Problem 12.1 The value ofis 3.1415962654. . . . . IfCisthecircumferenceofacircleandr isitsradius,determine the value ofr/Cto four signicant digits.Solution:C = 2r rC=12= 0.159154943.To four signicant digits we haverC= 0.1592Problem 12.2 Thebaseof natural logarithmsis e =2.718281828 . . .(a) Expresse to ve signicant digits.(b) Determine the value ofe2to ve signicant digits.(c) Use the value of e you obtained in part (a) to deter-mine the value ofe2to ve signicant digits.[Part (c) demonstratesthehazardofusingrounded-offvalues in calculations.]Solution: The value of e is:e = 2.718281828(a) To ve signicant gures e = 2.7183(b) e2to ve signicant gures ise2= 7.3891(c) Usingthevaluefrompart (a) wende2= 7.3892whichisnot correct in the fth digit.Problem 12.3 Amachinist drillsacircularholeinapanel with a nominal radius r = 5 mm. The actual radiusof the hole is in the range r = 5 0.01 mm. (a) To whatnumber of signicant digits can you express the radius?(b) To what number of signicant digits can you expressthe area of the hole?5 mmSolution:(a) The radius is in the range r1 = 4.99 mm to r2 = 5.01 mm. Thesenumbers are not equal at the level of three signicant digits, butthey are equal if they are rounded off to two signicant digits.Two:r = 5.0 mm(b) The area of the hole is in the range fromA1 = r21 = 78.226 m2to A2 = r22 = 78.854 m2. These numbers are equal only if roundedto one signicant digit:One:A = 80 mm2Problem 12.4 The opening in the soccer goal is 25 ftwide and 8 ft high, so its area is 24 ft 8 ft = 192 ft2.What is its area in m2to three signicant digits?Solution:A = 192 ft2

1 m3.281 ft

2= 17.8 m2A = 17.8 m21c 2008PearsonEducation South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from thepublisherprior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Problem 12.5 The Burj Dubai, scheduled for comple-tion in 2008, will be the worlds tallest building with aheight of 705 m. The area of its ground footprint will be8000m2. ConvertitsheightandfootprintareatoU.S.customary units to three signicant digits.Solution:h = 705 m

3.281 ft1 m

= 2.31 103ftA = 8000 m2

3.218 ft1 m

2= 8.61 104ft2h = 2.31 103ft, A = 8.61 104ft2Problem 12.6 Supposethat youhavejust purchasedaFerrari F355coupeandyouwant toknowwhetheryoucanuseyour set of SAE(U.S. CustomaryUnits)wrenches to work on it. You have wrenches with widthsw = 1/4 in, 1/2 in, 3/4 in, and 1 in, and the car has nutswithdimensions n = 5 mm, 10 mm, 15 mm, 20 mm,and25 mm. Deningawrenchtot if wisnomorethan 2% larger thann, which of your wrenches can youuse?nSolution: Convert themetricsizentoinches, andcomputethepercentage difference between the metric sized nut and the SAEwrench. The results are:5 mm

1 inch25.4 mm

= 0.19685.. in,

0.19685 0.250.19685

100= 27.0%10 mm

1 inch25.4 mm

= 0.3937.. in,

0.3937 0.50.3937

100 = 27.0%15 mm

1 inch25.4 mm

= 0.5905.. in,

0.5905 0.50.5905

100 = +15.3%20 mm

1 inch25.4 mm

= 0.7874.. in,

0.7874 0.750.7874

100 = +4.7%25 mm

1 inch25.4 mm

= 0.9843.. in,

0.9843 1.00.9843

100 = 1.6%A negative percentage implies that the metric nut is smaller than theSAEwrench;apositivepercentagemeansthatthenutis largerthenthe wrench. Thus within the denition of the 2% t, the 1 in wrenchwill t the 25 mm nut. The other wrenches cannot be used.2c 2008PearsonEducation South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from thepublisherprior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Problem 12.7 Suppose that the height of Mt. Everestis known to be between 29,032 ft and 29,034 ft. Basedon this information, to how many signicant digits canyou express the height (a) in feet? (b) in meters?Solution:(a) h1 = 29032 fth2 = 29034 ftThe two heights are equal if rounded off to four signicant digits.The fth digit is not meaningful.Four: h = 29,030 ft(b) In meters we haveh1 = 29032 ft

