Halliday - Fun...h) e solutions - halliday&resnick fundamphys 9th solman, Notas de estudo de Engenharia Elétrica

Baixe Halliday - Fun...h) e solutions - halliday&resnick fundamphys 9th solman e outras Notas de estudo em PDF para Engenharia Elétrica, somente na Docsity! www. EL L http://www .elsolucionario.blogspot.com IONARIO. LIBROS UNIVERISTARIOS Y SOLUCIONARIOS DE MUCHOS DE ESTOS LIBROS LOS SOLUCIONARIOS CONTIENEN TODOS LOS EJERCICIOS DEL LIBRO RESUELTOS Y EXPLICADOS DE FORMA CLARA VISITANOS PARA DESARGALOS GRATIS. Chapter 1 1. Various geometric formulas are given in Appendix E. (a) Expressing the radius of the Earth as R=(637x10m)(10ºkm/m) = 6.37 x 10º km, its circumference is s= 27R =27(6.37 x 10º km) = 4.00x10º km. (b) The surface area of Earth isA= 41Rº = 41 (6.37 x 10º km =5.10x 10º km”. . 47 3 47 3 3 12 3 (c) The volume of Earthis V= 985 (6.37x10' km) =1.08x 10º km”. 2. The conversion factors are: 1 gry = 1/10 line, 1 line=1/12 inch and 1 point = 1/72 inch. The factors imply that 1 gry = (1/10)(1/12)72 points) = 0.60 point. Thus, 1 gry? = (0.60 point)? = 0.36 point?, which means that 0.50 gry?= 0.18 point”. 3. The metric prefixes (micro, pico, nano, ...) are given for ready reference on the inside front cover of the textbook (see also Table 1-2). (a) Since 1 km= 1x 10 mand Im= 1x 10º um, lkm= 10 m= (10º m)(10º um/m) =10 um. The given measurement is 1.0 km (two significant figures), which implies our result should be written as 1.0 x 10º um. (b) We calculate the number of microns in | centimeter. Since | cm = 102 m, lem = 102m = (102m)(10º umfm) = 10º um. We conclude that the fraction of one centimeter equal to 1.0 um is 1.0 x 10. (c) Since 1 yd = (3 1t(0.3048 m/ft) = 0.9144 m, 1 www.ELSOLUCIONARIO.bl .com 2 CHAPTER 1 L0yd = (0.9tm)(10º um/m) =9.1x 10º um. 4. (a) Using the conversion factors 1 inch = 2.54 cm exactly and 6 picas = 1 inch, we obtain 0.80 cm= (0.80 cm) linch | 6 Pois | 1.9 picas. 2.54 em À linch (b) With 12 points = 1 pica, we have 0.80 em = (0.80 cm) | -Jinch |) 6 picas |/ 12 points | (o js, 2.54 cm )l 1inch 1 pica 5. Given that 1 furlong = 201.168 m, Irod=5.0292 m and 1 chain = 20.117 m, we find the relevant conversion factors to be 1rod 1.0 furlong = 201.168 m = (201.168 1) ——— = 40 rods, 8 é 2 so292 and 1 chain 1.0 furlong =201.168 m = (201.168 nf) ——— =10 chains. uuons ( 20 sorr7 ae TO chai Note the cancellation of m (meters), the unwanted unit. Using the given conversion factors, we find (a) the distance d in rods to be d = 4.0 furlongs =(4.0 fuslongs) 40208 - 160 rods, 1 fufong (b) and that distance in chains to be d = 4.0 furlongs = (4.0 fuflongs O hains 40 chains. 1 fuflong 6. We make use of Table 1-6. (a) We look at the first (“cahiz”) column: 1 fanega is equivalent to what amount of cahiz? We note from the already completed part of the table that 1 cahiz equals a dozen fanega. Thus, 1 fanega =5 cahiz, or 8.33 x 10? cahiz. Similarly, “1 cahiz = 48 cuartilla” (in the already completed part) implies that 1 cuartilla = = cahiz, or 2.08 x 10? cahiz. Continuing in this way, the remaining entries in the first column are 6.94 x 10 and 3.47x10º. www.ELSOLUCIONARIO.bl .com (b) In the second (“fanega”) column, we find 0.250, 8.33 x 102, and 4.17 x 10? for the last three entries. (c) In the third (“cuartilla”) column, we obtain 0.333 and 0.167 for the last two entries. (d) Finally, in the fourth (“almude”) column, we get 3 = 0.500 for the last entry. (e) Since the conversion table indicates that 1 almude is equivalent to 2 medios, our amount of 7.00 almudes must be equal to 14.0 medios. (f) Using the value (1 almude = 6.94 x 10º cahiz) found in part (a), we conclude that 7.00 almudes is equivalent to 4.86 x 107? cahiz. (g) Since each decimeter is 0.1 meter, then 55.501 cubic decimeters is equal to 0.055501 mº 0r 55501 em”, Thus, 7.00 almudes = 19º fanega = 10º (55501 cm?) = 3.24 x 10º cm”, 7. We use the conversion factors found in Appendix D. 1 acre ft = (43,560 ft?) -ft = 43,560 ft Since 2 in. = (1/6) ft, the volume of water that fell during the storm is V=(26 km?)(1/6 ft) = (26 km?)(3281ft/km) (1/6 ft) = 4.66x10" fé. Thus, 7 3 = — 466 x10 ft = 11x 10? acre: ft. 4.3560 x 10! ft” /acre- ft 8. From Fig. 1-4, we see that 212 S is equivalent to 258 W and 212 - 32 = 1805S is equivalent to 216 — 60 = 156 Z. The information allows us to convert S to W or Z. (a) In units of W, we have 50.08 = (50058) (E )- 60.8 W (b) In units of Z, we have 500 8= (5008) 62]. 4337 1808 9. The volume of ice is given by the product of the semicircular surface area and the thickness. The area of the semicircle is A = 17º/2, where r is the radius. Therefore, the www.ELSOLUCIONARIO.bl .com 6 CHAPTER 1 which yields the result 1 = 1557.80644887275 s (though students who do this calculation on their calculator might not obtain those last several digits). (c) Careful reading of the problem shows that the time-uncertainty per revolution is +3x10"s . We therefore expect that as a result of one million revolutions, the uncertainty should be (+3x10")(1x10%)= +3x 10"! s. 17. None of the clocks advance by exactly 24 h in a 24-h period but this is not the most important criterion for judging their quality for measuring time intervals. What is important is that the clock advance by the same amount in each 24-h period. The clock reading can then easily be adjusted to give the correct interval. If the clock reading jumps around from one 24-h period to another, it cannot be corrected since it would impossible to tell what the correction should be. The following gives the corrections (in seconds) that must be applied to the reading on each clock for each 24-h period. The entries were determined by subtracting the clock reading at the end of the interval from the clock reading at the beginning. CLOCK Sun. Mon. Tues. Wed. Thurs. | Fri. -Mon. -Tues. -Wed. -Thurs. -Fhi. -Sat. A -16 -16 -15 —17 -15 -15 B =3 +5 -10 +5 +6 = c -58 -58 -58 -58 -58 -58 D +67 +67 +67 +67 +67 +67 E +70 +55 +2 +20 +10 +10 Clocks C and D are both good timekeepers in the sense that each is consistent in its daily drift (relative to WWF time); thus, C and D are easily made “perfect” with simple and predictable corrections. The correction for clock C is less than the correction for clock D, so we judge clock C to be the best and clock D to be the next best. The correction that must be applied to clock A is in the range from 15 s to 17s. For clock B it is the range from -5 s to +10s, for clock E it is in the range from -70s to -2.s. After C and D, A has the smallest range of correction, B has the next smallest range, and E has the greatest range. From best to worst, the ranking of the clocks is C, D, A, B, E. 18. The last day of the 20 centuries is longer than the first day by (20 century) (0.001 s/century) = 0.02 s. The average day during the 20 centuries is (O + 0.02)/2 = 0.01 s longer than the first day. Since the increase occurs uniformly, the cumulative effect T is www.ELSOLUCIONARIO.bl .com T = (average increase in length of a day)(number of days) o. (e ] (= day day y = 7305 s )eoo y) or roughly two hours. 19. When the Sun first disappears while lying down, your line of sight to the top of the Sun is tangent to the Earth's surface at point 4 shown in the figure. As you stand, elevating your eyes by a height h, the line of sight to the Sun is tangent to the Earth's surface at point B. Line of sight to ff top of he Sun First su Discame Sum / Second sunset Center of Earih Let d be the distance from point B to your eyes. From the Pythagorean theorem, we have P+r=(r4+n="+2rh+h or d?=2rh+h?, where r is the radius of the Earth. Since r >> A, the second term can be dropped, leading to d? = 2rh. Now the angle between the two radii to the two tangent points A and B is 6, which is also the angle through which the Sun moves about Earth during the time interval t= 11.1 s. The value of 9 can be obtained by using 0 t 360º 24h' This yields (360)(L 1.18) =" CTADO O -qo462sº. (24 h)(60 min/h)(60 s/min) Using d=rtan0, wehave dº =rº tan?0=2rh,or 2h r=—— tan“ 0 Using the above value for Oand h = 1.7 m, we have r=5.2x 10º m. www.ELSOLUCIONARIO.bl .com 8 CHAPTER 1 20. (a) We find the volume in cubic centimeters 3 - =731x10 cm lin ss . 193 gal = (193 gal) (Si )(pem lgal and subtract this from 1 x 10 cm? to obtain 2.69 x 10º cm”. The conversion gal > in” is given in Appendix D (immediately below the table of Volume conversions). (b) The volume found in part (a) is converted (by dividing by (100 cm/m)) to 0.731 mê, which corresponds to a mass of (1000 kg/m) (0.731 m?)= 731 kg using the density given in the problem statement. At a rate of 0.0018 kg/min, this can be filled in — Blke 4 06x10min=0.77 y 0.0018 kg/min after dividing by the number of minutes in a year (365 days)(24 h/day) (60 min/h). 21. If Mg is the mass of Earth, m is the average mass of an atom in Earth, and N is the number of atoms, then Mg = Nm or N = Mz/m. We convert mass m to kilograms using Appendix D (1 u= 1.661 x 107 kg). Thus, no Me 5.98 x 10! kg —— DE =90x10º. m (40 u)(1661x 107 kg/u) í 22. The density of gold is 0328 1932 gem. lem =. ? V (a) We take the volume of the leaf to be its area A multiplied by its thickness z. With density 0 = 19.32 g/cm? and mass m = 27.63 g, the volume of the leaf is found to be v=2=1430em'. o We convert the volume to SI units: www.ELSOLUCIONARIO.bl .com q 3 lkg 100 cm =(7.87 g/em')|) —E- || | =7870kg'm'. º ( e! Moo) Im “ If we ignore the empty spaces between the close-packed spheres, then the density of an individual iron atom will be the same as the density of any iron sample. That is, if M is the mass and Vis the volume of an atom, then -26 Mo 90 ke gg mp), Pp 7.87x10 kg/m (b) We setV= 4nR'/3, where R is the radius of an atom (Appendix E contains several geometry formulas). Solving for R, we find 1/3 B (3(LISxI0? m R= (4) = 3(1.18x 10? m?) =141x 10º m. 41 41 The center-to-center distance between atoms is twice the radius, or 2.82 x 101º m. 28. If we estimate the “typical” large domestic cat mass as 10 kg, and the “typical” atom (in the cat) as 10u x 2 x 10% kg, then there are roughly (10 kg)/( 2 x 107% kg) = 5 x 10% atoms. This is close to being a factor of a thousand greater than Avogadro”s number. Thus this is roughly a kilomole of atoms. 