Solution Manual For College Physics With Masteringphysics, 7th Edition

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CHAPTER 1MEASUREMENT AND PROBLEM SOLVINGRemind students that their answers to odd-numbered exercises may be slightly different from those given herebecause of rounding. Refer to Problem-Solving Hint: The "Correct" Answer in Chapter 1.Multiple Choice:1.(c).2.(b).3.(c).4.(b). The gram is the mass of one cm3of water, so the kilogram (1000g) and tonne (1000 kg) are all related to avolume of water. Only the pound is not related to a volume of water.5.(b).6.(b).7.(a).8.(d). A quart is slightly less than a liter, 2000 μL is 2 mL, and 2000 mL is 2 L, so 2000 mL has the greatest volume.9.(c). Micro-is 10-6, centi-is 10-2, nano-is 10-9, and milli-is 10-3, so nano-is the smallest.10.(d).11.(d).12.(c).13.(a). The kg is a unit of mass and the lb is a unit of force. At the surface of the Earth, 1 kg isequivalentto 2.2 lb,which means that a 1-kg object weighs 2.2 lb.14.(c). Because the μL is the smallest volume of those listed, you will need more of them to make up any given volume,and hence alarger numberof μL.

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2College Physics Seventh Edition: Instructor Solutions Manual15.(a).16.(c).17.(b).18.(a).19.(c).20.(d).Conceptual Questions:1.Because there are no more fundamental units. The units of all quantities can be expressed in terms of thefundamental, or base, units.2.Weight depends on the force of gravity, which can vary with location.3.The mean solar day replaced the original definition. No, because this has been replaced by atomic clocks.4.One major difference is the decimal versus duodecimal basis. Another difference is that SI basic units are meters,kilograms and seconds, whereas the British system uses feet, pounds and seconds.5.No, because 3 cm is over an inch. Ladybugs are on the order of several mm long. Yes, a 10-kg salmon would weighon the order of 22 pounds, which is typical for a medium sized fish like that.6.This is by definition31 L1000 mL and 1 L1000 cm .==7.The metric ton is actually a misnomer since it is not a weight unit but a mass unit, defined as the mass of 1 cubicmeter of water. But3l m1000 L=and 1 L of water has a mass of 1 kg. So one metric ton is equal to 1000 kg.8.No, it only tells if the equation is dimensionally correct. You could be missing (or have extra) dimensionlessnumbers suchas ½ or π.9.No, unit analysis can only tell if it is dimensionally correct. Dimensionless factors such as π may be missing.10.By putting in units and solving for those of the unknown quantity.

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Chapter 1Measurement and Problem Solving311.πis dimensionless and therefore also unitless because it is defined as the ratio of two lengths, the circumference tothe diameter of a circle.12.No, they are not the same. An equivalent statement is not dimensionally correct.13.Yes, whether you multiply or divide should be consistent with unit analysis for the final answer.14.Give him 2.54 cm and he’ll take 1.61 km, since 2.54 cm = 1.00 in. and 1.61 km = 1.00 mi.15.To provide an estimate of the accuracy of a quantity.16.No, there is always one doubtful digit, the last digit.17.For (a) and (b), the result should have the least number of significant figures. For (c) and (d), the result should havethe least number of decimal places.18.Because 5 is midway between the upper and lower extremes of 9 and 1.19.See the six steps as listed in Chapter 1.20.No, since an order of magnitude calculation is only an estimate.21.The accuracy of the answer is expected to be within a factor of 10 of the correct answer.22.Approximate the area of skin-covered body parts using familiar geometric shapes. For example, use a sphere for thehead and cylinders for the arms, legs, and torso.23.Since a liter is close to a quart and there are four quarts in a gallon, this volume is about 75 gallons which isnotreasonable for a car (but might be for a large truck or other large vehicle such as an RV).24.No,since 30 km/h ≈ 19 mi/h < 25 mi/h.

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4College Physics Seventh Edition: Instructor Solutions ManualExercises:1.The decimal system (base 10) has a dime worth 10¢ and a dollar worth 10 dimes, or 100¢. By analogy, aduodecimal system would have a dime worth 12¢ and a dollar worth 12 “dimes,” or $1.44 in decimal dollars. Thena penny would be1144of a dollar.2.(a) Different ounces are used for volume and weight measurements. 16 oz = 1 pt is a volume measure and 16 oz = 1lb is a weight measure.(b) Two different poundunits are used. Avoirdupois lb= 16 oz‚troy lb = 12 oz.3.(a) 61 MB40,000,000 bytes10bytes=40 MB(b) 1 L0.5722 mL1000 mL=-45.72210L(c) 100 cm2.684 m1 m=268.4 cm(d)1 kilobuck5, 500 bucks1000 bucks=5.5 kilobucks4.That is because1 nautical mile = 6076 ft= 1.15 mi. A nautical mile is larger than a (statute) mile.5.1 g water(25 cm)(35 cm)(55 cm)31 cmwater= 48,100 g =48 kg6.Letxbe the length of each side of the cube.=33.788 L1000 cm3(1 qt)4 qt1 Lx= 947 cm3()==1/3947 cm9.82 cmx7.(a) (20 cm)3(1 L/1000 cm3) =8.0 L(b)()===31 m31000 kg/m(8.0 L)8.0 kg1000 LmV

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Chapter 1Measurement and Problem Solving58.(Length)=(Length)+(Length)(Time)(Time)=(Length)+ (Length) .9.(d).10.m2= (m)2= m2.11.Sinceax2is in meters,a=mm2=1/m.Sincebxis in meters,b=mm=dimensionless.cis inm.12.Yes, since [m3] = [m]3= [m3].13.No.V= 4r3/3 = 4(8r3)/24 = 4(2r)3/24 =d3/6. So it should beV=d3/6.14.Sincep=v2, the unit of pressureis (kg/m3)(m/s)2=kg/(ms2).No, this does not prove that this relationship is physically correct,because there might be a coefficient in theequation.15.Yes, becausem2=12m(m+m) =m2+m2.16.(a) SinceF=ma, newton = (kg)(m/s2) =kg·m/s2.(b)Yes, because (kg)m2/s2m= kg·m/s2(F=mv2/r).17.(a)The unit of angular momentum is (kg)(m/s)(m) =kgm2/s(b)The unit ofL22mr2is(kgm2/s)2kgm2=22kg ms, which is the unit of kinetic energy,K.(c) The unit of moment of inertia is (kg)(m)2=kgm2.18.(a) SinceE=mc2, the units of energy = (kg)(m/s)2=kgm2/s2.(b)Yes, because (kg)(m/s2)(m) = kgm2/s2(E=mgh).19.130 ft = (130 ft)1 m3.28 ft=39.6 m.