1 m3.281 ft

= 8848.52 mh2 = 29034 ft

1 m3.281 ft

= 8849.13 mThesetwoheightsareequal ifroundedofftothreesignicantdigits. The fourth digit is not meaningful.Three:h = 8850 mProblem 12.8 Themaglev(magneticlevitation)trainfrom Shanghai to the airport at Pudong reaches a speedof 430 km/h. Determine its speed (a) in mi/h; (b) ft/s.Solution:(a) v = 430kmh

0.6214 mi1 km

= 267 mi/h v = 267 mi/h(b) v = 430kmh

1000 m1 km

1 ft0.3048 m

1 h3600 s

= 392 ft/sv = 392 ft/sProblem 12.9 In the 2006 Winter Olympics, the mens15-kmcross-countryskiingracewas wonbyAndrusVeerpalu of Estonia in a time of 38 minutes, 1.3 seconds.Determine his average speed (the distance traveled divi-ded by the time required) to three signicant digits (a) inkm/h; (b) in mi/h.Solution:(a) v =15 km

38 + 1.360

min

60 min1 h

= 23.7 km/h v = 23.7 km/h(b) v = (23.7 km/h)

1 mi1.609 km

= 14.7 mi/h v = 14.7 mi/h3c 2008PearsonEducation South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from thepublisherprior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Problem 12.10 ThePorsches engineexerts 229 ft-lb(foot-pounds) of torque at 4600 rpm. Determine the valueof the torque in N-m (Newton-meters).Solution:T = 229 ft-lb

1 N0.2248 lb

1 m3.281 ft

= 310 N-m T = 310 N-mProblem 12.11 The kinetic energy of the man in ActiveExample 12.1 is dened by12mv2, wherem is his massandvis his velocity. Themans mass is 68 kgand heismovingat 6 m/s, sohiskineticenergyis12(68 kg)(6 m/s)2= 1224 kg-m2/s2. What is his kinetic energy inU.S. Customary units?Solution:T = 1224 kg-m2/s2

1 slug14.59 kg

1 ft0.3048 m

2= 903 slug-ft2/sT = 903 slug-ft2/sProblem 12.12 The acceleration due to gravity at sealevel in SI units isg = 9.81 m/s2. By converting units,use this value to determine the acceleration due to grav-ity at sea level in U.S. Customary units.Solution: Use Table 1.2. The result is:g = 9.81

ms2

1 ft0.3048 m

= 32.185 . . .

fts2

= 32.2

fts2

Problem 12.13 Afurlong per fortnight isafacetiousunit of velocity, perhaps madeupbya student as asatirical comment onthebewilderingvarietyof unitsengineers must deal with. A furlong is 660 ft (1/8 mile).A fortnight is 2 weeks (14 nights). If you walk to classat 2 m/s, what is your speed in furlongs per fortnight tothree signicant digits?Solution:v = 2 m/s

1 ft0.3048 m

1 furlong660 ft

3600 shr

24 hr1 day

14 day1 fortnight

v = 12,000 furlongsfortnightProblem 12.14 Determinethecross-sectional areaofthe beam (a) in m2; (b) in in2.120 mm xy40 mm40 mm40mm200 mmSolution:A = (200 mm)22(80 mm)(120 mm) = 20800 mm2(a) A = 20800 mm2

1 m1000 mm

2= 0.0208 m2A = 0.0208 m2(b) A = 20800 mm2

1 in25.4 mm

2= 32.2 in2A = 32.2 in24c 2008PearsonEducation South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from thepublisherprior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Problem 12.15 The cross-sectional area of theC1230 American Standard Channel steel beam isA =8.81 in2. What is its cross-sectional area in mm2?xyASolution:A = 8.81 in2

25.4 mm1 in

2= 5680 mm2Problem 12.16 A pressure transducer measures a valueof 300 lb/in2. Determinethevalueof thepressureinpascals. A pascal (Pa) is one newton per meter squared.Solution: Convert the units using Table 12.2 and the denition ofthe Pascal unit. The result:300

lbin2

4.448 N1 lb

12 in1 ft

2

1 ft0.3048 m

2= 2.0683 . . . (106)