29. The mass in kilograms is (289 piculs) (ep Nei 1octee o pen O sTToE Ipicul Igin 1Itahil 1 chee lhoon which yields 1.747 x 10º g or roughly 1.75x 10º kg. 30. To solve the problem, we note that the first derivative of the function with respect to time gives the rate. Setting the rate to zero gives the time at which an extreme value of the variable mass occurs; here that extreme value is a maximum. (a) Differentiating m(t) = 5.001º* — 3.00 + 20.00 with respect to 1 gives dm 4002 -3.00. dt The water mass is the greatest when dm/dt=0, orat t= (4.00/3.00)"2 = 4.21. www.ELSOLUCIONARIO.bl .com 12 CHAPTER 1 (b) Att=4.21s, the water mass is m(t = 4.218) =5.00(4.21)º* — 3.00(4.21) +20.00= 23.2 g. (c) The rate of mass change at 1 = 2.005 is =[4.00(2.00)*º 3.00] g/s= 0.48 g/s = o4g&. Ike 60s 1=2.008 s 1000g I min =2.89x 10º kg/min. dm dt (d) Similarly, the rate of mass change at += 5.005 is dm 1 =[4.00(5.007? = 3.00 | g/s= 0.101 g/s= -0.1018.— 8. t 1=200s s 1000g Imin =-6.05x 10º kg/min. 31. The mass density of the candy is m 00200g | 500 ma” 4.00x10 g/mm” = 4.00x 10º kg/cm”. .O mm If we neglect the volume of the empty spaces between the candies, then the total mass of the candies in the container when filled to height h is M= Ah, where A= (14.0 cm)(17.0 cm) = 238 cm? is the base area of the container that remains unchanged. Thus, the rate of mass change is given by a = aten) = pat = (4.00x 10* kg/em*)(238 cm?)(0.250 cm/s) r r r = 0.0238 kg/s = 1.43 kg/min. 32. The total volume V of the real house is that of a triangular prism (of height A = 3.0 m and base area A = 20 x 12 = 240 m?) in addition to a rectangular box (height hº = 6.0 m and same base). Therefore, v= Sua ena-[S 0 )A-1800m (a) Each dimension is reduced by a factor of 1/12, and we find 3 Vin = (1800 m?) (5) 2 10 mº. www.ELSOLUCIONARIO.bl .com 13 (b) In this case, each dimension (relative to the real house) is reduced by a factor of 1/144. Therefore, V miniature 3 = (1800 m?) 7] = 60x 10º m'. 33. In this problem we are asked to differentiate between three types of tons: displacement ton, freight ton and register ton, all of which are units of volume. The three different tons are given in terms of barrel bulk, with 1 barrel bulk = 0.1415 m” = 4.0155 U.S. bushels using 1 mº = 28.378 U.S. bushels. Thus, in terms of U.S. bushels, we have 4.0155 U.S. bushels 1 displacement ton = (7 barrels bulk) d ) = 28.108 U.S. bushels 1 barrel bulk 1 freight ton = (8 barrels bulk) x (Pora d moteis) =32.124 U.S. bushels 1 barrel bulk 1 register ton = (20 barrels bulk) x 4.0155 U.S. bushels = 80.31 U.S. bushels 1 barrel bulk (a) The difference between 73 “freight” tons and 73 “displacement” tons is AV = 73(freight tons — displacement tons) = 73(32.124 U.S. bushels — 28.108 U.S. bushels) = 293.168 U.S. bushels x 293 U.S. bushels (b) Similarly, the difference between 73 “register” tons and 73 “displacement” tons is AV = 73(register tons — displacement tons) = 73(80.31 U.S. bushels — 28.108 U.S. bushels) = 3810.746 U.S. bushels x 3.81x 10” U.S. bushels 34. The customer expects a volume V, = 20 x 7056 in? and receives V> = 20 x 5826 inº, the difference being AV =V,-V, = 24600 inº, or 3 AV = (24600 in) | 2Stem | [IL linch 1000 cm ) = 403L where Appendix D has been used. 35. The first two conversions are easy enough that a formal conversion is not especially called for, but in the interest of practice makes perfect we go ahead and proceed formally: www.ELSOLUCIONARIO.bl .com 16 CHAPTER 1 This means that the driver mistakenly believes that the car should need 18.8 U.S. gallons. (b) Using the conversion factor found above, the actual amount required is equivalent to 1.20095 U.S. gallons 1U.K. gallon 40. Equation 1-9 gives (to very high precision!) the conversion from atomic mass units to kilograms. Since this problem deals with the ratio of total mass (1.0 kg) divided by the mass of one atom (1.0 u, but converted to kilograms), then the computation reduces to simply taking the reciprocal of the number given in Eq. 1-9 and rounding off appropriately. Thus, the answer is 6.0 x 10%, V'=(18.75 U.K. gutons( | 22.5 U.S. gallons. 41. Using the (exact) conversion | in = 2.54 cm = 0.0254 m, we find that = (0.3048 m tft=12in.=(12in.)x 9.0254m lin and 1 ft? = (0.3048 m)” = 0.0283 m? for volume (these results also can be found in Appendix D). Thus, the volume of a cord of wood is V = (8 f)x(4 f)x(4 ft)=128 fé. Using the conversion factor found above, we obtain 0.0283 mº fé V=Icord= 128 ft = (128 nl Jrases mº which implies that Im” = ! cord = 0.276 cord x 0.3 cord. 3.625 42. (a) In atomic mass units, the mass of one molecule is (16 + 1 + 1)u = 18 u. Using Eg. 1-9, we find 1.6605402 x 107 kg lu I8u= (su) [ =30x10%kg. (b) We divide the total mass by the mass of each molecule and obtain the (approximate) number of water molecules: 21 nadéxiOo sx 1g!6 3.0x 10? 43. A million milligrams comprise a kilogram, so 2.3 kg/week is 2.3 x 10º mg/week. Figuring 7 days a week, 24 hours per day, 3600 second per hour, we find 604800 seconds are equivalent to one week. Thus, (2.3 x 109 mg/week)/(604800 s/week) = 3.8 mg/s. 44. The volume of the water that fell is www.ELSOLUCIONARIO.bl .com 17 V= (26 km?) (20 in)= (26 km?) (1900 m) (2.0 in.) (eram) = (26x 10º m?) (0.0508 m) =13x10º mº. We write the mass-per-unit-volume (density) of the water as: p= = i0 kg/m. The mass of the water that fell is therefore given by m = pV: m= (1x 10" kg/m”) (1.3x 10º m')=1.3x10º kg. 45. The number of seconds in a year is 3.156 x 10”. This is listed in Appendix D and results from the product (365.25 day/y) (24 h/day) (60 min/h) (60 s/min). (a) The number of shakes in a second is 10º; therefore, there are indeed more shakes per second than there are seconds per year. (b) Denoting the age of the universe as 1 u-day (or 86400 u-sec), then the time during which humans have existed is given by 10º - 197 = 104 u-day, 86400 u - sec =86u-sec. lu-day which may also be expressed as (10 u- day) 46. The volume removed in one year is V=(75x10'm)(26m) 22x 10 m 1km 1000 m 3 which we convert to cubic kilometers: V = (2 x 107 m') = 0020 km”. 47. We convert meters to astronomical units, and seconds to minutes, using www.ELSOLUCIONARIO.bl .com 18 CHAPTER 1 1000 m = 1 km LAU =1.50 x 10º km 60s =Imin. Thus, 3.0 x 10º m/s becomes post 1 km oii) [65] -02 AU/min. s 1000 m ) (1.50 x 10º km ) min 48. Since one atomic mass unit is Lu = 1.66x10* g (see Appendix D), the mass of one mole of atoms is about m = (1.66x10* g)(6.02x 10”) = Ig. On the other hand, the mass of one mole of atoms in the common Eastern mole is Therefore, in atomic mass units, the average mass of one atom in the common Eastern mole is m. Joe 66:10 g=10u. N, 6.02x10º A 49. (a) Squaring the relation 1 ken = 1.97 m, and setting up the ratio, we obtain lken? 197º mº - — = 3,88. Im” Im (b) Similarly, we find 3 3 3 1 ken - 1.97 m -765. Im Im (c) The volume of a cylinder is the circular area of its base multiplied by its height. Thus, ar h=7(3.00) (5.50) = 156 ken'. (d) If we multiply this by the result of part (b), we determine the volume in cubic meters: (155.5)(7.65) = 1.19 x 10 w?. 50. According to Appendix D, a nautical mile is 1.852 km, so 24.5 nautical miles would be 45.374 km. Also, according to Appendix D, a mile is 1.609 km, so 24.5 miles is 39.4205 km. The difference is 5.95 km. 51. (a) For the minimum (43 cm) case, 9 cubits converts as follows: www.ELSOLUCIONARIO.bl .com Chapter 2 1. The speed (assumed constant) is v = (90 km/h)(1000 m/km) /(3600 s/h) = 25 m/s. Thus, in 0.50 s, the car travels a distance d = vt = (25 m/s)(0.50 s) = 13 m. 2. (a) Using the fact that time = distance/velocity while the velocity is constant, we find 73.2 m+73.2m ae O Bm , 32m =1.74 mis. 122m/s * 3.05m (b) Using the fact that distance = vt while the velocity v is constant, we find — (122 m/s)(608) +(305 m/s)(60 8) ve 1205 =214 m/s. (c) The graphs are shown below (with meters and seconds understood). The first consists of two (solid) line segments, the first having a slope of 1.22 and the second having a slope of 3.05. The slope of the dashed line represents the average velocity (in both graphs). The second graph also consists of two (solid) line segments, having the same slopes as before — the main difference (compared to the first graph) being that the stage involving higher-speed motion lasts much longer. 3. Since the trip consists of two parts, let the displacements during first and second parts of the motion be Ax; and Ax;, and the corresponding time intervals be At, and At», respectively. Now, because the problem is one-dimensional and both displacements are in the same direction, the total displacement is Ax = Ax, + Ax», and the total time for the trip is At= At, + Afp. Using the definition of average velocity given in Eg. 2-2, we have da AntAm Vive — = . WE Ar Aj+AL To find the average speed, we note that during a time At if the velocity remains a positive constant, then the speed is equal to the magnitude of velocity, and the distance is equal to the magnitude of displacement, with d =| Ax|=vAt. 2 www.ELSOLUCIONARIO.bl .com 22 CHAPTER 2 (a) During the first part of the motion, the displacement is Ax, = 40 km and the time interval is = Ok) (33h (30 km/h) Similarly, during the second part the displacement is Ax, = 40 km and the time interval is 1-4 ogyn * (60km/h) The total displacement is Ax = Ax, + Axo = 40 km + 40 km = 80 km, and the total time elapsed is At = Ati + At, = 2.00 h. Consequently, the average velocity is v.= AX 60h 49 ymyh. Co A (20h) (b) In this case, the average speed is the same as the magnitude of the average velocity: s,, =40 km/h. (c) The graph of the entire trip is shown below; it consists of two contiguous line segments, the first having a slope of 30 km/h and comecting the origin to (Ati, Axi) = (1.33 h, 40 km) and the second having a slope of 60 km/h and connecting (Ati, Axi) to (Aí, Ax) = (2.00 h, 80 km). Ax (km) (20h, 80 km) At(h) 4. Average speed, as opposed to average velocity, relates to the total distance, as opposed to the net displacement. The distance D up the hill is, of course, the same as the distance down the hill, and since the speed is constant (during each stage of the motion) we have speed = D/t. Thus, the average speed is Dip + Dao 2D tpm DD Vup Vaown which, after canceling D and plugging in vp = 40 km/h and vaown = 60 km/h, yields 48 km/h for the average speed. 