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6College Physics Seventh Edition: Instructor Solutions Manual20.(a)The answer is(4)centimeter, as it is the smallest unit among those listed.(b) Since 1 ft = 30.5 cm, 6.00 ft = (6.00 ft)30.5 cm1 ft=183 cm.21.40000 mi = (40000 mi)1609 m1 mi= 64400000 m.So64400000 m1.75 m=37000000 times.22.(a) Since 1 gal = 3.785 L < 4 L, or ½ gal < 2 L, ½ gal holds(3)lesssoda.(b) 0.5 gal = (0.5 gal)3.785 L1 gal= 1.89 L.2 L−1.89 L = 0.11 L.So2 L by 0.11 L more.23.(a) 300 ft = (300 ft)1 m3.28 ft= 91.5 m.160 ft = (160 ft)1 m3.28 ft= 48.8 m.So the dimensions are91.5 m by 48.8 m.(b) 11 in. = (11.0in.)2.54 cm1 in.= 27.9 cm.11.25 in. = (11.25 in.)2.54 cm1 in.= 28.6 cm.So the length is27.9 cm to 28.6 cm.24.From Exercise 1.23, themetricfieldis larger.Acurrent= (91.4 m)(48.8 m) = 4.46103m2.Ametric= (100 m)(54 m) = 5.4103m2.So the difference is 5.4103m2–4.46103m2=229.410m.25.31 qt3.788 L1000 cm1 g(1 pt)32 pts4 qt1 L1 cm=474 g26. 1 gal4 qt(10 L)3.788 L1 gal=10.6 qt27.  2 yd3 ft1 m(175 fathoms)1 fathom1 yd3.281 ft=320 m28.763 mi/h = (763 mi/h)1609 m1 mi1 h3600 s=341 m/s.

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Chapter 1Measurement and Problem Solving7(b) 300 ft = (300 ft)1 m3.28 ft= 91.46 m.So the time is91.46 m341 m/s=0.268 s.29.(a) 1 km/h = (1 km/h)1000 m1 km1 h3600 s= 0.8 m/s < 1 m/s.1 ft/s = (1 ft/s)1 m3.28 ft= 0.30 m/s < 1 m/s.1 mi/h = (1 mi/h)1609 m1 mi1 h3600 s= 0.45 m/s < 1 m/s.So(1)1 m/srepresents the greatest speed.(b) 15.0 m/s = (15.0 m/s)1 mi1609 m3600 s1 h=33.6 mi/h.30.(a) 10 mi/h = (10 mi/h)1.609 km1 mi=16 km/h for each 10 mi/h.(b) 70 mi/h = (70 mi/h)1.609 km1 mi=1.1102km/h.31.(a)1 kg = 2.2 lb (equivalent). So 170 lb = (170 lb)1 kg2.2 lb=77.3kg.(b)The density of water is 1000 kg/m3.=mV,V=m=377.3 kg1000 kg / m=30.0773 m or about 77.3L.Using liter avoids the small decimals.32.The circumference of theMoonofdiameter3500 km, isdd=(3500km) =1.1104km.Sothe answer isyes.1.0105km1.1104km=9.1 times.33.In one day, there are 2460 min = 1440 min.So the volume of blood pumped per day is(60 beats/min)(1440 min/day)(75 mL/beat) =6.5106mL/day =6.5103L/day.

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8College Physics Seventh Edition: Instructor Solutions Manual34.(a) 2 fl. oz = (2 fl. oz)473 mL16 fl. oz=59.1 mL.(b) 100 g = (100 g)14.5 oz411 g=3.53 oz.35.(250 mL/min)(4.5106/mm3)1 min60 s1 L10−3mL106mm31 L=1.91010/s.36.18 in. = (18 in.)2.54 cm1 in.= 45.7 cm.5 ft, 6 in. = 66 in. = (66 in.)2.54 cm1 in.= 167.6 cm.So the growth per year is167.6 cm–45.7 cm20=6.1 cm.37.The Earth rotates 360° in 24 hr (= 2460 min = 1440 min), so in one minute it will rotate 1/1440 of 360°, which is0.250° =15.0 min of arc.38.(a) 13.6 g/cm3= (13.6 g/cm3)1 kg1000 g1003cm1 m=1.36104kg/m3.(b)=mV,m=V= (13.6 g/cm3)(0.250 L)1000 cm31 L= 3.40103g =3.40 kg.39.(a)The volume is equal toV=Ah=r2h =(125 m)2(10 ft)(0.305 m/ft) =1.5105m3.(b)Thewaterdensity of is 1000 kg/m3.=mV,m=V= (1000 kg/m3)(1.5105m3) =1.5108kg.(c)One kg is equivalent to 2.2 lb.1.5108kg = (1.5108kg)2.2 lb1 kg=3.3108lb.40.L= 300 cubits = (300 cubits)0.5 yd1 cubit3 ft1 yd1 m3.28 ft= 137 m.W= 50.0 cubits = 22.9 m,H= 30.0 cubits = 13.7 m.So the dimensions are137 m22.9 m13.7 m.(b)V=LWH= (137 m)(22.9 m)(13.7 m) =4.30104m3.41.50500m = (50500m)1 cm10000m=5.05 cm=5.0510−1dm=5.0510−2m.42.0.001 m or 1 mm.

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Chapter 1Measurement and Problem Solving943.(a)4.(b)3.(c)5.(d)2.44.(a)1.0 m.(b)8.0 cm.(c)16 kg.(d)1.510−2s.45.(a)96(b)0.0021(c)9400(d)0.0003446.(b)and(d);(a)has fourand(c)hassix.47.A=LW= (0.274 m)(0.222 m) =6.0810−2m2.48.V=LWH=(1.3 m)(3.281 ft/m)(1.05 m)(3.281 ft/m)(0.67 m)(3.281 ft/m) =32 ft3.49.(a) The smallest division is(2)cm, as the last digit is estimated.(b)A=LW= (1.245 m)(0.760 m) =0.946 m2.50.(a)(2)Three, since the height has only three significant figures.(b) The area is the sum of that of the top, the bottom, and the side. The side of thecan is a rectangle with a length equal to the circumference and width equal to theheight of the can.A=+24d+24d+Ch=24d+24d+ (d)h=(12.559 cm)24+(12.559 cm)24+(12.559 cm)(5.62 cm) =469cm2.51.(a) 12.634 + 2.1 =14.7.(b) 13.5−2.134 =11.4.(c)(0.25 m)2=0.20 m2.(d)2.37/3.5=0.82.52.(a)The answer is(1)zero, since 38 m has zero decimal places.(b) 46.9 m + 5.72 m–38 m =15 m.53.(a)v=xt=8.5 m2.7 s= 3.1 m/s,p=mv= (0.66 kg)(3.1 m/s) =2.0 kg·m/s.