Nm2

= 2.07(106) PaProblem 12.17 A horsepower is 550 ft-lb/s. A watt is1N-m/s. Determinehowmanywattsaregeneratedbytheenginesofthepassengerjet if theyareproducing7000 horsepower.Solution:P = 7000 hp

550 ft-lb/s1 hp

1 m3.281 ft

1 N0.2248 lb

= 5.22 106WP = 5.22 106WProblem 12.18 Distributed loads on beams are expres-sedinunitsofforceperunit length. Ifthevalueofadistributed load is 400 N/m, what is its value in lb/ft?.Solution:w = 400 N/m

0.2248 lb1 N

1 m3.281 ft

= 27.4 lb/ft w = 27.4 lb/ft5c 2008PearsonEducation South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from thepublisherprior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Problem 12.19 Themoment ofinertiaoftherectan-gular area about thexaxis is given by the equationI =13bh3.Thedimensionsoftheareaareb = 200 mmandh =100 mm. Determinethevalueof I tofour signicantdigits in terms of (a) mm4; (b) m4; (c) in4.hbxySolution:(a) I =13(200 mm)(100 mm)3= 66.7 106mm4(b)I = 66.7 106mm4

1 m1000 mm

4= 66.7 106m4(c) I = 66.7 106mm4

1 in25.4 mm

4= 160 in4Problem 12.20 In Example 12.3, instead of Einsteinsequation consider the equationL = mc, where the massm is in kilograms and the velocity of lightc is in meterspersecond. (a) What aretheSIunitsof L?(b) Ifthevalueof LinSIunitsis12,whatisthevalueinU.S.Customay base units?Solution:(a) L = mc Units(L) = kg-m/s(b) L = 12 kg-m/s

0.0685 slug1 kg

3.281 ft1 m

= 2.70 slug-ft/sL = 2.70 slug-ft/sProblem 12.21 The equation =MyIis used in the mechanics of materials to determinenormal stresses in beams.(a) When this equation is expressed in terms of SI baseunits,Mis in newton-meters (N-m), y is in meters(m),andI isinmeterstothefourthpower(m4).What are the SI units of?(b) If M = 2000 N-m, y = 0.1 m, and I = 7 105m4, what is the value of in U.S. Customarybase units?Solution:(a) =MyI=(N-m)mm4=Nm2(b) =MyI=(2000 N-m)(0.1 m)7 105m4

1 lb4.448 N

0.3048 mft

2= 59,700 lbft2Problem 12.22 The acceleration due to gravity on thesurface of the moon is 1.62 m/s2. (a) What would be themass of the C-clamp in Active Example 12.4 be on thesurface of the moon? (b) What would the weight of theC-clamp in newtons be on the surface of the moon?Solution:(a) The mass does not depend on location. The mass in kg is0.0272 slug

14.59 kg1 slug

= 0.397 kg mass = 0.397 kg(b) The weight on the surface of the moon isW = mg = (0.397 kg)(1.62 m/s2) = 0.643 N W = 0.643 N6c 2008PearsonEducation South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from thepublisherprior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Problem 12.23 The1 ft 1 ft 1 ft cubeof ironweighs 490 lb at sea level. Determine the weight in new-tons of a 1 m 1 m 1 m cube of the same materialat sea level.1 ft1 ft1 ftSolution: The weight density is =490 lb1 ft3The weight of the 1 m3cube is:W = V =

490 lb1 ft3

(1 m)3

1 ft0.3048 m

3

1 N0.2248 lb

= 77.0 kNProblem 12.24 The area of the Pacic Ocean is64,186,000 square miles and its average depth is 12,925 ft.Assume that the weight per unit volume of ocean wateris64 lb/ft3. Determinethemassof thePacicOcean(a) in slugs; (b) in kilograms.Solution: The volume of the ocean isV = (64,186,000 mi2)(12,925 ft)

5,280 ft1 mi

2= 2.312 1019ft3(a) m = V =

64 lb/ft332.2 ft/s2

(2.312 1019f t3) = 4.60 1019slugs(b) m = (4.60 1019slugs)