5. Using x = 3t — 4? + É with SI units understood is efficient (and is the approach we www.ELSOLUCIONARIO.bl .com 23 will use), but if we wished to make the units explicit we would write x=( m/st- (4 m/s)P + (m/s) We will quote our answers to one or two significant figures, and not try to follow the significant figure rules rigorously. (a) Pluggingint= | syieldsx=3-4+1=0. (b) Witht=2s we getx = 3(2)- 42) +(2)'= 2 m. (c) Witht=3s we havex=0m. (d) Plugging int=4sgivesx=12m. For later reference, we also note that the position att = 0 is x= 0. (e) The position at t = O is subtracted from the position at t = 4's to find the displacement Ax = 12 m. (f) The position at t = 2s is subtracted from the position at t = 4s to give the displacement Ax = 14 m. Eq. 2-2, then, leads to Ax lám o Vs SS 7 m/s. Fo A 2s (g) The position of the object for the interval O <t <4 is plotted below. The straight line drawn from the point at (1, x) = (2 s,-2 m)to (45, 12 m) would represent the average velocity, answer for part (f). x(m) (45, 12m) 6. Huber's speed is vo = (200 m)/(6.509 s) =30.72 m/s = 110.6 km/h, where we have used the conversion factor 1 m/s = 3.6 km/h. Since Whittingham beat Huber by 19.0 km/h, his speed is v; = (110.6 km/h + 19.0 km/h) = 129.6 km/h, or 36 m/s (1 km/h = 0.2778 m/s). Thus, using Eq. 2-2, the time through a distance of 200 m for Whittingham is www.ELSOLUCIONARIO.bl .com 26 CHAPTER 2 (c) Since x> L, the direction of the shock wave is downstream. 13. (a) Denoting the travel time and distance from San Antonio to Houston as T and D, respectively, the average speed is DOS km +00 km XT!) o am Savgi = T T which should be rounded to 73 km/h. (b) Using the fact that time = distance/speed while the speed is constant, we find D D Sue => Dis Di” 683 kmlh 55 km/h * 90 km/h which should be rounded to 68 km/h. (c) The total distance traveled (2D) must not be confused with the net displacement (zero). We obtain for the two-way trip s, «Pk. “ue Do, Do 72.5 kmlh * 68.3 km/h (d) Since the net displacement vanishes, the average velocity for the trip in its entirety is zero. (e) In asking for a sketch, the problem is allowing the student to arbitrarily set the distance D (the intent is not to make the student go to an atlas to look it up); the student can just as easily arbitrarily set T instead of D, as will be clear in the following discussion. We briefly describe the graph (with kilometers-per-hour understood for the slopes): two contiguous line segments, the first having a slope of 55 and comnecting the origin to (t,, x1) = (T/2, 55T/2) and the second having a slope of 90 and comecting (t,, x1) to (T; D) where D = (55 + 90)T/2. The average velocity, from the graphical point of view, is the slope of a line drawn from the origin to (T; D). The graph (not drawn to scale) is depicted below: x www.ELSOLUCIONARIO.bl .com 27 14. Using the general property Lexp(bx) = bexp(bx), we write de. (to), e ra (4) v=>—D = dt dt dt If a concern develops about the appearance of an argument of the exponential (—t) apparently having units, then an explicit factor of 1/T where T = 1 second can be inserted and carried through the computation (which does not change our answer). The result of this differentiation is v=I6(1- De” with t and v in SI units (s and m/s, respectively). We see that this function is zero when t = 1 s. Now that we know when it stops, we find out where it stops by plugging our result t = 1 into the given function x = 16te” with x in meters. Therefore, we find x = 5.9 m. 15. We use Eq. 2-4 to solve the problem. (a) The velocity of the particle is v=E.4 (4124431) =—12+6r. dt dt Thus, att = 1s, the velocity is v = (-12 + (6)(1)) =—6 m/s. (b) Since v < 0, itis moving in the —x direction att= 1 s. (c)Att=1s, the speed is |V = 6 m/s. (d) For0<t<2s, |v| decreases until it vanishes. For 2 < t<3s, |v| increases from zero to the value it had in part (c). Then, |v is larger than that value for t>3s. (e) Yes, since v smoothly changes from negative values (consider the t = 1 result) to positive (note that as t > + 00, we have v —> + 09). One can check that v = O when t=2s. (f) No. In fact, from v = —12 + 6t, we know thatv> O fort>2s. 16. We use the functional notation x(t), v(t), and a(?) in this solution, where the latter two quantities are obtained by differentiation: dx(t) o dv(t — =-12t and a(t)= MV ). dt t ——+= —12 dt vlt)= with SI units understood. (a) From v(t) = O we find it is (momentarily) at rest att = 0. www.ELSOLUCIONARIO.bl .com 28 CHAPTER 2 (b) We obtain x(0) = 4.0 m. (c) and (d) Requiring x(t) = O in the expression x(t) = 4.0 — 6.0 leads to t= +0.82.s for the times when the particle can be found passing through the origin. (e) We show both the asked-for graph (on the left) as well as the “shifted” graph that is relevant to part (f). In both cases, the time axis is given by -3 <t <3 (SI units understood). x x 100: 100: (f) We arrived at the graph on the right (shown above) by adding 20t to the x(t) expression. (g) Examining where the slopes of the graphs become zero, it is clear that the shift causes the v = O point to correspond to a larger value of x (the top of the second curve shown in part (e) is higher than that of the first). 17. We use Eq. 2-2 for average velocity and Eq. 2-4 for instantaneous velocity, and work with distances in centimeters and times in seconds. (a) We plug into the given equation for x for t = 2.00 s and t = 3.00 s and obtain x, = 21.75 cm and x; = 50.25 cm, respectively. The average velocity during the time interval 2.00 < t < 3.00s is vo = Ax — 50.25 cm— 21.75 em “EA 3005-2005 which yields vavg = 28.5 cm/s. (b) The instantaneous velocity is v== 45t, which, at time t = 2.005, yields v = (4.5)(2.00) = 18.0 cm/s. (c) Att =3.00s, the instantaneous velocity is v = (4.513.002 =40.5 cm/s. (d) Att = 2.50 s, the instantaneous velocity is v = (4.5)(2.502 = 28.1 cm/s. (e) Let tm stand for the moment when the particle is midway between x» and x; (that is, when the particle is at xy = (x2 + x3)/2 = 36 cm). Therefore, x,=975 + 151, > t,=259%6 in seconds. Thus, the instantaneous speed at this time is v = 4.5(2.596)? = 30.3 cm/s. www.ELSOLUCIONARIO.bl .com 31 direction of motion during 5 min < t < 10 min is taken to be the positive x direction. We also use the fact that Ax=vAt' when the velocity is constant during a time interval Af”. (a) The entire interval considered is At = 8 —- 2 = 6 min, which is equivalent to 360 s, whereas the sub-interval in which he is moving is only At' =8-5=3min=180s. His position at t = 2 minis x = O and his position at t = 8 minis x=vAf'= (2.2X180) = 396m.. Therefore, - 396m-0 Vae = =110m/s. ne 360 s (b) The man is at rest at t = 2 min and has velocity v = +2.2 m/s at t= 8 min. Thus, keeping the answer to 3 significant figures, —22m/s-0 as =00061 m/s”. e 360 s (c) Now, the entire interval considered is At = 9 - 3 = 6 min (360 s again), whereas the sub-interval in which he is moving is Af=9-5=4min=2405s). His position at t=3 minis x = O and his position at t = 9 min isx=vAf' =(2.2)(240)=528 m. Therefore, -528m-0 Vae = =147 m/s. ne 360 s (d) The man is at rest at t = 3 min and has velocity v = +2.2 m/s at t = 9 min. Consequently, dave = 2.2/360 = 0.00611 m/s? just as in part (b). (e) The horizontal line near the bottom of this x-vs-t graph represents the man standing at x = O for O <t < 300 s and the linearly rising line for 300 < t < 600 s represents his constant-velocity motion. The lines represent the answers to part (a) and (c) in the sense that their slopes yield those results. x (e) (a) t o 500 The graph of v-vs-t is not shown here, but would consist of two horizontal “steps” (one atv=0Ofor0<t<300s and the next at v = 2.2 m/s for 300 < + < 600 s). The indications of the average accelerations found in parts (b) and (d) would be dotted lines connecting the “steps” at the appropriate t values (the slopes of the dotted lines representing the values of dave). www.ELSOLUCIONARIO.bl .com 32 CHAPTER 2 22. In this solution, we make use of the notation x(t) for the value of x at a particular t. The notations v(t) and a(t) have similar meanings. (a) Since the unit of cf is that of length, the unit of c must be that of length/time?, or m/s? in the SI system. (b) Since bê has a unit of length, b must have a unit of length/time?, or m/s). (c) When the particle reaches its maximum (or its minimum) coordinate its velocity is zero. Since the velocity is given by v = dx/dt = 2ct — 3b?, v = O occurs for t = O and for pn Ze 2GOm/S) ph “3% M20m/S) Fort=0,x=x,=0andfort=1.0s,x=1.0m> xo. Since we seek the maximum, we reject the first root (t = 0) and accept the second (t = 1s). (d) In the first 4 s the particle moves from the origin to x = 1.0 m, turns around, and goes back to x4s)=(30m/sX40s) — (20m/s'X40s)'=-80m. The total path length it travels is 10m + 1.0m +80m=82m. (e) Its displacement is Ax = x2 — x1, where x, = O and x, = -80 m. Thus, Ar=-80 m. The velocity is given by v = 2ct — 3b? = (6.0 m/s?)t — (6.0 m/s*)º. (f) Plugging in t= 1 s, we obtain v(L s)= (6.0 m/s (1.0 s) — (6.0 m/s (1.0 s? =0. (g) Similarly, v(2 s)= (6.0 m/S”)(2.0 s)— (6.0 m/s)(2.0s = —12m/s. (bh) v(38)=(6.0 m/s*)(3.0 s)—(6.0 m/s')(3.0 8) = —36 m/s. (1) (48) =(60 m/s” Y40 s)-(6.0m/s*X4.0 5) =-72 m/s . The acceleration is given by a = dv/dt = 2c — 6b = 6.0 m/s? — (12.0 m/S*t. (7) Pluggingint=1s, we obtain a(ls)=6.0 m/s” — (12.0 m/S)(L.O s)= —6.0 m/s”. (9) a(29)=6.0 M/S? — (12.0 m/SY2.0 s)= —18 m/s”. www.ELSOLUCIONARIO.bl .com 33 (D a(3 9)=6.0 m/s? — (12.0 m/s")(3.0 5) =-30 m/s”. (m) a(45)=6.0 m/s? — (12.0 m/S 4.0 5)= —42 m/s?. 23. Since the problem involves constant acceleration, the motion of the electron can be readily analyzed using the equations in Table 2-1: v=vo+at (2-11) x-n=ugtãar (2-15) vV=y+2a(x—x,) (2-16) The acceleration can be found by solving Eq. (2-16). With v,=1.50x10/ m/s, v=5.70x10º m/s, xo = O and x = 0.010 m, we find the average acceleration to be vivi GTA mis = (1.5x10º mis) Cd 2(0.010 m) =1.62x10" m/s”. 24. In this problem we are given the initial and final speeds, and the displacement, and are asked to find the acceleration. We use the constant-acceleration equation given in 2.3 Eq. 2-16,v = vo + 2a(x — xo). (a) Given that v,=0, v=1.6m/s, and Ax=5.0um, the acceleration of the spores during the launch is a- vv om) = =2.56x10º m/s? = 2.6x10!g 2x 2(5.0x10º m (b) During the speed-reduction stage, the acceleration is o Om) og 0 mpi =-13x102g Z — X10x10" m) The negative sign means that the spores are decelerating. 25. We separate the motion into two parts, and take the direction of motion to be positive. In part 1, the vehicle accelerates from rest to its highest speed; we are given vo = 0; v=20 m/s and a = 2.0 m/s. In part 2, the vehicle decelerates from its highest speed to a halt; we are given vo = 20 m/s; v = O and a = —1.0 m/s? (negative because the acceleration vector points opposite to the direction of motion). (a) From Table 2-1, we find t, (the duration of part 1) from v = vo + at. In this way, 20=0+2.0r, yields t, = 10 s. We obtain the duration t, of part 2 from the same equation. Thus, O = 20 + (-1.0)t leads to t, = 20s, and the totalist=t; + =30s. (b) For part 1, taking xo = O, we use the equation = vê + 2a(x — xo) from Table 2-1 www.ELSOLUCIONARIO.bl .com 36 CHAPTER 2 traveled during t; and tp are the same so that they total to 2(10.59 m) = 21.18 m. This implies that for a distance of 190 m — 21.18 m = 168.82 m, the elevator is traveling at constant velocity. This time of constant velocity motion is -16882m 3515, 508 m/s 3 Therefore, the total time is 8.33 s + 33.21 s = 41.55. 