average velocity of zero . (b) 2. 6 The average velocity over any time interval is ? x x f ? xi = ? t t f ? ti ? x 10. 0 m ? 0 v= = = 5. 00 m s ? t 2. 00 s ? 0 v= (a) (b) (c) (d) (e) v= v= v= v= ? x 5. 00 m ? 0 = = 1. 25 m s ? 4. 00 s ? 0 ? x 5. 00 m ? 10. 0 m = = ? 2. 50 m s ? t 4. 00 s ? 2. 00 s ? x ? 5. 00 m ? 5. 00 m = = ? 3. 33 m s ? t 7. 00 s ? 4. 00 s 0? 0 ? x x2 ? x1 = = = 0 ? t t 2 ? t1 8. 00 s ? 0 2. 7 (a) (b) 1h ? Displacement = ? x = (85. 0 km h ) ( 35. 0 min ) ? ? ? + 130 km = 180 km ? 60. 0 min ? 1h ? The total elapsed time is ? t = ( 35. 0 min + 15. 0 min ) ? ? ? + 2. 00 h = 2. 83 h ? 60. 0 min ? so, v= ? x 180 km = = 63. 6 km h ? t 2. 84 h http//helpyoustudy. info Motion in One Dimension 25 2. 8 The average velocity over any time interval is ? x x f ? xi = ? t t f ? ti ? x 4. 0 m ? 0 v= = = + 4. 0 m s ? t 1. 0 s ? 0 ? ? 2 . 0 m ? 0 v= = = ? 0. 50 m s ? t 4. 0 s ? 0 v= (a) (b) (c) (d) v= v= ? x 0 ? 4. 0 m = = ? 1. 0 m s ? t 5. 0 s ? 1. 0 s ? x 0? 0 = = 0 ? t 5. 0 s ? 0 2. 9 The plane starts from rest ( v0 = 0 ) and maintains a constant acceleration of a = +1. 3 m s 2 . Thus, we ? nd the distance it will travel before reaching the required hoax speed ( v = 75 m s ) , from 2 v 2 = v0 + 2a ( ? x ) , as ? x = 2 v 2 ? v0 ( 75 m s ) ? 0 = = 2. 2 ? 10 3 m = 2. 2 km 2 2a 2 (1. 3 m s ) 2 Since this distance is less than the length of the runway, the plane takes off safely. 2. 10 (a) The time for a car to make the cutting is t = cars to omplete the same 10 mile trip is ? t = t1 ? t 2 = (b) ? x ? x ? 10 mi 10 mi ? ? 60 min ? ? =? ? ? = 2. 3 min v1 v2 ? 55 mi h 70 mi h ? ? 1 h ? ?x . Thus, the difference in the times for the two v When the faster car has a 15. 0 min lead, it is ahead by a distance equal to that traveled by the slow er car in a time of 15. 0 min. This distance is given by ? x1 = v1 ( ? t ) = ( 55 mi h ) (15 min ). The faster car pulls ahead of the slower car at a rate of vrelative = 70 mi h ? 55 mi h = 15 mi h Thus, the time required for it to get distance ? x1 ahead is ? t = ? x1 = vrelative ( 55 mi h ) (15 min ) 15. 0 mi h = 55 minFinally, the distance the faster car has traveled during this time is ? x2 = v2 ( ? t ) = 2. 11 (a) ( 70 mi h ) ( 55 min ) ? ? 1h ? ? = 64 mi ? 60 min ? From v 2 = vi2 + 2a ( ? x ) , with vi = 0 , v f = 72 km h , and ? x = 45 m, the acceleration of the f cheetah is found to be km ? ? 10 3 m ? ? 1 h 72 ? 0 h ? ? 1 km ? ? 3 600 s v 2 ? vi2 f a= = = 4. 4 m s 2 2 ( ? x ) 2 ( 45 m ) continued on next page 2 http//helpyoustudy. info 26 Chapter 2 (b) The cheetahs displacement 3. 5 s after starting from rest is 1 1 2 ? x = vi t + at 2 = 0 + ( 4. 4 m s 2 ) ( 3. 5 s ) = 27 m 2 2 2. 12 (a) (b) (c) (d) 1 = v2 = ( ? x )1 + L = = + L t1 ( ? t )1 t1 ( ? x )2 ? L = = ? L t 2 ( ? t )2 t 2 ( ? x ) total ( ? x )1 + ( ? x )2 + L ? L 0 = = = 0 = t1 + t 2 t1 + t 2 t1 + t 2 ( ? t ) total +L + ? L total distance traveled ( ? x )1 + ( ? x )2 2L = = = ( ave. speed )trip = t1 + t 2 t1 + t 2 t1 + t 2 ( ? t ) total vtotal = The total time for the trip is t total = t1 + 22 . 0 min = t1 + 0. 367 h , where t1 is the time washed-out travel at v1 = 89. 5 km h. Thus, the distance traveled is ? x = v1 t1 = vt total, which gives 2. 13 (a) (89. 