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10College Physics Seventh Edition: Instructor Solutions Manual(b)p=mxt=(0.66 kg)(8.5 m)2.7 s=2.1 kg·m/s.(c)No, the results are not the same. The difference comes fromroundingdifference.54.According tothePythagorean theorem,c=a2+b2=(37 m)2+(42.3 m)2=56 m.55.Since 1 m = 100 cm, (1 m)3=(100 cm)3or 1 m3= 106cm3.=mV,m=V= (0.10 g/cm3)(1 m3)106cm31 m3=1.0105g =100 kg.56.() 33.281 ft1 cord3.0 m1 m8.0 ft4.0 ft4.0 ft=0.83 cord57.(a) The percentage is(18 g)(9 cal/g)310 cal= 0.52 =52%.(b) Total fat =18 g0.28=64 g;saturated fat =7 g0.35=20 g.58.(a) One sheet has two pages. 860 pages have430 sheets.The average thickness per sheet is3.75 cm430 sheets=8.7210−3cm.(b)1 cm100 sheets=about10−2cm.59.==343EmmVR=()=2433365.9810kg5.410kg/m46.410m360.(a) From the sketch, it is clear that the stadium is(4)south of west, relative to yourhouse.(b) Consider the right triangle on the bottom of the sketch. The two sidesperpendicular to each other are 500 m each.Using Pythagorean theorem,d=(500 m)2+(500 m)2=707 m.500 m1500 m1000 md

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Chapter 1Measurement and Problem Solving1161.According tothePythagorean theorem, (1.0 m)2= (0.50 m)2+d2.Sod=(1.0 m)2−(0.50 m)2=0.87 m.62.The12-in.pizza is a better buy. A better buy gives you morearea(more pepperoni) per dollar,and the area of apizza depends on the square of the diameter.For the 9.0 in.:(4.5 in.)2$7.95=8.0 in.2/dollar.For the 12 in.:(6.0 in.)2$13.50=8.4 in.2/dollar.63.The area of a 12-in. pizza is πR2= π(6.0 in.)2= 113 in.2The area of two 8-in. pizzas is 2(πR2) = 2π(4.0 in.)2= 101in.2This is not such a good deal!64.For the center circle:A=r2=(0.640cm)2= 1.3 cm2.For the outer ring:A=(r22–r21) =[(1.78 cm)2–(1.66 cm)2] = 1.3cm2.Soit is thesame area for both,1.3 cm2if calculated to two significant figures.65.t=xv=31 mi75 mi/h= 0.41 h =25 min.66.One liter has 1000cm3,and each cm3has 1000 mm3,so 1 liter has 1.0106mm3.(1 cm3) = (10 mm)3= 1000 mm3The total number of white cells in 5.0 L of blood is(7 000 /mm3)(5106mm3) =3.51010white cells.The total number of platelets in 5.0 L of blood is(250 000 /mm3)(5106mm3) =1.31012platelets.67.(a)The number of hairs lost in a month is (65 hairs/day)(30 days) =1950 hairs.(b)15% bald means 85% with hair. So in one day, the “bald is beautiful” person loses(0.85)(65 hairs) = 55 hairs.In one year, the total is (365)(55 hairs) =2.0104hairs.0.50 m0.50 m1.0 md

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12College Physics Seventh Edition: Instructor Solutions Manual68.(a) Sinced= (13 mi) tan 25and tan 25< 1 (tan 45= 1),dis(1)less than13 mi.(b)d= (13 mi) tan 25=6.1 mi.69.(a) It will be(3)less than the 190 mi/h. More time spent at lower speeds, so affect average speed to be belowaverage of all speeds.(b) The time intervals for each lap is, respectively,Lap 1:t1=2.5 mi160 mi/h= 0.015625 h;Lap 2:t2=2.5 mi180 mi/h= 0.013889 h;Lap 3:t3=2.5 mi200 mi/h= 0.012500 h;Lap 4:t4=2.5 mi220 mi/h= 0.011364 h.So the total time isttotal=t1+t2+t3+t4= 0.053378 h.Therefore the average speed for four laps is4(2.5 mi)0.053378 h=187 mi/h.70.118 mi = (118 mi)1609 m1 mi= 105m,307 mi = (307 mi)1609 m1 mi= 105m,279 ft = (279 ft)1 m3.28 ft= 102m.SoV=LWD(105m)(105m)(102m) =about1012m3.71.(a)The answer is(2)between 5and 7.The sketch below illustrateswhyitis the answer.(b)The elevation for the 2.0 km segment is(2.0 km) sin 5=0.174 km.The horizontal distance for the 2.0 km segment is(2.0 km) cos 5= 1.99 km.The elevation for the 3.0 km segment is(3.0 km) sin 7= 0.366 km.The horizontal distance for the 3.0 km segment is(3.0 km) cos 7= 2.98 km.So the tangent of the net angle of rise is tan=0.174 km + 0.366 km1.99 km + 2.98 km= 0.109.Therefore= tan−1(0.109) =6.2.72.d=xtan 30= (50 m−x) tan 40= (50 m) tan 40−xtan 40.x=(50 m)tan 40tan 30+ tan 40= 29.6 m.Sod= (29.6 m) tan 30=17 m.73.Expressing the area of the horse pasture,AH, in terms ofhgivesd3040x50 m−xd2513 mi

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Chapter 1Measurement and Problem Solving1321(200 m)2(200 m)233=−=−HhhAhhhThe total area of his lot is==21 200 m32(200 m)sin 60(200 m)224TotA. The area for the horse pasture,AH, is1/3 of the total area of the triangular lot, giving2213(200 m)(200 m)343−=hhSolving forhgivesh= 314.6 m andh= 31.78 m. Sincehcannot be larger than the side of the triangle, we discardthe first solution, givingh=31.8 m.74.(a) The volume of the drilled hole isV=r2L=(0.0100 m)2(8.00 in.)(0.0254 m/in.) = 6.3810−5m3.The density of water is 1000 kg/m3, so the density of lead is (11.4)(1000 kg/m3) = 1.14104kg/m3.=mV,m=V= (1.14104kg/m3)(6.3810−5m3) =0.727 kg.(b) The total volume of the brick is(2.00 in.)(0.0254 m/in.)(4.00 in.)(0.0254 m/in.)(8.00 in.)(0.0254 m/in.) = 1.04910−3m3.The percentage of the original lead remaining in the brick is1.04910−3m3−6.3810−5m31.04910−3m3=93.9%.(c) The mass of the plastic ismP= [(2)(1000 kg/m3)](6.3810−5m3) = 0.1276 kg.The mass of the lead brick with the hole drilled ismL= (1.14104kg/m3)(1.04910−3m3−6.3810−5m3) = 11.23 kg.Therefore the overall density is11.23 kg + 0.1276 kg1.04910−3m3=1.08104kg/m3.75.(a) Callingrthe radius of the inner surface and using the fact thatVr= 0.900VTot, we have=3344(0.900)(20.0 cm)33r, which givesr=19.3 cm.(b)š==224(20.0 cm)1.074 (19.3 cm)outerinnerAA, or the outer area is 7% larger than the inner area.