30. We choose the positive direction to be that of the initial velocity of the car (implying that a < O since it is slowing down). We assume the acceleration is constant and use Table 2-1. (a) Substituting vo = 137 km/h = 38.1 m/s, v = 90 km/h = 25 m/s, and a = -5.2 m/s? into v = vo + at, we obtain p= 2ôm/s-38m/s | 5 25s. -52 m/s (b) We take the car to be at x = O when the brakes “(em are applied (at time t = 0). Thus, the coordinate of go the car as a function of time is given by S0; x=(38m/s)t + a(-s2 mis)f 40 : : . Na 2 in SI units. This function is plotted from t = O to t 0 =2.5 s on the graph to the right. We have not shown the v-vs-t graph here; it is a descending straight line from vo to v. O) 31. The constant acceleration stated in the problem permits the use of the equations in Table 2-1. (a) We solve v = vo + at for the time: =%o (0x10m/9) 4, gps a 9.8m/s” which is equivalent to 1.2 months. (b) We evaluate x= x, + wt+4at”, with xo =0. The result is x= > (9.8 m/s? ) (3.1x10ºs? =4.6x10º m. Note that in solving parts (a) and (b), we did not use the equation” = vw, +2a(x— x,). This equation can be employed for consistency check. The final velocity based on this www.ELSOLUCIONARIO.bl .com 37 equation is v="/v; + 2a(x— x9) = /0+ 2(9.8 m/s” )(4.6x 10º m— 0) =3.0x10" m/s, which is what was given in the problem statement. So we know the problems have been solved correctly. 32. The acceleration is found from Eq. 2-11 (or, suitably interpreted, Eq. 2-7). , [km (1020 km/h) 1000 m/ km 23600 s/h a. SD ompamis. Ar 145 In terms of the gravitational acceleration g, this is expressed as a multiple of 9.8 m/s? as follows: 202.4 m/s? =| |g=21lg. q 9.8 m/s? ): 8 33. The problem statement (see part (a)) indicates that a = constant, which allows us to use Table 2-1. (a) We take xo = 0, and solve x = vot + La? (Eq. 2-15) for the acceleration: a = 2(x — vol/?. Substituting x = 24.0 m, vo = 56.0 km/h = 15.55 m/s and t = 2.00 s, we find Ax-wt) 2(240m- (15.55m/s) (2.005)) aà DUO AO E ; = -3.56m/s?, P (2.008 or |a| =3.56 m/s”. The negative sign indicates that the acceleration is opposite to the direction of motion of the car. The car is slowing down. (b) We evaluate v = vo + at as follows: v=155S m/s (356 m/s”) (2.00 5)=843 m/s which can also be converted to 30.3 km/h. 34. Let d be the 220 m distance between the cars at t = 0, and v, be the 20 km/h = 50/9 m/s speed (corresponding to a passing point of x, = 44.5 m) and v, be the 40 km/h =100/9 m/s speed (corresponding to a passing point of x, = 76.6 m) of the red car. We have two equations (based on Eq. 2-17): d-x=vot + Zan? where t=1/v d-xw=vob +jat? where =X, /w, www.ELSOLUCIONARIO.bl .com 38 CHAPTER 2 We simultaneously solve these equations and obtain the following results: (a) The initial velocity of the green car is vo = — 13.9 m/s. or roughly — 50 km/h (the negative sign means that it's along the —x direction). (b) The corresponding acceleration of the car is a = — 2.0 m/s? (the negative sign means that it's along the —x direction). 35. The positions of the cars as a function of time are given by 1, Lo x(D)=xot+>at =(-350m)+—af (D =X, qu ( ) qu X(D=*,o +v,t= (270 m)- (20 m/s where we have substituted the velocity and not the speed for the green car. The two cars pass each other at t=12.0s when the graphed lines cross. This implies that (270 m)= (20 m/s)(12.0 5) = 30 m=(-35.0 m)+5a (20 sy which can be solved to give a, = 0.90 m/s”. 36. (a) Equation 2-15 is used for part 1 of the trip and Eq. 2-18 is used for part 2: Axi = Voir ti+ Zan where a, = 2.25 m/s? and Axy = 2“ m Av=wt— : at? where a, =—(0.75 m/s? and Am = Ao m In addition, vo: = v2= O. Solving these equations for the times and adding the results givest=ti+tp =56.65. (b) Equation 2-16 is used for part 1 of the trip: v= (vo) + 201Ax =0+ 202589) = 1013 mys? which leads to v = 31.8 m/s for the maximum speed. 37. (a) From the figure, we see that xo = —2.0 m. From Table 2-1, we can apply X-—X0=Vot+ 1a? with t = 1.0 s, and then again with t = 2.0 s. This yields two equations for the two unknowns, vo and a: www.ELSOLUCIONARIO.bl .com 41 40. We take the direction of motion as +x, so a = 5.18 m/s, and we use SI units, so vo = 55(1000/3600) = 15.28 m/s. (a) The velocity is constant during the reaction time 7, so the distance traveled during itis d,=wT- (15.28 m/s) (0.75 s) = 11.46 m. We use Eq. 2-16 (with v = 0) to find the distance dp traveled during braking: (15.28 m/ v=y+2ad, > d, (5.18 mA which yields dp = 22.53 m. Thus, the total distance is d, + dp = 34.0 m, which means that the driver is able to stop in time. And if the driver were to continue at vo, the car would enter the intersection in t = (40 m)/(15.28 m/s) = 2.6 s, which is (barely) enough time to enter the intersection before the light turns, which many people would consider an acceptable situation. (b) In this case, the total distance to stop (found in part (a) to be 34 m) is greater than the distance to the intersection, so the driver cannot stop without the front end of the car being a couple of meters into the intersection. And the time to reach it at constant speed is 32/15.28 = 2.1 s, which is too long (the light turns in 1.8 s). The driver is caught between a rock and a hard place. 41. The displacement (Ax) for each train is the “area” in the graph (since the displacement is the integral of the velocity). Each area is triangular, and the area of a triangle is 1/2( base) x (height). Thus, the (absolute value of the) displacement for one train (1/2)(40 m/s)(5 s) = 100 m, and that of the other train is (1/2)(30 m/s)(4 s) = 60 m. The initial “gap” between the trains was 200 m, and according to our displacement computations, the gap has narrowed by 160 m. Thus, the answer is 200 — 160 = 40 m. 42. (a) Note that 110 km/h is equivalent to 30.56 m/s. During a two-second interval, you travel 61.11 m. The decelerating police car travels (using Eq. 2-15) 51.11 m. In light of the fact that the initial “gap” between cars was 25 m, this means the gap has narrowed by 10.0 m — that is, to a distance of 15.0 m between cars. (b) First, we add 0.4 s to the considerations of part (a). During a 2.4s interval, you travel 73.33 m. The decelerating police car travels (using Eg. 2-15) 58.93 m during that time. The initial distance between cars of 25 m has therefore narrowed by 14.4 m. Thus, at the start of your braking (call it to) the gap between the cars is 10.6 m. The speed of the police car at to is 30.56 — 5(2.4) = 18.56 m/s. Collision occurs at time t when Xyou = Xpolice (We choose coordinates such that your position is x = O and the police car”s position is x = 10.6 mat). Eq. 2-15 becomes, for each car: Xpolice — 10.6 = 18.56(1 — to) — 5 (St — to) Xyou= 30.56(€ — 10) = 3 (5Xt— 10? www.ELSOLUCIONARIO.bl .com 42 CHAPTER 2 Subtracting equations, we find 10.6= (30.56 -18.56t—to) > 0883s=t-—to. At that time your speed is 30.56 + a(t — to) = 30.56 — 5(0.883) = 26 m/s (or 94 km/h). 43. In this solution we elect to wait until the last step to convert to SI units. Constant acceleration is indicated, so use of Table 2-1 is permitted. We start with Eq. 2-17 and denote the train”s initial velocity as v, and the locomotive”s velocity as v, (which is also the final velocity of the train, if the rear-end collision is barely avoided). We note that the distance Ax consists of the original gap between them, D, as well as the forward distance traveled during this time by the locomotive v,t . Therefore, vv Ar D+rvt D., 2 t t ros We now use Eq. 2-11 to eliminate time from the equation. Thus, v+VE D => +V 2 (v,-v)/a which leads to vo+v, VE>V, 1, 2 a=|[+—L-y, ==——(v,-v). 2 D 2D . Hence, a=-—— fog *m q6km | - pogsg km/h? 20.676 km) h h which we convert as follows: a=(12888 hm/ nº) (1000 ( lh ) cos m/s 1km 3600 s so that its magnitude is 0.994 m/s? A graph is shown here for the case where a collision is just x avoided (x along the vertical axis is in meters and £ along the horizontal axis is in seconds). The top 800 (straight) line shows the motion of the locomotive and the bottom curve shows the motion of the passenger train. 400 600 The other case (where the collision is not quite ?º avoided) would be similar except that the slope of 4 f the bottom curve would be greater than that of the 10 2 30 top line at the point where they meet. 44. We neglect air resistance, which justifies setting a = -g = 9.8 m/s? (taking down as the —y direction) for the duration of the motion. We are allowed to use Table 2-1 (with Ay replacing Ax) because this is constant acceleration motion. The ground level www.ELSOLUCIONARIO.bl .com 43 is taken to correspond to the origin of the y axis. (a) Using y=vt-4gf”, with y = 0.544 mand t = 0.200 s, we find y+8ÊIZ 0544 m+ (9.8m/s8)(0.2005)?/2 . f . 0.200s =3.70m/s. Vo (b) The velocity at y = 0.544 mis v =w-gt = 370 m/s-(9.8 m/s?) (0.200) = 1.74 m/s. (c) Using v=v; —2gy (with different values for y and v than before), we solve for the value of y corresponding to maximum height (where v = 0). = CIM o eggm 28 U98m/s?) , Thus, the armadillo goes 0.698 — 0.544 = 0.154 m higher. 45. In this problem a ball is being thrown vertically upward. Its subsequent motion is under the influence of gravity. We neglect air resistance for the duration of the motion (between “launching” and “landing”), so a = —g =-9.8 m/s? (we take downward to be the —y direction). We use the equations in Table 2-1 (with Ay replacing Ax) because this is a = constant motion: p= 81 (2-11) 1 330 = Wi er (2-15) vV=v 28-39) (2-16) We set yo = O. Upon reaching the maximum height y, the speed of the ball is momentarily zero (v = 0). Therefore, we can relate its initial speed vo to y via the equation0=v?=v; —2g). The time it takes for the ball to reach maximum height is given by v=v,—gt=0, or t=vo/g . Therefore, for the entire trip (from the time it leaves the ground until the time it returns to the ground), the total flight timeis T=2t=2w,/g. (a) At the highest point v=0 and vw, =2gy. Since y=50 m we find w= 287 = 29.8 m/s” )(50 m) =31.3 m/s. (b) Using the result from (a) for vo, we find the total flight time to be www.ELSOLUCIONARIO.bl .com 46 CHAPTER 2 the same as the velocity of the balloon, vo = +12 m/s, and that its initial coordinate is Yo =+80 m. (a) We solve y=Y)+vot-+gt” for time, with y = O, using the quadratic formula (choosing the positive root to yield a positive value for 1). + +28 2h + Ja2 mis? + 2(9.8m/s”)(80m) g 9.8 m/s? =54s (b) If we wish to avoid using the result from part (a), we could use Eq. 2-16, but if that is not a concern, then a variety of formulas from Table 2-1 can be used. For instance, Eq. 2-11 leads to v=w-gt=12m/s—(9.8m/s”)(5.447 s)=—41.38 m/s Tts final speed is about 41 m/s. 50. The y coordinate of Apple 1 obeys y — yo: =— 5 gê where y=0 when t=20s. This allows us to solve for yo1, and we find yo1 = 19.6 m. The graph for the coordinate of Apple 2 (which is thrown apparently at t = 1.0 s with velocity v>) is pa = vilt= 10) - dg (1 10? where yo2 = Yo1 = 19.6 m and where y = O when t = 2.25 s. Thus, we obtain |w] = 9.6 m/s, approximately. 