5 km h ) t1 = ( 77. 8 km h ) ( t1 + 0. 367 h ) = ( 77. 8 km h ) t1 + 28. 5 km or, (89. 5 km h ? 77. km h ) t1 = 28. 5 km From which, t1 = 2 . 44 h for a total time of t total = t1 + 0. 367 h = 2. 81 h (b) The distance traveled during the trip is ? x = v1 t1 = vt total, giving ? x = v ttotal = ( 77. 8 km h ) ( 2. 81 h ) = 219 km 2. 14 (a) At the end of the race, the tortoise has been moving for time t and the rabbit for a time t ? 2 . 0 min = t ? 120 s. The speed of the tortoise is vt = 0. 100 m s, and the speed of the hare is vh = 20 vt = 2 . 0 m s. The tortoise travels distance xt, which is 0. 20 m larger than the distance xh traveled by the hare. Hence, xt = xh + 0. 20 m which becomes or vt t = vh ( t ? 120 s ) + 0. 0 m ( 0. 100 m s ) t = ( 2 . 0 m s ) ( t ? 120 s ) + 0. 20 m t = 1. 3 ? 10 2 s This gives the time of the race as (b) 2. 15 xt = vt t = ( 0. 100 m s ) (1. 3 ? 10 2 s ) = 13 m The maximum allowed time to complete the trip is t total = total distance 1600 m ? 1 km h ? = ? ? = 23. 0 s required average speed 250 km h ? 0. 278 m s ? The time spent in the ? rst half of the trip is t1 = half distance 800 m ? 1 km h ? = ? ? = 12 . 5 s v1 230 km h ? 0. 278 m s ? continued on next page http//helpyoustudy. info Motion in One Dimension 27 Thus, the maximum time that can be spent on the second half of the trip is t 2 = t total ? 1 = 23. 0 s ? 12 . 5 s = 10. 5 s and the required average speed on the second half is v2 = 2. 16 (a) ? 1 km h ? half distance 800 m = = 76. 2 m s ? ? = 274 km h t2 10. 5 s ? 0. 278 m s ? In order for the trailing jock to be able to catch the leader, his speed (v1) must be greater than that of the leading athletic supporter (v2), and the distance between the leading athlete and the ? nish line must be great enough to give the trailing athlete suf? cient time to make up the de? cient distance, d. During a time t the leading athlete will travel a distance d2 = v2 t and the trailing athlete will travel a distance d1 = v1t .Only when d1 = d2 + d (where d is the initial distance the trailing athlete was behind the leader) will the trailing athlete have caught the leader. Requiring that this condition be satis? ed gives the elapsed time required for the second athlete to overstep the ? rst d1 = d2 + d giving or v1t = v2 t + d or t = d ( v1 ? v2 ) (b) v1t ? v2 t = d (c) In order for the trailing athlete to be able to at least tie for ? rst place, the initial distance D between the leader and the ? nish line must be greater than or equal to the distance the leader c an travel in the time t calculated above (i. e. , the time required to get over the leader).That is, we must require that D ? d2 = v2 t = v2 ? d ( v1 ? v2 ) ? ? ? or D? v2 d v1 ? v2 2. 17 The instantaneous velocity at any time is the slope of the x vs. t graph at that time. We compute this slope by using two points on a straight segment of the curve, one point on each side of the point of interest. (a) (b) (c) (d) vt=1. 00 s = vt=3. 00 s = 10. 0 m ? 0 = 5. 00 m s 2 . 00 s ? 0 ( 5. 00 ? 10. 0 ) m = ? 2 . 50 m s ( 4. 00 ? 2 . 00 ) s ( 5. 00 ? 5. 00 ) m vt=4. 50 s = = 0 ( 5. 00 ? 4. 00 ) s 0 ? ( ? 5. 00 m ) vt=7. 50 s = = 5. 00 m s (8. 00 ? 7. 00 ) s http//helpyoustudy. info 28 Chapter 2 2. 18