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14College Physics Seventh Edition: Instructor Solutions Manual76.(a)tan 30° =y/50 km→y= (50 km) tan 30° =28.9 km(b) cos 30° = (50 km)/d→d= (50 km)/cos 30° =57.7 km(c) Callingthe angle subtended byyfrom point C, which is 70 km from point A, we have: tan=y/(70 km),which gives=22.4° north of due west.77.(a) At constant speed, the distancextraveled isx = vt, which givest = x/v. Sincevis the same in both cases, takingthe ratio oftfor both trips gives====circumfcircumfcircumfdiamdiamdiamdiamdiamxtxxvxtxxttcircumf=πtdiam=π(30.0 s) =94.2 s(b) The diameter of the pool isx=vt= (0.500 m/s)(30.0 s) = 15.0 m, its radius isR= 7.50 m and its depth isd=1.50 m. The volume isV= (πR2)d=π(7.50 m)2(1.50 m) = 265 m3. Converting this result to gallons gives=34-331 L1 gal(265 m )7.0010 gal3.788 L10m78.(a)m =V= (9.0 g/cm3)(1.0 cm)(2.0 cm)(4.0 cm) = 72 g =0.072 kg(b) Callingxthe length of each side of the cube and using the fact thatV = m/gives==33144 g9.0 g/cmmx→x=2.5 cm50 kmEdBy30°AEast

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CHAPTER 2KINEMATICS: DESCRIPTION OF MOTIONMultiple ChoiceQuestions:1.(a).2.(d). Choice (a) can never be true; choices (b) and (c) are sometimes true; only choice (d) is always true.3.(c).4.(c).5.(c). Since the magnitude of the displacement vector is a distance and the magnitude of the velocity vector is thespeed, choice (c) is correct.6.(d).7.(d).8.(c). A negative acceleration only means that the acceleration is pointing in the negative direction. If an object ismoving in the positive x-direction, the velocity of the object decreases. But if it is moving in the negative x-direction, its velocity will increase so it will speed up.9.(d). Any change in either magnitude or direction results in a change in velocity. The brake and gearshift change themagnitude, and the steering wheel changes the direction.10.(b). Since acceleration is the rate of change of the velocity, a constant acceleration implies a constant rate of changeof the velocity, making (b) the correct choice.11.(c). The speed of a decelerating object is decreasing, which can only happen if the acceleration is opposite to thevelocity.12.(c). In both cases, the ratio of the velocity change to the time interval for that change is the same, which means theyhave the same magnitude acceleration.13.(c).14.(d). The graph of x as a function of t is a parabola, depending on the square of the time.15.(a). Since0=+,vvat+++===00(0)1.222vvatvat16.(d).

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16College Physics Seventh Edition: Instructor Solutions Manual17.(d). Free fall is motion under the influence of gravity alone, and the acceleration is g. The initial velocity does notaffect the acceleration.18.(c). It accelerates at 9.80 m/s2, so it increases its speed by 9.80 m/s during each second.19.(a). The acceleration is not zero; it is 9.80 m/s2downward.20.(c). It always acceleratesat 9.80 m/s2downward.Conceptual Questions:1.Yes, for a round-trip. No; distance is always greater than or equal to the magnitude of displacement.2.No final position can be given. It may be anywhere from 0 to 750 m from the start.3.The distance traveled is greater than or equal to 300 m. The object could travel a variety of ways as long as it endsup at 300 m north. If the object travels straight north, then the minimum distance is 300 m.4.No, this is generally not the case. The average velocity can be zero (e.g. a round trip), while the average speed isnever zero.5.Yes, this is possible. The jogger can jog in the opposite direction during part of the jog (negative instantaneousvelocity) as long as the overall jog is in the forward direction (positive average velocity).6.Yes, although the speed of the car is constant, its velocity is not, because of the change in direction. A change invelocity means that there is acceleration, and the velocity (a vector quantity) can change by either changingdirection, magnitude or both.7.Not necessarily. The change in velocity is the key. If a fast-moving object does not change its velocity, itsacceleration is zero. However, if a slow-moving object changes its velocity, it will have some non-zero acceleration.8.Not necessarily. A negative acceleration can cause an increase in speed if the velocity is also negative (that is, if thevelocity is in the same direction as the acceleration).9.In part (a), the object accelerates uniformly first, maintains constant velocity (zero acceleration) for a while, andthen accelerates uniformly at the same rate as in the first segment. In (b), the object accelerates uniformly.10.The final velocity isovsince an equal amount of time is spent on acceleration and deceleration and both of thesehave the same magnitude.

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Chapter 2Kinematics: Description of Motion1711.If (assuming uniform accelerations) we apply the formulav2=vo2+ 2a(x–x0), we see that both cars have thesame initial speed (v0= 0) and the same final speedv, so the quantitya(x–x0) must be the same for both of them.Since car B travels twice as far as car A, its acceleration must therefore be half as large as that of car A, which tellsus that=AB2aa. Another way to view this problem is the following. Both cars have the sameaveragespeed buttravel unequal distances. Car A will take less time to reach the line because it has less distance to travel. Since thechangein velocities are the same, car A will have a higher rate of change of velocity, thus A’s acceleration is greaterthan B’s. Since car A travels half the distance as B at the same average speed as A, it will take half as long to finishas B. Thus A will have twice the acceleration as B.12.It is zero because the velocity is constant.13.Not necessarily because evenif the acceleration is negative,the object can still have positive velocity (meaning it isslowing) and the result could be a positive value forx.14.Consider the displacement()oxx−as one quantity; there are four quantities involved in each of the kinematicequations (Eqs. 2.8, 2.10, 2.11, and 2.12). All but one of the four must be known before one can solve for anyunknown.15.Yes, if the displacement is negative meaning the object accelerates to the left.16.When it reaches the highest point, its velocity is zero (velocity changes from up to down, so it is zeroat that instant),but its acceleration is29.80 m/sdownward because the velocity is changing direction, signifying acceleration (dueto gravity).17.No, since one value of the instantaneous velocity doesnottell you if the velocity is changing. It could be zero justfor an instant and not zero either before or after that instant, thus it could be changing and the object could beaccelerating. You need two values of instantaneous velocity to determine if an object is accelerating.18.The ball moves at constant velocity because there is no gravitational acceleration in deep space. If an object’sacceleration is zero, thenvis a constant, in magnitude and direction, but is not necessarily zero.19.Since the first stone has been accelerating downward for a longer time, it will always have a higher speed and thusas time goes by it will have fallen further and thus the gap between them (y) will increase.20.First, the gravitational acceleration on the Moon is only1/6of that on the Earth, org=g6.MEHence droppedobjects take longer to reach the surface than on the Earth, and tossed objects will go higher and stay in flight longerthan on Earth. Secondly, there is no air resistance on the Moon, which means that all objects, regardless of massand/or shape, will accelerate at the same rate, whereas this is only an approximation that works well for smallmassive objects on the Earth.