51. (a) With upward chosen as the +y direction, we use Eg. 2-11 to find the initial velocity of the package: v=v+at > 0=vw-(98mS(205) which leads to vo = 19.6 m/s. Now we use Eq. 2-15: Ay = (19.6 m/S)(2.0 8) + 5 (9.8 m/S*)(2.0 8) = 20 m. We note that the “2.0 s” in this second computation refers to the time interval 2< 1< 4 in the graph (whereas the “2.0 s” in the first computation referred to the O < t < 2 time interval shown in the graph). (b) In our computation for part (b), the time interval (“6.0 s”) refers to the 2 <t < 8 portion of the graph: Ay= (19.6 m/s)(6.08) + + (-9.8 m/SX6.0 5) =-59m, or |Ay-59m. www.ELSOLUCIONARIO.bl .com 47 52. The full extent of the bolt's fall is given by 1 y-yo=5 8º where y — yo = —90 m (if upward is chosen as the positive y direction). Thus the time for the full fall is found to be t = 4.29 s. The first 80% of its free-fall distance is given by-72=-g T/2, which requires time T = 3.83 s. (a) Thus, the final 20% of its fall takes t — T = 0.45 s. (b) We can find that speed using v=—gt. Therefore, |v| = 38 m/s, approximately. (c) Similarly, vana=— gt > |Vjnal=42 m/s. 53. The speed of the boat is constant, given by v, = d/t. Here, d is the distance of the boat from the bridge when the key is dropped (12 m) and t is the time the key takes in falling. To calculate t, we put the origin of the coordinate system at the point where the key is dropped and take the y axis to be positive in the downward direction. Taking the time to be zero at the instant the key is dropped, we compute the time t when y = 45 m. Since the initial velocity of the key is zero, the coordinate of the key is givenby y=1gf”.Thus, (= [DZ agg5, 8 9.8m/s” Therefore, the speed of the boat is 12m = =40m/s. 3.03s Vo 54. (a) We neglect air resistance, which justifies setting a = —g = -9.8 mis? (taking down as the —y direction) for the duration of the motion. We are allowed to use Eq. 2-15 (with Ay replacing Ax) because this is constant acceleration motion. We use primed variables (except t) with the first stone, which has zero initial velocity, and unprimed variables with the second stone (with initial downward velocity —vo, so that vo is being used for the initial speed). SI units are used throughout. 1, Ay =0(1)-58º av =()lt-D->e(t-1) Since the problem indicates Ay" = Ay = —43.9 m, we solve the first equation for t (finding t = 2.99 s) and use this result to solve the second equation for the initial speed of the second stone: www.ELSOLUCIONARIO.bl .com 48 CHAPTER 2 = o , 43.9 m=(-v)(1.995)- +(98 m/s?) (1.99) which leads to vo = 12.3 m/s. (b) The velocity of the stones are given by 1 = MAy) d(Ay) =" =—8t, v,=—>"+=—wy-g(t-1 1% 8 1% o 8-1) The plot is shown below: Vs) os 1.15 2 25 3 ts -5 -10 as Ss 55. During contact with the ground its average acceleration is given by ad avg At where Av is the change in its velocity during contact with the ground and At=20.0x10” sis the duration of contact. Thus, we must first find the velocity of the ball just before it hits the ground (y = 0). (a) Now, to find the velocity just before contact, we take t = O to be when it is dropped. Using Eq. (2-16) with y =15.0m, we obtain v=—Jwi-2g(9- y9))=—/0-2(9.8 m/s? )(0-15 m) =—17.15 m/s where the negative sign is chosen since the ball is traveling downward at the moment of contact. Consequently, the average acceleration during contact with the ground is n = Av 0-CITimis) =857 m/s? “Car 200x10ºs www.ELSOLUCIONARIO.bl .com 51 (b) The time in free fall (up to the moment drop 1 lands) for drop 3 is t; = t; -0.426s =0.213s. Its position at the moment drop 1 strikes the floor is 1 p=-qe6= --08 m/s? YX0.2135)? =-0.222 m, or about 22 cm below the nozzle. 60. To find the “launch” velocity of the rock, we apply Eq. 2-11 to the maximum height (where the speed is momentarily zero) v=w-8t > 0=w-—(9.8m/s')(2.55) so that vo = 24.5 m/s (with +y up). Now we use Eq. 2-15 to find the height of the tower (taking yo = O at the ground level) v= Joc + jar > v = 0=(245mls)(1.55) = (9.8 ml? (1.55). Thus, we obtain y = 26 m. 61. We choose down as the +y direction and place the coordinate origin at the top of the building (which has height H). During its fall, the ball passes (with velocity vi) the top of the window (which is at y1) at time fi, and passes the bottom (which is at y>) at time t>. We are told y> — y, = 1.20 m and o — t, = 0.125 s. Using Eq. 2-15 we have , vo Lo, 2 Jd =Milh = t+ sh =h) which immediately yields 1.20 m — 4(9.8m/sº)(0.125s o 0.125s = 8.99 m/s. Y From this, Eq. 2-16 (with vo = 0) reveals the value of y;: It reaches the ground (y; = H) at t. Because of the symmetry expressed in the problem (“upward flight is a reverse of the fall”) we know that t; — tp = 2.00/2 = 1.00 s. And this means t; — t, = 1.00 s + 0.125 s = 1.125 s. Now Eq. 2-15 produces 1 2 = (tb = + 8(h 1) 35— 4.12 m=(8.99 m/s)(1.125 +98 m/s?) (1.125 5) www.ELSOLUCIONARIO.bl .com 52 CHAPTER 2 which yields y; = H = 20.4 m. 62. The height reached by the player is y = 0.76 m (where we have taken the origin of the y axis at the floor and +y to be upward). (a) The initial velocity vo of the player is vw= 287 =/2(9.8 m/s?) (0.76 m) = 3.86 m/s. This is a consequence of Eq. 2-16 where velocity v vanishes. As the player reaches y; =0.76m-0.15m=0.61 m, his speed v, satisfies vw, —v; = 2gy,, which yields v=v — 28), = ((3.86m/s? — 2(9.80 m/s?)(0.61m) = 1.71 m/s . The time t, that the player spends ascending in the top Ay; = 0.15 m of the jump can now be found from Eq. 2-17: 2(0.15 m) Ay = A 1.7im/s + 0 (u+v)s > 6- =0.175s 1 2 which means that the total time spent in that top 15 cm (both ascending and descending) is 2(0.175 s) = 0.35 s = 350 ms. (b) The time t, when the player reaches a height of 0.15 m is found from Eq. 2-15: Lis 1 na 015 m=vt, -z gt, =(3.86 m/s)t, “208 m/s , which yields (using the quadratic formula, taking the smaller of the two positive roots) to = 0.041 s = 41 ms, which implies that the total time spent in that bottom 15 cm (both ascending and descending) is 2(41 ms) = 82 ms. 63. The time t the pot spends passing in front of the window of length L = 2.0 m is 0.25 s each way. We use v for its velocity as it passes the top of the window (going up). Then, with a = —g = -9.8 m/s? (taking down to be the —y direction), Eq. 2-18 yields L=vt ! P > L1 f v 3 g v 772 gt. The distance H the pot goes above the top of the window is therefore (using Eq. 2-16 with the final velocity being zero to indicate the highest point) 2 (Lit-gtl2) (200m/025s-(9.80m/s')0.255)/2 nel 8 pl ( - 10255) Dos. 28 28 2(9.80 m/s?) www.ELSOLUCIONARIO.bl .com 53 64. The graph shows y = 25 m to be the highest point (where the speed momentarily vanishes). The neglect of “air friction” (or whatever passes for that on the distant planet) is certainly reasonable due to the symmetry of the graph. (a) To find the acceleration due to gravity gp on that planet, we use Eg. 2-15 (with +y up) von ut os, > 25m-0=(0)(25s)t+5e, (255) so that gp = 8.0 mis. (b) That same (max) point on the graph can be used to find the initial velocity. (1,+0)(255) 1 1 =aluto > 25 m-0=> Therefore, vo = 20 m/s. 65. The key idea here is that the speed of the head (and the torso as well) at any given time can be calculated by finding the area on the graph of the heads acceleration versus time, as shown in Eq. 2-26: area between the acceleration curve v—M, . . no and the time axis, from £, to f, (a) From Fig. 2.14a, we see that the head begins to accelerate from rest (vo= 0) at fo = 110 ms and reaches a maximum value of 90 m/s” at ti = 160 ms. The area of this region is area = >(160-110)* 10ºs-(90 m/s? )=2.25 m/s which is equal to vi, the speed at t;. (b) To compute the speed of the torso at t/=160 ms, we divide the area into 4 regions: From O to 40 ms, region A has zero area. From 40 ms to 100 ms, region B has the shape of a triangle with area area, = >(00600 sX(50.0 m/s?) = 1.50 m/s. From 100 to 120 ms, region C has the shape of a rectangle with area area. = (0.0200 s) (50.0 m/s?) = 1.00 m/s. From 110 to 160 ms, region D has the shape of a trapezoid with area area, = =(0,0400 s) (50.0 + 20.0) m/s? = 1,40 m/s. Substituting these values into Eq. 2-26, with vo= O then gives www.ELSOLUCIONARIO.bl .com 56 CHAPTER 2 Thus, the difference in the speed is Av=v, =0.82 m/s—0.26 m/s = 0.56 m/s. unhelmeted ” Vhelmeted 68. This problem can be solved by noting that velocity can be determined by the graphical integration of acceleration versus time. The speed of the tongue of the salamander is simply equal to the area under the acceleration curve: v=area= a0º S)(100 m/s?) + s0º S)(100 m/s? + 400 m/s?) + ado? s)(400 m/s?) =5.0m/s. 69. Since v=dx/dt (Eq. 2-4), then Ax=[ vdt, which corresponds to the area under the v vs f graph. Dividing the total area 4 into rectangular (base x height) and triangular (+ base x height) areas, we have A =A + A + A, +A, 2<1<10 10<r<12 12<1c16 = 28) + (8)(8) + (oro + 1044) + (44) O<r<2 with SI units understood. In this way, we obtain Ax = 100 m. 70. To solve this problem, we note that velocity is equal to the time derivative of a position function, as well as the time integral of an acceleration function, with the integration constant being the initial velocity. Thus, the velocity of particle 1 can be written as ut “(600º +3.00+2.00) = 12.01+3.00. dt dt Similarly, the velocity of particle 2 is v, = Voo + Jazdr =20.0+ Je-8.00nar = 20.0- 4.00. The condition that v,=v, implies 12.01+3.00= 20.04.00” => 4.007 +12.0t-17.0=0 which can be solved to give (taking positive root) t= (-3+426)/2= 1.05s. Thus, the velocity at this time is v, = v, = 12.0(1.05)+3.00= 15.6 m/s. 71. (a) The derivative (with respect to time) of the given expression for x yields the “velocity” of the spot: vD=9-5? www.ELSOLUCIONARIO.bl .com 57 with 3 significant figures understood. It is easy to see that v = O when t = 2.00 s. (b) Att=2s,x=9(2)— KO) = 12. Thus, the location of the spot when v = O is 12.0 cm from left edge of screen. (c) The derivative of the velocity is a = — 5 t, which gives an acceleration of -9.00 cm/m? (negative sign indicating leftward) when the spot is 12 em from the left edge of screen. (d) Since v > O for times less than t = 2 s, then the spot had been moving rightward. (e) As implied by our answer to part (c), it moves leftward for times immediately after t=2s. In fact, the expression found in part (a) guarantees that for allt>2,v< 0 (that is, until the clock is “reset” by reaching an edge). (f) As the discussion in part (e) shows, the edge that it reaches at some t > 2 s cannot be the right edge; it is the left edge (x = 0). Solving the expression given in the problem statement (with x = 0) for positive t yields the answer: the spot reaches the left edge at t= [12 s= 3.465. 