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18College Physics Seventh Edition: Instructor Solutions ManualExercises:1.Displacement is the change in position.Therefore the magnitude of the displacement for half a lap is300 m.For a full lap (the car returns to its starting position), the displacement iszero.2.vav=d/t, whered= 80 km + 50 km = 130 kmt1=d1/v1= (80 km)/(100 km/h) = 0.800 ht2=d2/v2= (50 km)/(75 km/h) = 0.667 ht=t1+t2= 0.800 h + 0.667 h = 1.467 hvav=d/t= (130 km)/(1.467 h) =89 km/h3.t=d/vav, where = 100 yd3 ft1 m9.0 s1 yd3.281 ftavv= 10.16 m/st= (100 m)/(10.16 m/s) =9.8 s4.(a)s=dt=(0.30 km)(1000 m/km)(10 min)(60 s/min)=0.50 m/s.(b)s1= 1.20s= 1.20(0.50 m/s) = 0.60 m/s.Sot=ds1=300 m0.60 m/s= 500 s =8.3 min.5.1 cc = 1 mL.This is analogous to average speed.t=ds=500 mL4.0 mL/min=125 min.6.s=dt=2(25 m)[2(0.50min)+ 4.0 min](60 s/min)=0.17m/s.7.The time going istgoing=d/v1= (300 km)/(75 km/h) = 4.00 hThe time returning istreturn= d/v2= (300 km)/(85 km/h) = 3.53 hThe average speed isvav=2d/ttotal= (600 km)/(4.00 h + 3.53 h + 0.50 h) =75 km/hThe average velocity iszerobecause the net displacement is zero.8.(a) The answer is(2)greater thanRbut less than 2R. For any right triangle, thehypotenuse is always greater than any one of the other two sides (R) and less than the sumof the sum of the other two sides (R+R= 2R).(b)d=22(50m)(50m)+=71 m.50 m50 md150 m150 m

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Chapter 2Kinematics: Description of Motion199.(a)===   150 km1.43 h = 1.4 hmi1 km65 h0.6214 midtv(b)===   150 km1.165 hmi1 km80 h0.6214 midtvThe time you wouldsaveis 1.43 h–1.165 h =0.27 hor about16 min.10.(a) The average velocity is(1)zero,because the displacement is zero for a complete lap.(b)s=dt=2rt=2(500 m)50 s=63 m/s.11.(a) The magnitude of the displacement is(3)between 40 m and 60 m,because any sideof a triangle cannot be greater than the sum of the other two sides. In this case, looking atthe triangle shown, the two sides perpendicular to each other are 20 m and 40 m,respectively. The magnitude of the displacement is the hypotenuse of the right triangle, soit cannot be smaller than the longer of the sides perpendicular to each other.(b)d=(40 m)2+(50m−30 m)2=45 m.=tan−−150 m30 m40 m=27west of north.12.(a)s=dt=2(7.1 m)2.4 s=5.9 m/s.(b)Average velocity iszero, because the ball is caught at the initial heightsodisplacement is zero.13.(a)s=dt=27 m + 21 m(30 min)(60 s/min)=2.7 cm/s.(b) The displacement isx=(27 m)2+(21 m)2= 34.2 m.v=xt=34.2 m(30 min)(60 s/min)=1.9 cm/s.14.(a)v=xt,soABv=1.0 m–1.0 m1.0 s–0=0;BCv=7.0 m–1.0 m3.0 s–1.0s=3.0 m/s;CDv=9.0 m–7.0 m4.5 s–3.0 s=1.3 m/s;DEv=7.0 m–9.0 m6.0 s–4.5s=−1.3 m/s;EFv=2.0 m–7.0 m9.0 s–6.0s=−1.7 m/s;FGv=2.0 m–2.0 m11.0 s–9.0 s=0;BGv=2.0 m−1.0 m11.0 s−1.0 s=0.10 m/s.(b)The motion of BC‚CD‚and DE are not uniform,since they are not straight lines.(c) The object changes its direction of motion at point D. So it has to stop momentarily,andv=0.50 md40 m30 m

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20College Physics Seventh Edition: Instructor Solutions Manual15.Uses=dtandv=xt.(a)0-2.0 ss=2.0 m–02.0 s–0=1.0 m/s;2.0 s-3.0 ss=2.0 m–2.0m3.0 s–2.0=0;3.0 s-4.5 ss=4.0 m–2.0m4.5 s–3.0 s=1.3 m/s;4.5 s-6.5 ss=4.0 m–(–1.5 m)6.5 s–4.5 s=2.8 m/s;6.5 s-7.5 ss=–1.5 m–(–1.5m)7.5 s–6.5 s=0;7.5 s-9.0 ss=0–(–1.5 m)9.0 s–7.5 s=1.0 m/s;(b)0-2.0 sv=2.0 m–02.0 s–0=1.0 m/s;2.0 s-3.0 sv=2.0 m–2.0m3.0 s–2.0=0;3.0 s-4.5 sv=4.0 m–2.04.5 s–3.0 s=1.3 m/s;4.5 s-6.5 sv=–1.5 m–4.0m6.5 s–4.5s=−2.8 m/s;6.5 s-7.5 sv=–1.5m–(–1.5 m)7.5 s–6.5 s=0;7.5 s-9.0 sv=0–(–1.5 m)9.0 s–7.5 s=1.0 m/s.(c)v1.0 s=0-2.0 ss0-2.0 s=1.0 m/s;v2.5 s=2.0 s-3.0 ss=0;v4.5 s=0since the object reverses its direction of motion;v6.0 s=4.5 s-6.5 ss=−2.8 m/s.(d)v4.5 s-9.0 s=0–4.0m9.0 s–4.5s=−0.89 m/s.16.(a) The magnitude of the displacement isd=(90.0 ft)2+(10.0 ft)2=90.6 ft.=110.0tan90.0−=6.3above horizontal.(b)v=90.6 ft at 632.5 s=36.2 ft/s at 6.3.(c) Average speed depends on the total path length, which is not given.The ball might take a curved path.17.(a) Att= 0,x=a=10 m(b)x=x4–x2=a–b(4.0 s)2–[a–b(2.0 s)2] = (0.50 m/s2)[(2.0 s)2–(4.0 s)2]x=−6.0 m(c) At the origin,x= 0: 0 =a–bt2==210 m0.50 m/satb=4.5 s18.(a)x= 3(2.0 s)2m–0 = 12 m, andt= 2.0 s, sovav=x/t= (12 m)/(2.0 s) =6.0 m/s(b)x= 3(4.0 s)2m–3(2.0 s)2m = 36 m, andt= 2.0 s, sovav=x/t= (36 m)/(2.0 s) =18 m/s333330.0 yd = 90.0 ft3333d3333333310.0 ft

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Chapter 2Kinematics: Description of Motion2119.d= 3.5 cm−1.5 cm = 2.0 cm.s=dt,t=2.0 cm2.0cm/mo=1 month.20.The minimum speed iss=dt=675 km7.00 h= 96.4 km/h =59.9 mi/h.No, she does not have to exceed the 65 mi/h speed limit.21.(a)See the sketch on the right.d=22(400km)(300km)+=500 km.=1300tan400−=37east of north.(b)t= 45 min + 30 min = 75 min = 1.25 h.v=xt=500 km37east of north1.25 h=400 km/h37east of north.(c)s=dt=400 km + 300km1.25h=560km/h.(d)Sincespeed involves total distancewhich is greater than the magnitude of the displacement, the average speedisnot equal to the magnitude of the average velocity.22.To the runner on the right, the runner on the left is running at a velocity of+4.50 m/s–(−3.50 m/s) = +8.00 m/s.So it takest=xv=100 m8.00 m/s=12.5 s.They meet at (4.50 m/s)(12.5 s) =56.3 m(relative torunner on left).23.15.0 km/h = (15.0 km/h)1000 m1 km1 h3600 s=4.167m/s,65.0 km/h = 18.06 m/s.Soa=vt=18.06 m/s–4.167m/s6.00 s=2.32m/s2.24.60 mi/h = (60 mi/h)1609 m1 mi1 h3600 s= 26.8 m/s.a=vt=26.8 m/s–03.9 s=6.9 m/s2.25.t=26.8 m/s–07.2 m/s2=3.7 s.26.(a) The direction of the acceleration vector is(2)opposite to velocityas the object slows down.dd1d2