72. We adopt the convention frequently used in the text: that "up" is the positive y direction. (a) At the highest point in the trajectory v = O. Thus, with t = 1.60 s, the equation v=vo- gt yields vo = 15.7 m/s. (b) One equation that is not dependent on our result from part (a) is y— yo = vt + 38º; this readily gives Ymax — Yo = 12.5 m for the highest ("max") point measured relative to where it started (the top of the building). (c) Now we use our result from part (a) and plug into y — yo = vot + 28º with t = 6.00 s and y = O (the ground level). Thus, we have 0 yo = (15.68 m/s)(6.00 s) — : (9.8 m/s*)(6.00 s)?. Therefore, yo (the height of the building) is equal to 82.3 m. 73. We denote the required time as t, assuming the light turns green when the clock reads zero. By this time, the distances traveled by the two vehicles must be the same. (a) Denoting the acceleration of the automobile as a and the (constant) speed of the truck as v then Ax= Eu l (0 ua which leads to www.ELSOLUCIONARIO.bl .com 58 CHAPTER 2 » M9.5m/s p-U- 219.5 mis). 86s. a 22 m/s Therefore, Ax=vt=(9.5m/s)(8.6s)=82m. (b) The speed of the car at that moment is Va =at=(2.2m/s' )(8.65)=19 m/s . 74. If the plane (with velocity v) maintains its present course, and if the terrain continues its upward slope of 4.3º, then the plane will strike the ground after traveling are SM 4655ma0.465 km. tan6 tan4.3º This corresponds to a time of flight found from Eq. 2-2 (with v = vavg since it is constant) — Ax 0465km = = 0,000358 h= 1.35. v. 1300km/h This, then, estimates the time available to the pilot to make his correction. 75. We denote t, as the reaction time and t, as the braking time. The motion during t, is of the constant-velocity (call it vo) type. Then the position of the car is given by 1, X=Wi, + vd, + Sat, 2 where vo is the initial velocity and a is the acceleration (which we expect to be negative-valued since we are taking the velocity in the positive direction and we know the car is decelerating). After the brakes are applied the velocity of the car is given by v = vo+ atp. Using this equation, with v = 0, we eliminate t, from the first equation and obtain We write this equation for each of the initial velocities: Lv H= Vol, — 5 2a and 1v m=vol,-——E. - “2a Solving these equations simultaneously for t, and a we get www.ELSOLUCIONARIO.bl .com 61 using Eq 2-16 again. Specifically, its velocity at that moment would be —10 m/s since it is still traveling in the —x direction when it crashes. If the computation of v had failed (meaning that a negative number would have been inside the square root) then we would have looked at the possibility that there was no collision and examined how far apart they finally were. A concern that can be brought up is whether the primed train collides before it comes to rest; this can be studied by computing the time it stops (Eq. 2-11 yields t = 20 s) and seeing where the unprimed train is at that moment (Eq. 2-18 yields x = 350 m, still a good distance away from contact). 79. The y coordinate of Piton 1 obeys y — yoi = — : g Ê where y=0whent=3.0s. This allows us to solve for yo1, and we find yo; = 44.1 m. The graph for the coordinate of Piton 2 (which is thrown apparently at t = 1.0s with velocity vi) is y=30=vi(t-10)- 5 g(1-1.02 where yo2 obtain [vi] yo1 + 10 = 54.1 m and where (again) y = O when t=3.0s. Thus we 17 m/s, approximately. 80. We take +x in the direction of motion. We use subscripts 1 and 2 for the data. Thus, v/=+30 m/s, v> = +50 m/s, and x» — x, = +160m. (a) Using these subscripts, Eq. 2-16 leads to vv) (50m/s) — (30mis) —2 1 =50 m/s. 24 —4,) 2(160 m) : a= (b) We find the time interval corresponding to the displacement x, — x, using Eg. 2-17: Am =x) 2(160m) vtv 30m/s+50m/s b-t= =40s. (c) Since the train is at rest (vo = 0) when the clock starts, we find the value of t, from Eq. 2-1: v=vo + aí, (d) The coordinate origin is taken to be the location at which the train was initially at rest (so xo = 0). Thus, we are asked to find the value of x. Although any of several equations could be used, we choose Eq. 2-17: x = +v)i =+(30 m/s)(605)=90 m. (e) The graphs are shown below, with SI units understood. www.ELSOLUCIONARIO.bl .com 62 CHAPTER 2 so, 200 81. Integrating (from t = 2 s to variable t = 4 s) the acceleration to get the velocity and using the values given in the problem leads to v=w+ f adt=v, +, (5.00)dt = v, + > 6oNº = 174 ISOS -2)=47m/s. ni 82. The velocity v at t = 6 (SI units and two significant figures understood) is v, +[adr. A quick way to implement this is to recall the area of a triangle G base x height). The result is v = 7 m/s + 32 m/s =39 m/s. 83. The object, once it is dropped (vo = 0) is in free fall (a = -g = -9.8 m/s? if we take down as the —y direction), and we use Eq. 2-15 repeatedly. (a) The (positive) distance D from the lower dot to the mark corresponding to a certain reaction time t is given by Ay=-D=-lgf”, or D = gên. Thus, fort, = 50.0 ms, (98 m/s?)(50.0x 10º À 3 00123m=1.23cm. 2 a (b) For to = 100 ms, D, -(ebmir) (100x10"5) = 0.049m=4D,. (9.8mis?) (150x107s) (c) For ts = 150 ms, D=""" 2 0077 = 0.11m=9D.. (9.8m/s?) (200x 105) (4) For ts =200 ms, D,=1 >> = 0.196m=16D, 98m/s?)(250x 10º 5 (e) Forts=250ms, D, = Câmre)(eso mto") =0306m=25D,. 84. We take the direction of motion as +x, take xo = O and use SI units, so v = 1600(1000/3600) = 444 m/s. www.ELSOLUCIONARIO.bl .com 63 (a) Equation 2-11 gives 444 = a(1.8) or a = 247 m/s?. We express this as a multiple of g by setting up a ratio: 247 mis? =|20D |g=25g. 9.8m/sº 8 8 (b) Equation 2-17 readily yields 1 1 1=5(4 +v)t =5(444 m/s)(1.8s)= 400 m. 85. Let D be the distance up the hill. Then total dist: traveled 2D average speed = tota cistancetraveieo CD DD * 25 km/h. total time of travel 20 km/h * 35 km/h 86. We obtain the velocity by integration of the acceleration: v-»; =[. (6.1-1.2r)dr”. Lengths are in meters and times are in seconds. The student is encouraged to look at the discussion in the textbook in $2-7 to better understand the manipulations here. (a) The result of the above calculation is v =w+6.11-0.67, where the problem states that vo = 2.7 m/s. The maximum of this function is found by knowing when its derivative (the acceleration) is zero (a = 0 when t = 6.1/1.2=5.1s) and plugging that value of t into the velocity equation above. Thus, we find v=18m/s. (b) We integrate again to find x as a function of t: x-x= | vdr'= [(w+6.17-06t)dt = wt+3057 - 020. With xo = 7.3 m, we obtain x= 83 m fort=6. This is the correct answer, but one has the right to worry that it might not be; after all, the problem asks for the total distance traveled (and x — xo is just the displacement). If the cyclist backtracked, then his total distance would be greater than his displacement. Thus, we might ask, "did he backtrack?" To do so would require that his velocity be (momentarily) zero at some point (as he reversed his direction of motion). We could solve the above quadratic equation for velocity, for a positive value of t where v = O; if we did, we would find that at t = 10.6 s, a reversal does indeed happen. However, in the time interval we are concerned with in our problem (0 < t < 6 s), there is no reversal and the displacement is the same as the total distance traveled. www.ELSOLUCIONARIO.bl .com 66 CHAPTER 2 m; using this and the value found in part (a), Eq. 2-2 produces 2x6) = 40) 15m-Im ne 5-1 4s =3.5 m/s. (e) From Eq. 2-7 and the values v(t) we read directly from the graph, we find = M$ uD ( 2mis-2mis À 0. ne 5-1 4s 91. Taking the +y direction downward and yo = 0, we have y=vot+L gr”, which (with vo =0) yields t=2y/g. (a) For this part of the motion, y; = 50 m so that p= PSOm = 325. 9.8 m/s” E (b) For this next part of the motion, we note that the total displacement is y, = 100 m. Therefore, the total time is n= [2000 45, - 9.8m/s” The difference between this and the answer to part (a) is the time required to fall through that second 50 m distance: At=t,-t,=4.55-3.28=13s. 92. Direction of +x is implicit in the problem statement. The initial position (when the clock starts) is xo = O (where vo = 0), the end of the speeding-up motion occurs at x, = 1100/2 = 550 m, and the subway train comes to a halt (vz = 0) at x, = 1100 m. (a) Using Eq. 2-15, the subway train reaches x; at = (550 = 280 (60 ss, a ViZms The time interval t, — t, turns out to be the same value (most easily seen using Eq. 2-18 so the total time is tp = 2(30.3) = 60.6 s. (b) Its maximum speed occurs at t; and equals v=votat =363m/s. (c) The graphs are shown below: www.ELSOLUCIONARIO.bl .com 67 x(m) 1200 1000 800 600 400 200 Hs) ã 1 mw 304050 607 !(8) 93. We neglect air resistance, which justifies setting a = —g = -9.8 m/s? (taking down as the —y direction) for the duration of the stones motion. We are allowed to use Table 2-1 (with Ax replaced by y) because the ball has constant acceleration motion (and we choose yo = 0). (a) We apply Eq. 2-16 to both measurements, with SI units understood. 2. Ly > w=w-28, O (5º) +28(34 +3)=w vúi=w-Zg > += We equate the two expressions that each equal v; and obtain 2 , 2 3. 5 + 2 +28()= 1º +29, > 28(3)=7vº which yields v=/24(4) = 885 m/s. (b) An object moving upward at A with speed v = 8.85 m/s will reach a maximum height y— ya = vg = 4.00 m above point A (this is again a consequence of Eq. 2-16, now with the “final” velocity set to zero to indicate the highest point). Thus, the top of its motion is 1.00 m above point B. 94. We neglect air resistance, which justifies setting a = —g = 9.8 m/s? (taking down as the —y direction) for the duration of the motion. We are allowed to use Table 2-1 (with Ay replacing Ax) because this is constant acceleration motion. The ground level www.ELSOLUCIONARIO.bl .com 68 CHAPTER 2 is taken to correspond to the origin of the y-axis. The total time of fall can be computed from Eq. 2-15 (using the quadratic formula). = qo ot vo 28Ay 1, Ay=vt-— gt VE bio 8 e with the positive root chosen. With y = 0, vo =0, and yo = h = 60 m, we obtain v2sh Mass, 8 8 t= Thus, “1.2 s earlier” means we are examining where the rockisatt=2.3s: 1 2 y=h=w(235)-—8(235) > y=34m where we again use the fact that A = 60 m and vo = 0. 