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22College Physics Seventh Edition: Instructor Solutions Manual(b) 40.0 km/h = (40 km/h)1000 m1 km1 h3600 s= 11.1 m/s.Soa=vt=0–11.1m/s5.0 s=−2.2 m/s2or−2.2 m/seach second.The negative sign indicates that the acceleration vector is inopposite direction of velocity.27.75 km/h = (75 km/h)1000 m1 km1 h3600 s=20.8 m/s,30 km/h =8.33m/s.Soa=vt=8.33m/s–20.8m/s6.0 s=−2.1m/s2.The negative sign indicates that the acceleration vector is in opposite direction of velocity.28.(a) Once the object is released, the only force acting on it is gravity, so its acceleration is29.80 m/sdownward.(b) At the instant the object is dropped, the height of the balloon isv2=vo2+ 2a(x–x0)(15 m/s)2= 0 + 2(3.0 m/s2)h→h= 37.5 mWhen the ball hits the ground, its vertical position is zero and its initial position was 37.5 m. Thereforev2=vo2+ 2a(x–x0)v2= (15 m/s)2+ 2(–9.8 m/s2)(0–37.5 m)→v=31 m/sdownward29.(a) When they meet,xandtare the same for both cars.(60 km/h)t= ½ (3.0 m/s2)t2[(60,000 m)/(3600 s)]t= 1.5 m/s2t2→t= 11.1 sThe distance down the road isx= ½at2= ½ (3.0 m/s2)(11.1 s)2=190 m(b) From part (a), we see thatt=11s.(c)v=v0+at= 0 + (3.0 m/s2)(11.1 s) =33 m/s30.v=v+vo2=vo+ 02,vo= 2v=−70.0 km/h=−19.4 m/s.a=v–vot=0–(–19.4 m/s)7.00 s=+2.78 m/s2.In this case, the positive 2.78 m/s2indicates deceleration because the velocity is negative.31.(a) Given:vo= 35.0 km/h = 9.72m/s,a= 1.50 m/s2,x= 200 m (takexo= 0).Find:v.v2=v2o+ 2a(x−xo) = (9.72m/s)2+ 2(1.5m/s2)(200 m) = 694m2/s2,v=26.3m/s.(b)v=vo+at,t=v–voa=26.3 m/s–9.72m/s1.50 m/s2=11.1s.32.Use the direction to the rightas the positive direction.

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Chapter 2Kinematics: Description of Motion23a=vt=11 m/s−(−35 m/s)0.095 s=4.8102m/s2.This is a very large acceleration due to the change in direction of the velocity and the short contact time.33.a0-4.0=vt=8.0 m/s–04.0 s–0=2.0 m/s2;a4.0-10.0=8.0 m/s–8.0m/s10.0 s–4.0s=0;a10.0-18.0=0–8.0 m/s18.0 s–10.0 s=−1.0 m/s2.The object accelerates at 2.0 m/s2first, moves with constant velocity, then decelerates at 1.0 m/s2.34.(a)a0-1.0 s=vt=0−01.0 s−0=0;a1.0 s-3.0 s=8.0 m/s−03.0 s−1.0 s=4.0 m/s2;a3.0 s-8.0 s=−12 m/s−8.0 m/s8.0 s−3.0 s=−4.0 m/s2;a8.0 s-9.0 s=−4 m/s−(−12.0 m/s)9.0 s−8.0 s=8.0 m/s2;a9.0 s-13.0 s=−4.0 m/s−4.0 m/s13.0 s−9.0 s=0.(b)Constant velocity of−4.0 m/s.35.(a) See the sketch on the right.(b) The acceleration is negativeasthe object slows down(assume velocity is positive).v=vo+at= 25 m/s + (−5.0 m/s2)(3.0 s)=10 m/s.(c)x=x1+x2+x3= (25 m/s)(5.0 s)+ (25 m/s)(3.0 s) +12(−5.0 m/s2)(3.0 s)2+(10 m/s)(6.0 s)=237.5 m =2.4102m.(d)s=dt=237.5 m14.0 s=17 m/s.36.72 km/h = (72 km/h)1000 m1 km1 h3600 s= 20 m/s.During deceleration,t1=va=0−20 m/s−1.0 m/s2= 20 s;x1=v1t1=20 m/s + 02(20 s) = 200 m.It would have taken the train200 m20 m/s= 10 s to travel 200 m.So it lost only 20 s–10 s = 10 s during deceleration.During acceleration,t2=20 m/s–00.50 m/s2= 40 s;x2=0 + 20 m/s2(40 s) = 400 m.Vvelocity (m/s)V5V10V15V25V20V15V10V5Vtime (s)

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24College Physics Seventh Edition: Instructor Solutions ManualIt would have taken the train400 m20 m/s= 20 s to travel 400 m. So it lost only 40 s–20 s = 20 s during acceleration.Therefore, the train lost 2 min + 10 s + 20 s =150 sin stopping at the station.37.The average velocity isv=xt=100 m4.5 s= 22.2 m/s.v=ovv2+=v2.So the final velocity must bev= 2(22.2 m/s) = 44.4 m/s.a=vt,t=va=44.4 m/s–09.0 m/s2= 4.9 s > 4.5 s.Sono, the driver did not do it.The acceleration must be44.4 m/s–04.5 s=9.9 m/s2.38.Given:vo= 0,a= 2.0 m/s2,t= 5.00 s.Find:vandx(takexo= 0).(a)v=vo+at= 0 + (2.0 m/s2)(5.0 s) =10 m/s.(b)x=xo+vot+12at2= 0 + 0(5.00 s) +12(2.0 m/s2)(5.0 s)2=25 m.39.Given:vo=25 mi/h = 11.2m/s,v= 0,x= 35 m (takexo= 0).Find:aandt.(a)v2=vo2+ 2a(x–xo),a=v2−vo22x=(0)2−(11.2m/s)22(35 m)=−1.79 m/s2=−1.8m/s2.The negative sign indicates that the acceleration vector is intheopposite direction ofthevelocity.(b)v=vo+at,t=v−voa=0−11.2m/s−1.79m/s2=6.3s.40.Given:vo= 60 km/h = 16.7m/s,v= 40 km/h = 11.1 m/s,x= 50 m (takexo= 0).Find:a.v2=vo2+ 2a(x−xo),a=v2−vo22x=(11.1 m/s)2−(16.7m/s)22(50 m)=−1.6m/s2.41.(a) Given:vo= 100 km/h = 27.78 m/s,a=−6.50 m/s2,x= 20.0 m (takexo= 0).Find:v.v2=vo2+ 2a(x−xo)= (27.78 m/s)2+ 2(−6.50 m/s2)(20.0 m) = 511.6 m2/s2,Sov= 22.62 m/s =81.4 km/h.(b)v=vo+at,t=v–voa=22.62 m/s–27.78 m/s–6.50 m/s2=0.794 s.42.(a)v=v0+at= 20 m/s + (–0.75 m/s2)(10 s) =13 m/s(b)x=v0t +½at2= (20 m/s)(10 s) + ½ (–0.75 m/s2)(10 s)2=160 m43.Given:vo= 250 km/h = 69.44 m/s,a=−8.25 m/s2,x= 175 m(Takexo= 0).Find:t.x=xo+vot+12at2,so13.7 sPoint ofReverse Thrust