95. (a) The wording of the problem makes it clear that the equations of Table 2-1 apply, the challenge being that vo, v, and a are not explicitly given. We can, however, apply x — xo = vot + ja? to a variety of points on the graph and solve for the unknowns from the simultaneous equations. For instance, 1 16m-0=v(2.08) + 5 a(2.0 s2 1 27m-0=v(3.08)+ 5 a(30 s2 lead to the values vo = 6.0 m/s and a = 2.0 m/s? (b) From Table 2-1, 1 1 x-xo=vt- af > 2]m-0=v30s)-5 (20 m/s*Y3.0 s)? which leads to v = 12 m/s. . . . . 1 (c) Assuming the wind continues during 3.0 < t< 6.0, we apply x — xo = vot + a? to this interval (where vo = 12.0 m/s from part (b)) to obtain 1 Av= (12.0 m/s)(3.0 8) + 5 (2.0 m/s)(3.0s)=45m. 96. (a) Let the height of the diving board be h. We choose down as the +y direction and set the coordinate origin at the point where it was dropped (which is when we start the clock). Thus, y = h designates the location where the ball strikes the water. Let the depth of the lake be D, and the total time for the ball to descend be T. The speed of the ball as it reaches the surface of the lake is then v = 2gh (from Eq. www.ELSOLUCIONARIO.bl .com n This is a result of Eq. 2-11 where speeds are used instead of the (negative-valued) velocities (so that final-velocity minus initial-velocity turns out to equal initial-speed minus final-speed); we also note that the acceleration vector for this part of the motion is positive since it points upward (opposite to the direction of motion — which makes it a deceleration). The total time of flight is therefore t, + tp = 175. (b) The distance through which the parachutist falls after the parachute is opened is given by qe (31m/s) 2a (224 In the computation, we have used Eq. 2-16 with both sides multiplied by —1 (which changes the negative-valued Ay into the positive d on the left-hand side, and switches the order of v; and v> on the right-hand side). Thus the fall begins at a height of h = 50 +d=290m. 101. We neglect air resistance, which justifies setting a = —g =-9.8 m/s? (taking down as the —y direction) for the duration of the motion. We are allowed to use Table 2-1 (with Ay replacing Ax) because this is constant acceleration motion. The ground level is taken to correspond to y = 0. (a) With yo = h and vo replaced with —vo, Eq. 2-16 leads to v= Cu) -280 —39) = vo + 28h. The positive root is taken because the problem asks for the speed (the magnitude of the velocity). (b) We use the quadratic formula to solve Eq. 2-15 for t, with vo replaced with —vo, =vot vo) 284y 1 Ay= -vt-=g > t= 2 8 where the positive root is chosen to yield t > 0. With y = O and yo = A, this becomes Vo + 2gh —- v, 8 (c) If it were thrown upward with that speed from height h then (in the absence of air friction) it would return to height h with that same downward speed and would therefore yield the same final speed (before hitting the ground) as in part (a). An important perspective related to this is treated later in the book (in the context of energy conservation). (d) Having to travel up before it starts its descent certainly requires more time than in part (b). The calculation is quite similar, however, except for now having +vo in the equation where we had put in —vo in part (b). The details follow: www.ELSOLUCIONARIO.bl .com 72 CHAPTER 2 with the positive root again chosen to yield t > 0. With y = O and yo = h, we obtain 2 NW+ 2gh + vo 8 102. We assume constant velocity motion and use Eq. 2-2 (with vaga = v > 0). Therefore, ax= var =| 30342 [1000m/km || (109,.1035)-84m. h | 3600s/h / www.ELSOLUCIONARIO.bl .com Chapter 3 1. The x and the y components of a vector à lying on the xy plane are given by a =acos0, a,=asin6 where a=|á | is the magnitude and 0 is the angle between à and the positive x axis. (a) The x component of à is givenby a, = acos0= (7.3 m)cos250º=-2.50 m. (b) Similarly, the y component is given by a,=asin0=(7.3 m)sin250º=-6.86 mx-6.9 m. The results are depicted in the figure below: (m) XxX (m) In considering the variety of ways to compute these, we note that the vector is 70º below the — x axis, so the components could also have been found from a,=-(7.3 m)cos70º=-2.50m, a,=-(7.3 m)sin70º=-6.86 m. Similarly, we note that the vector is 20º to the left from the — y axis, so one could also achieve the same results by using a,=-(7.3 m)sin20º=-2.50 m, a,=-—(7.3 m)cos20º=-6.86 m. As a consistency check, we note that da +02 =V(-2.50m) +(-6.86m) =7.3m 73 www.ELSOLUCIONARIO.bl «com 76 CHAPTER 3 7. The displacement of the fly is illustrated in the figure below: A coordinate system such as the one shown (above right) allows us to express the displacement as a three-dimensional vector. (a) The magnitude of the displacement from one corner to the diagonally opposite corner is d=|d|= Vw ++ Substituting the values given, we obtain a=|dj=Nw+P+hº = (370 m) +(4.30m) +(3.00m) = 6.42 m. (b) The displacement vector is along the straight line from the beginning to the end point of the trip. Since a straight line is the shortest distance between two points, the length of the path cannot be less than the magnitude of the displacement. (c) It can be greater, however. The fly might, for example, crawl along the edges of the room. Its displacement would be the same but the path length would be (+w+h=11.0 m. (d) The path length is the same as the magnitude of the displacement if the fly flies along the displacement vector. (e) We take the x axis to be out of the page, the y axis to be to the right, and the z axis to be upward. Then the x component of the displacement is w = 3.70 m, the y component of the displacement is 4.30 m, and the z component is 3.00 m. Thus, d = (3.70 m)i +(4.30 m)j+(3.00 m)k . An equally correct answer is gotten by interchanging the length, width, and height. www.ELSOLUCIONARIO.bl .com 7 1 I 1 1 1 (f) Suppose the path of the fly is as shown by the dotted lines on the upper diagram. Pretend there is a hinge where the front wall of the room joins the floor and lay the wall down as shown on the lower diagram. The shortest walking distance between the lower left back of the room and the upper right front corner is the dotted straight line shown on the diagram. Its length is 2 Luas r+h)+0 = [(310m+3.00m) + (430m? =7.96m. To show that the shortest path is indeed given by Lj; + We write the length of the path as L=yy+w +=) +. The condition for minimum is given by dL y E du A little algebra shows that the condition is satisfied when y=Iw/(w+h), which gives 2 P 2 P o 2 2 Lain Sa4lW (1+mo)* h (1:55) Nemo +. Any other path would be longer than 7.96 m. 8. We label the displacement vectors Á, B, and É (and denote the result of their vector sum as 7 ). We choose east as the i direction (+x direction) and north as the j direction (+y direction). All distances are understood to be in kilometers. (a) The vector diagram representing the motion is shown next: www.ELSOLUCIONARIO.bl .com 78 CHAPTER 3 north B west “> cast à =G1 km)j C Á south B (24kmi C=(-52km)j (b) The final point is represented by F=-Ã+B+C=(24kmji+(-21 km) whose magnitude is p= 4(-24km+(-2.1km) 3.2 km. (c) There are two possibilities for the angle: O=tan! [EM ator 21º. —Le m We choose the latter possibility since 7 is in the third quadrant. It should be noted that many graphical calculators have polar «>» rectangular “shorteuts” that automatically produce the correct answer for angle (measured counterclockwise from the +x axis). We may phrase the angle, then, as 221º counterclockwise from East (a phrasing that sounds peculiar, at best) or as 41º south from west or 49º west from south. The resultant F is not shown in our sketch; it would be an arrow directed from the “tail” of À to the “head” of Cc. 9. All distances in this solution are understood to be in meters. (a) d+b=[4.0+(-1.0)]i+ [(-3.0)+1.0]) +(1.0+4.0)k = (3.01-2.0]+5.0h) m. (b) à-5=[4.0-(1.0)Ji+ [(-3.0)-1.0]]+ (1.0- 4.0)k = (5.01-4.0]-3.0k) m. (c) The requirement ã-b+e=0 leads to E=b-à, which we note is the opposite of what we found in part (b). Thus, & = (-5.01 + 4.0) + 3.0) m. 10. The x, y, and z components of 7 = c+ d are, respectively, ()r=c+d =74m+44 m=I2m, (b)r,=c,+d,=-3.8m-2.0m=-5.8m,and www.ELSOLUCIONARIO.bl .com 81 15. It should be mentioned that an efficient way to work this vector addition problem is with the cosine law for general triangles (and since ab, and F form an isosceles triangle, the angles are easy to figure). However, in the interest of reinforcing the usual systematic approach to vector addition, we note that the angle b makes with the +x axis is 30º +105º = 135º and apply Eq. 3-5 and Eg. 3-6 where appropriate. (a) The x component of 7 is r;= (10.0 m) cos 30º + (10.0 m) cos 135º = 1.59 m. (b) The y component of 7 is r, = (10.0 m) sin 30º + (10.0 m) sin 135º = 12.1 m. (c) The magnitude of 7 is r=[|7|=(1.59m)+(12.1m) =122m. (d) The angle between 7 and the +x direction is tan [(12.1 m)/(1.59 m)] = 82.5º. 16.(a) a+b = (3.01+4.0)) m+(5.01- 2.0) m= (8.0 m)i+(2.0 m)). (b) The magnitude of à + bis [a+b E (80 m)? +(2.0m) =82m. (c) The angle between this vector and the +x axis is tan "[(2.0 m)/(8.0 m)] = 14º. (d) b-ã=(5.01-2.0) m-(3.01+4.0)) m= (2.0 m)i- (6.0 m)). (e) The magnitude of the difference vector b-áâis b-ã E V(2.0m)+(-60m) =6.3m. (f) The angle between this vector and the +x axis is tan![(-6.0 m)/(2.0 m)] = —72º. The vector is 72º clockwise from the axis defined by i. 17. Many of the operations are done efficiently on most modern graphical calculators using their built-in vector manipulation and rectangular <> polar “shortcuts.” In this solution, we employ the “traditional” methods (such as Eg. 3-6). Where the length unit is not displayed, the unit meter should be understood. (a) Using unit-vector notation, www.ELSOLUCIONARIO.bl .com 82 CHAPTER 3 = (50 m) cos(30º)i + (50 m) sin(30º)j = (50 m)cos (195º) 1+ (50 m) sin (195º) j E = (50 m)cos(315º) 1+ (50 m)sin (315º) j The magnitude of this result is (30.4 Mm) +(-23.3m) =38m. (b) The two possibilities presented by a simple calculation for the angle between the vector described in part (a) and the +x direction are tan HC 23.2 m)/(30.4 m)] = —37.5º, and 180º + (—37.5º) = 142.5º. The former possibility is the correct answer since the vector is in the fourth quadrant (indicated by the signs of its components). Thus, the angle is —37.5º, which is to say that it is 37.5º clockwise from the +x axis. This is equivalent to 322.5º counterclockwise from +x. (c) We find à-b+2=[433-(-48.3)+35.4]i-[25-(—12.9)+(- 35.4)]j= (1275+2.60) m in unit-vector notation. The magnitude of this result is la-b+cl=(127 m) +(2.6 m) = 1.30x10º m. (d) The angle between the vector described in part (c) and the +x axis is tan (2.6 m/127 m)x1.2º. (e) Using unit-vector notation, d is given by d=a+b-E=(-40414474))m, which has a magnitude of +/(-40.4 m)? +(47.4m) =62 m. (f) The two possibilities presented by a simple calculation for the angle between the vector described in part (e) and the +x axis are tan (47.4/(-40.4))=-50.0º, and 180º+(-50.0º)= 130º . We choose the latter possibility as the correct one since it indicates that d is in the second quadrant (indicated by the signs of its components). 