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Chapter 2Kinematics: Description of Motion25175 m = 0 + (69.44 m/s)t+12(−8.25 m/s2)t2.Reduce to quadratic equation,4.125t2–69.44t+ 175 = 0.Solving,t=3.09 sand 13.7 s.The 13.7 s answer is physically possible but not likely in reality. After 3.09 s, it is 175 m from where the reversethrust was applied, but the rocket keeps traveling forward while slowing down. Finally it stops. However, if thereverse thrust is continuously applied (which is possible, but not likely), it will reverse its direction andgoback to175 m from the point where the initial reverse thrust was applied; a process that would take 13.7 s.44.(a) Given:Car A:aA= 3.00 m/s2,vo= 2.50 m/s,t= 10 s.Car B:aB= 3.00 m/s2,vo= 5.00 m/s,t= 10 s.Find:x(takingxo= 0).Fromx=xo+vot+12at2,xA= 0 + (2.50 m/s)(10 s) +12(3.00 m/s)2(10 s)2= 175 m,xB= 0 + (5.00 m/s)(10 s) +12(3.00 m/s)2(10 s)2= 200 m.Sox=xB−xA= 200 m−175 m =25 m.(b) Fromv=vo+at,vA= 2.50 m/s + (3.00 m/s)(10 s) = 32.5 m/s,vB= 5.00 m/s + (3.00 m/s)(10 s) = 35.0 m/s.Socar Bis faster.45.If the acceleration is less than 4.90 m/s2, then there is friction.Given:vo= 0,x= 15.00 m (takexo= 0),t= 3.0 s.Find:a.x=xo+vot+12at2,15.00 m = 0 +12a(3.0 s)2.a= 3.33 m/s2. So the answer isno, the incline is not frictionless.46.(a)(3)The object willtravel in the +x-direction and then reverseits direction. This is because the object has initialvelocity in the +x-direction, and it takes time for the object to decelerate, stop, and then reverse direction. We takexo= 0.Given:vo= 40 m/s,a=−3.5 m/s2,x= 0 (“returns to the origin”).Find:tandv.(b)x=xo+vot+12at2,0 = 0 + (40 m/s)t+12(−3.5 m/s2)t2.Reduce to quadratic equation:1.75t2–40t= 0.Solving,t= 0 or 22.9 s.Thet= 0 answer corresponds to the initial time. So the answer ist=23 s.(c)v=vo+at= 40 m/s + (−3.5 m/s2)(22.9 s) =−40 m/s =40 m/sin the−x-direction.47.Given:vo= 330 m/s,v= 0,x= 25cm = 0.25m (Takexo= 0).Find:a.v2=v2o+ 2a(x−xo),a=v2–v2o2x=(0)2–(330 m/s)22(0.25m)=−2.2105m/s2.

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26College Physics Seventh Edition: Instructor Solutions ManualThe negative sign here indicates that the acceleration vector is in the opposite direction ofthevelocity.48.40 km/h = (40 km/h)1000 m1 km1 h3600 s= 11.11 m/s.Duringthereactiontime, the car travels a distance ofd= (11.11 m/s)(0.25 s) = 2.78 m.So the car really has only 13 m–2.78 m = 10.2 m to come to rest.Let’s calculate the stopping distance of the car. We takexo= 0.Given:vo= 11.1 m/s,v= 0,a=−8.0 m/s2.Find:x. (Takexo= 0.)v2=v2o+ 2a(x−xo),x=v2–v2o2a=0–(11.1 m/s)22(–8.0 m/s2)= 7.70 m.So it takes the car only2.78 m + 7.70 m =10.5 m(< 13 m)to stop.Yes, the car will stop before hitting the child.49.Repeat the calculation of Exercise 2.48.d= (11.1 m/s)(0.50 s) = 5.55 m.5.55 m + 7.70 m =13.3 m> 13 m.No, the car will not stop before hitting the child.50.Given:vo= 350 m/s,v= 210 m/s,x= 4.00 cm = 0.0400 m (takexo= 0).Find:t.x=xo+vt=vo+v2t,t=2xvo+v=2(0.0400 m)350 m/s + 210 m/s=1.4310−4s.51.(a) For constant acceleration, thevvs.tplot is a straight line.Point p has coordinates of (0,vo) and point q has coordinates of(t,vo+at). The distance from point q to point o is thereforeat. Thearea under the curve is the area of the triangle12(at)tplus the area of the rectanglevot.SoA=vot+12at2=x−xo. (Herex−xois displacement.)(b) The total area consists of two triangles from 0 to 4.0 s and from 10.0 s to 18.0 s and a rectangle from4.0 s to 10.0 s.x−xo=A=12(4.0 s–0)(8.0 m/s) + (10.0 s–4.0 s)(8.0 m/s) +12(18.0 s–10.0 s)(8.0 m/s) =96 m.52.(a)(3)t1>t2.Sincetheobject is accelerating, it will spend less time in traveling the second 3.00 m.(b)For the first 3.00 m:Given:vo=0,a= 2.00 m/s2,x=3.00 m(takexo= 0).Find:t.x=xo+vot+12at2= 0 + 0 +12at2,t1=2xa=2(3.00 m)2.00 m/s2=1.73 s.At the end of the first 3.00 m, the velocity of the object isv=vo+at= 0 + (2.00 m/s2)(1.73 s) = 3.46 m/s.This isthenthe initial velocity for the second 3.00 m.vqatvopo0tt

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Chapter 2Kinematics: Description of Motion27For the second 3.00 m:Given:vo= 3.46 m/s,a= 2.00 m/s2,x= 3.00 m.Find:t.x=xo+vot+12at2,3.00 m = 0 + (3.46 m/s)t2+12(2.00 m/s2)t22.Reducing to quadratic equation,t2+ 3.46t−3.00 = 0.Solving,t2=0.718 sor−4.18 s.53.(a) At the end of phase 1, the change in velocity isv1−0 =v1. At the end of phase 2, the change in velocity isv2−v1. Since the object is accelerating, it spends less time in phase 2 than in phase 1. Since the change in velocity isequal to acceleration times the time, the change in velocity is greater in phase 1 than in phase 2. Orv1>v2−v1.That is 2v1>v2.Therefore,v1>12v2. The answer is(3)v1>12v2.(b) Forphase 1:vo= 0,a= 0.850 m/s2,x= 50.0 m (takexo= 0).Find:v.v2=v2o+ 2a(x−xo) = 02+ 2(0.850 m/s2)(50.0 m) = 85.0 m2/s2,v1=9.22 m/s.For phase 2:vo= 9.22 m/s,a= 0.850 m/s2,x= 50.0 m (takexo= 0).Find:v.v2=v2o+ 2a(x−xo) = (9.22 m/s)2+ 2(0.850 m/s2)(50.0 m) = 170 m2/s2,v2=13.0 m/s.Sov1= 9.22 m/s >12v2=12(13.0 m/s) = 6.50 m/s.54.Takexo= 0.During acceleration:vo= 0,a= 1.5 m/s2,t= 6.0 s.x1=xo+vot+12at2= 0 + 0 +12(1.5 m/s2)(6.0 s)2= 27 m,

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28College Physics Seventh Edition: Instructor Solutions Manualv=vo+at= 0 + (1.5 m/s2)(6.0 s) = 9.0 m/s.During constant velocity:x2= (9.0 m/s)(8.0 s) = 72 m.Sov=xt=27 m + 72 m14 s=7.1 m/s.55.(a)v(8.0 s) =−12 m/s;v(11.0 s) =−4.0 m/s.(b) Use the result of Exercise 2.51a. The total area consists of a rectangle from 0 to 1.0 s, a triangle from 1.0 s to 5.0s, a trapezoid from 5.0 s to 11.0 s, and a triangle from 6.0 s to 9.0 s with baseline at–4.0 m/s.x−xo=A= 0 +12(5.0 s–1.0 s)(8.0 m/s) +(11.0 s–6.0 s)+(11.0 s–5.0 s)2(–4.0 m/s)+12(9.0 s–6.0 s)[(–12.0 m/s)–(–4.0 m/s)] =−18 m.(c) The total distance (not displacement) is the addition of the absolute values of the areas.d=Ai= 0 +12(5.0 s–1.0 s)(8.0 m/s) +(11.0 s–6.0s)+(11.0 s–5.0 s)2(4.0 m/s)+12(9.0 s–6.0 s)[(12.0 m/s–4.0 m/s] =50 m.56.(a)v2=vo2+ 2a(x−xo) ,x−xo=v2−vo22a=02−vo22a=−20a.2vTakingxo= 0, so (x−xo) =xis proportional tovo2. Ifvodoubles, thenxbecomes 4 times as large.The answer is then(3)4x.(b)x2x1=22ov21ov=602402= 2.25.Sox2= 2.25x1= 2.25 (3.00 m) =6.75 m.57.(a)Given:a= 3.00 m/s2,t= 1.40 s,x= 20.0 m (takexo= 0).Find:vo.x=xo+vot+12at2,20.0 m = 0 +vo(1.40 s) +12(3.00 m/s2)(1.40 s)2.Solving,vo=12.2 m/s.v=vo+at= 12.2 m/s + (3.00 m/s2)(1.40 s) =16.4 m/s.(b)Given:vo= 0,a= 3.00 m/s2,v= 12.2 m/s.Find:x(takexo= 0).v2=vo2+ 2a(x−xo),x−xo=v2−vo22a=(12.2 m/s)2−022(3.00 m/s2)=24.8 m.(c)v=vo+at,t=v−voa=12.2 m/s−03.00 m/s2=4.07 s.58.75.0mi/h = 33.5 m/s.(a) Given:vo=33.5 m/s,a=−1.00 m/s2,x= 100 m(takexo= 0).Find:v.v2=vo2+ 2a(x−xo)= (33.5 m/s)2+ 2(−1.00 m/s2)(100 m) = 922 m2/s2.Sov=30.4 m/s.

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Chapter 2Kinematics: Description of Motion29(b)The initial velocity on dryconcreteis then 30.4 m/s.Consider on dry concrete.Given:vo= 30.4 m/s,a=−7.00 m/s2,v= 0 m.Find:x.v2=vo2+ 2ax,x=v2−vo22a=02−(30.4 m/s)22(−7.00 m/s2)=66.0 m.So the total distance is 100 m + 66.0 m =166 m.(c) Usev=vo+at.On ice:t1=v−voa=30.4 m/s−33.5 m/s−1.00 m/s2= 3.10 s,On dry concrete:t2=0−30.4 m/s−7.00 m/s2= 4.34 s.So the total time is 3.10 s + 4.34 s =7.44 s.59.(a) Given:vo= 0,t= 2.8 s.Find:v(takeyo= 0).v=vo–gt= 0–(9.80 m/s2)(2.8 s) =−27 m/s.(b)y=yo+vot–12gt2= 0 + 0−12(9.80 m/s2)(2.8 s)2=–38 m.60.(a) We takeyo= 0.y=yo+vot−12gt2=−12gt2.Soyis proportional to the time squared.Therefore twice the time means(3)fourtimesthe height.Given:vo= 0,t= 1.80 s.Find:yAandyB.(b)yA=−12(9.80 m/s2)(1.80 s)2=−15.9 m.So the height of cliff A above the water is15.88 m =15.9 m.yB=yA4=15.88m4=3.97m.61.(a) A straight line(linear), slope =−g.(b) A parabola.62.Given:vo= 0,y=−0.157 m (takeyo= 0).Find:t.y−yo=vot−12gt2=−12gt2,t=2y−g=2(−0.157 m)−9.80 m/s2= 0.18 s < 0.20 s.

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30College Physics Seventh Edition: Instructor Solutions ManualIt takes less than the average human reaction time for the dollar bill to fall.So the answer isnonot a good deal.63.If the ball is in the air for twice as long on the second toss as it is on the first toss, the time for it to fall from itsmaximum height on the second toss will be twice as long as the time for it to fall from its maximum height on thefirst toss. Realizing thatt2= 2t1, the maximum heights reached are=21112hgt===22211221122(2 )2hgtgtgtTaking the ratio of the heights gives==2212111224hgthgtTherefore21= 4hh, so it must be tossed4 times as high.64.Given:vo= 15 m/s,v= 0 (maximum height).Find:y. (Takeyo= 0.)v2=vo2−2g(y−yo),y=vo2−v22g=(15 m/s)2−(0)22(9.80 m/s2)=11 m.65.From Exercise 2.64,y=vo2−v22g=(15 m/s)2−(0)22(1.67 m/s2)=67 m.66.Takingyo= 0,y=yo+vot–12gt2= 0 + 0−12gt2=–12gt2,sot=–2yg.Fory=−452 m,t= 9.604 s;fory=−443 m,t= 9.508 s.Sot= 9.604 s–9.508 s =0.096 s.67.First convert 100 km/h to m/s, giving 100 km/h = 27.78 m/s. Now use the formulav2=v02+ 2a(x–x0) and solve forx.(27.78 m/s)2= 0 + 2(9.80 m/s2)x→x=39.4 m68.Given:vo= 6.0 m/s,y=−12 m (takeyo= 0).Find:tandv.(a)y=yo+vot−12gt2,−12 m = 0 + (6.0 m/s)t–12(9.80 m/s2)t2.Or 4.9t2–6.0t–12 = 0.Solving,t=2.3 sor–1.1 s. The negative time is discarded.(b)v=vo−gt= 6.0 m/s–(9.80 m/s2)(2.29 s) =−16 m/s.69.(a) When the ball rebounds, it is a free fall with an initial upward velocity. At the maximum height, the velocity iszero. Takingyo= 0,v2=v2o–2g(y−yo),y=v2o–v22g.Soymax=v2o2g.Therefore, the height depends on the initial velocity squared. 95% = 0.95 and 0.952= 0.90 < 0.95.

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