18. If we wish to use Eg. 3-5 in an unmodified fashion, we should note that the angle between C and the +x axis is 180º + 20.0º = 200º. (a) The x and y components of B are given by By= C,— As = (15.0 m) cos 200º — (12.0 m) cos 40º = 23.3 m, B,=C,-Ay= (15.0 m) sin 200º — (12.0 m) sin 40º =—12.8m. www.ELSOLUCIONARIO.bl .com 83 Consequently, its magnitude is |B|= (23.3 m) +(c12.8m) =266m. (b) The two possibilities presented by a simple calculation for the angle between B and the +x axis are tan[(—12.8 m)/(-23.3 m)] = 28.9º, and 180º + 28.9º = 209º. We choose the latter possibility as the correct one since it indicates that B sin the third quadrant (indicated by the signs of its components). We note, too, that the answer can be equivalently stated as — 151º. 19. (a) With 1 directed forward and j directed leftward, the resultant is (5.001 + 2.00) m. The magnitude is given by the Pythagorean theorem: «(5.00 m)?+(2.00 m)? = 5.385 m x 5.39m. (b) The angle is tan "(2.00/5.00) a 21.8º (left of forward). 20. The desired result is the displacement vector, in units of km, A = (5.6 km), 90º (measured counterclockwise from the +x axis), or Ã= (5.6 km), where 5 s the unit vector along the positive y axis (north). This consists of the sum of two displacements: during the whiteout, B= (7.8 km), 50º, or B=(7.8 km)(cos50%+sin50º ))= (5.01 km)i+(5.98 km) and the unknown C. Thus, A= B+C. (a) The desired displacement is given by C= À-B=(-5.01 km) i-(0.38 km) j. The magnitude is 4/(-5.01 km)? +(-0.38 km)? = 5.0 km. (b) The angle is tan '[(-0.38 km)/(-5.01 km)]=4.3º, south of due west. 21. Reading carefully, we see that the (x, y) specifications for each “dart” are to be interpreted as (Ax, Ay) descriptions of the corresponding displacement vectors. We combine the different parts of this problem into a single exposition. (a) Along the x axis, we have (with the centimeter unit understood) 30.0 + b, — 20.0 — 80.0 = —140, which gives by =-—70.0 cm. (b) Along the y axis we have www.ELSOLUCIONARIO.bl .com 86 CHAPTER 3 their respective second parts: B is 0.80 m at 30º north of cast and D is the unknown. The final position of Beetle 1 is A+B= (0.5 mji+ (0.8 m)(cos30º | +sin30º 5) = (1.19 m) 1+(0.40 m)j. The equation relating these is A+B=C+D, where C = (1.60 m)(cos50.0ºi+sin50.0º)) = (1.03 mji+ (1.23 m)j (a) We find D= A+B-C=(0.16 m)i+(-0.83 m)), and the magnitude is D = 0.84 m. (b) The angle is tan'(-0.83/0.16)=—79º, which is interpreted to mean 79º south of east (or 11º east of south). 29. Let 1, = 2.0cm be the length of each segment. The nest is located at the endpoint of segment w. (a) Using unit-vector notation, the displacement vector for point A is d,= Wb+9+7 += In(cos 60% + sin60º 5) + (ly 3) + y(cos120ºi+ sin! 20º 5) + (4, 5) =(0+ 3) À. Therefore, the magnitude of d, is |d, |= (2+3)(2.0 cm) = 7.5 cm. (b) The angle of d, is 6= tan “(d,,/d,)=tan!(0)=90º. (c) Similarly, the displacement for point B is do=W+V+j+pD+ô = (cos 60º] + sin 602) +(1, 1)+ lu(cos 60º] + sin60? 7) + Iy(cos 30%] + sin30º 5)+ (1, 1) = (0+43/ Dl i+(3/2+N Dl j. Therefore, the magnitude of d, is [ds |= /(2+43/27 +(3/24 3) = (2.0cm)(4.3)=8.6cm. (d) The direction of d, is www.ELSOLUCIONARIO.bl .com 87 2+43/2 d O,=tan” [5 = ad) =tan!(1.13)=48º. Bx 30. Many of the operations are done efficiently on most modem graphical calculators using their built-in vector manipulation and rectangular <> polar “shortcuts.” In this solution, we employ the “traditional” methods (such as Eg. 3-6). (a) The magnitude of à isa= (4.0 m) +(-3.0m) =5.0m. (b) The angle between à and the +x axis is tan! [(-3.0 m)/(4.0 m)] = -37º. The vector is 37º clockwise from the axis defined by 1. (c) The magnitude of b is b= (6.0 m) +(8.0 m) =10 m. (d) The angle between b and the +x axis is tan "(8.0 m)/(6.0 m)] = 53º. (e) d+b=(4.0 m+6.0 m) i+[(-30 m)+8.0 m]j =(10 mi +(5.0 m)j. The magnitude of this vector is [4+b E (10 m)+(5.0 m) =1Im; we round to two significant figures in our results. (£) The angle between the vector described in part (e) and the +x axis is tan [(5.0 my/(10 m)]=27º. (g) b-ã= (6.0 m-4.0 m) 1+[8.0 m—(-3.0 m)]j= (2.0m) i+ (1! m)j. The magnitude of this vector is [b-à | (2.0 m) +(11m) =11 m, which is, interestingly, the same result as in part (e) (exactly, not just to 2 significant figures) (this curious coincidence is made possible by the fact that à 1 b). (h) The angle between the vector described in part (g) and the +x axis is tan (1 m)/(2.0 m)] = 80º. (i) a-b=(4.0m-6.0 m) i+[(-3.0 m)-8.0m]j =(-2.0m) i+(-11 m)j.The magnitude of this vector is [db E (2.0 mP+(Cllm)=IIm. ()) The two possibilities presented by a simple calculation for the angle between the vector described in part (i) and the +x direction are tan [CI m/(-2.0 m)] = 80º, and 180º + 80º = 260º. The latter possibility is the correct answer (see part (k) for a further observation related to this result). www.ELSOLUCIONARIO.bl .com 88 CHAPTER 3 (k) Since a — b= E Db- a), they point in opposite (anti-parallel) directions; the angle between them is 180º. 31. (a) As can be seen from Figure 3-30, the point diametrically opposite the origin (0,0,0) has position vector a i+a j + a É and this is the vector along the “body diagonal.” (b) From the point (a, 0, 0), which corresponds to the position vector a 1, the diametrically opposite point is (0, a, a) with the position vectora j+ ak. Thus, the vector along the line is the difference -ai+aj +ak. a (c) If the starting point is (0, a, 0) with the corresponding position vector aj, the diametrically opposite point is (a, 0, a) with the position vector a i+ak. Thus, the vector along the line is the difference a i- aj +ak. (d) Tf the starting point is (a, a, 0) with the corresponding position vector a i+a 5 the diametrically opposite point is (0, 0, a) with the position vector a k. Thus, the vector along the line is the difference —a i- aj +ak. (e) Consider the vector from the back lower left corner to the front upper right comer. It isai+ ajra k. We may think of it as the sum of the vector ai parallel to the x axis and the vector aj +ak perpendicular to the x axis. The tangent of the angle between the vector and the x axis is the perpendicular component divided by the parallel component. Since the magnitude of the perpendicular component is Va? +a? =av2 and the magnitude of the parallel component is a, tan O = (av2)ta = 2. Thus 6 = 547º. The angle between the vector and each of the other two adjacent sides (the y and z axes) is the same as is the angle between any of the other diagonal vectors and any of the cube sides adjacent to them. () The length of any of the diagonals is given by Va? + a? +a? = 3. www.ELSOLUCIONARIO.bl .com 91 3€ (24x 8)=3(7.00i-8.00])-(44.01+16.05 +34.0k) =3[(7.00)(44.0)+(-8.00)(16.0) + (0)(34.0)]= 540. 39. From the definition of the dot product between À and B . A-B= ABcosO + we have coso- AB AB With A=6.00, B=7.00and À-B=14.0, cos0= 0.333, or 0=70.5º. 40. The displacement vectors can be written as (in meters) d, = (4.50 m)(cos 63º )+ sin 63ºf) = (2.04 m) j+ (4.01 m)k d, = (1.40 m)(cos 30ºi-+sin 30ºf) = (1.21 m)i+(0.70 m)k. (a) The dot product of d, and d, is d,:d, = (2.04]+4.01f) (1.211 40.708) = (4.01%)-(0.70k) = 2.81m?. (b) The cross product of d, and d, is dx d, = (2.04]+4.01k)x (1.212 +0.706) = (2.049). 21N-Éo) + (2.040.701 + (4.0D(1.21)j =(1431+4.86j-2.48)m?. (c) The magnitudes of d, and d, are d = (2.04 m)? +(4.01m) = 4.50 m d,=(1.21m)+(0.70 m) =1.40m. Thus, the angle between the two vectors is 8-cos! dd, - cos! — 28im -63.5º. dd, (4.50 m)(1.40 m) 41. Since ab cos 6= ab; + a;b, + a.b., www.ELSOLUCIONARIO.bl .com 92 CHAPTER 3 ab +ab, + ab, cos = cosg ab The magnitudes of the vectors given in the problem are [= (6.007 + (3.002 + (3.002 =5.20 b =|b|= [2.002 + (1.00? + (3.00 =3.74. The angle between them is found from (3.00) (2.00) + (3.00) (1.00) + (3.00)(3.00) (5.20) (3.74) cosy = = 0.926. The angle is 9= 22º. As the name implies, the scalar product (or dot product) between two vectors is a scalar quantity. It can be regarded as the product between the magnitude of one of the vectors and the scalar component of the second vector along the direction of the first one, as illustrated below (see also in Fig. 3-18 of the text): à-b=ab cosó = (a)(bcos 6) 42. The two vectors are written as, in unit of meters, d=4.045.0j)=d,ji+d,j, d;=-3.0+4.0)= di+d,j) (a) The vector (cross) product gives d xd, = (dd, - dd, = [(4.0)(4.0)-(5.0)(-3.0)]k=31 k (b) The scalar (dot) product gives d d,= dd, +d,d,, = (4.0(-3.0)+(5.0)(4.0) = 8.0. (c) (d+d,)-d,=d, d+ d2=8.0+(3.0" +(4.02 = 33. www.ELSOLUCIONARIO.bl .com 93 (d) Note that the magnitude of the d, vector is "/16+25 = 6.4. Now, the dot product is (6.4)(5.0)cos6 = 8. Dividing both sides by 32 and taking the inverse cosine yields O = 75.5º. Therefore the component of the d, vector along the direction of the d, vector is 6.4cos0s 1.6. 43. From the figure, we note that c L b , which implies that the angle between c and the +x axis is 0+ 90º. In unit-vector notation, the three vectors can be written as a a, b=bi+b,)=(bcoso)i+(bsin9)j E=ci+c,)=[ecos(9+90º)li + [csin(o +90] The above expressions allow us to evaluate the components of the vectors. (a) The x-component of à isa;=acos0º=a=3.00m. (b) Similarly, the y-compometof à isa,=asin0º=0. (c) The x-component of b is by =b cos 30º = (4.00 m) cos 30º = 3.46 m, (d) and the y-component is b, = b sin 30º = (4.00 m) sin 30º = 2.00 m. (e) The x-component of c isc;=c cos 120º = (10.0 m) cos 120º =-5.00 m, (f) and the y-component is c, = c sin 30º = (10.0 m) sin 120º = 8.66 m. (8) The fact that E = pã + qb implies c= ci+ cj = pla d)+ q(bi+ bd) =(pa, + qbdi+ qb, or c-pa+gb c=qb Substituting the values found above, we have -5.00m= p (3.00m)+q (3.46m) 8.66 m= q (2.00 m). Solving these equations, we find p = 6.67. (h) Similarly, q = 4.33 (note that it's easiest to solve for q first). The numbers p and q have no units. www.ELSOLUCIONARIO.bl .com