14.59 kg1 slug

= 6.71 1020kgProblem 12.25 The acceleration due to gravity atsea level is g = 9.81 m/s2. The radius of the earthis 6370 km. The universal gravitational constant isG = 6.67 1011N-m2/kg2. Use this information todetermine the mass of the earth.Solution: Use Eq. (12.3)a =GmER2. Solve for the mass,mE =gR2G=(9.81 m/s2)(6370 km)2

103mkm

26.67(1011)

N-m2kg2

= 5.9679 . . . (1024) kg = 5.97(1024) kgProblem 12.26 Aperson weighs 800Nsea level.The radius of the earth is 6372 km. What force is exertedon the person by the gravitational attraction of the earthif heis inaspacestationinorbit 322 km abovethesurface of the earth?Solution: Use Eq. (12.5).W=mg

REr

2=

WEg

g

RERE +H

2=WE

+

2=( 0)(0. ) =Problem 12.27 The acceleration due to gravity on thesurface of the moon is 1.62 m/s2. The moons radius isRM = 1738 km.(a) What istheweight innewtonsonthesurfaceofthe moon of an object that has a mass of 10 kg?(b) Using the approach described in Example 12.5, de-termine the force exerted on the object by the grav-ityofthemooniftheobject islocated1738 kmabove the moons surface.Solution:(a) W = mgM = (10 kg)(1.26 m/s2) = 12.6 N W = 12.6 N(b) Adapting equation 1.4 we haveaM = gM

RMr

2. The force isthenF = maM = (10 kg)(1.62m/s2)

1738 km1738 km+1738 km

2= 4.05 NF = 4.05 N7637280 9519 262 N6372 322c 2008PearsonEducation South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from thepublisherprior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Problem 12.28 If an object is near the surface of theearth, the variation of its weight with distance from thecenter of the earth can often be neglected. The acceler-ationduetogravityatsealevelisg = 9.81 m/s2.Theradius of the earth is 6370 km. The weight of an objectat sea level ismg, wherem is its mass. At what heightabovetheearthdoestheweightoftheobjectdecreaseto 0.99mg?Solution: Use a variation of Eq. (12.5).W = mg

RERE +h

2= 0.99 mgSolve for the radial height,h = RE

10.99 1

= (6370)(1.0050378 1.0)= 32.09 . . . km = 32,100 m = 32.1 kmProblem 12.29 TheplanetNeptunehas anequatorialdiameter of 49,532 km and its mass is 1.0247 1026kg.If the planet is modeled as a homogeneous sphere, whatis the acceleration due to gravity at its surface? (The uni-versal gravitational constant is G = 6.67 1011h-m2/kg2.)Solution:We have:W = GmNmr2N=

GmNr2

m gN = GmNr2NNote that the radius of Neptune is rN =12(49,532 km)= 24,766 kmThus gN =

6.67 1011N-m2kg2

1.0247 1026kg(24766 km)2

1 km1000 m

2= 11.1 m/s2gN = 11.1 m/s2Problem 12.30 At apoint betweentheearthandthemoon, the magnitudeofthe forceexerted on anobjectby the earths gravity equals the magnitude of the forceexertedontheobject bythemoonsgravity. What isthedistancefromthecenteroftheearthtothat pointto three signicant digits? The distance from the centeroftheearthtothecenterofthemoonis383,000 km,and the radius of the earth is 6370 km. The radius of themoon is 1738 km, and the acceleration due to gravity atits surface is 1.62 m/s2.Solution: Let rEp be the distance from the Earth to the point wherethe gravitational accelerations are the same and letrMp be the distancefrom the Moon to that point. Then, rEp+rMp = rEM = 383,000 km.The fact that the gravitational attractions by the Earth and the Moonat this point are equal leads to the equationgE

RErEp

2= gM

RMrMp

2,whererEM = 383,000 km. Substitutingthecorrectnumerical valuesleads to the equation9.81

ms2

6370 kmrEp

2= 1.62

ms2

1738 kmrEM rEp

2,where rEp is the only unknown. Solving, we get rEp = 344,770 km =345,000 km.8c 2008PearsonEducation South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from thepublisherprior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Problem 13.1 In Example 13.2, suppose that the vehi-cleisdroppedfromaheight h = 6m. (a) What isthedownwardvelocity 1 s afteritisreleased? (b) Whatisits downward velocity just before it reaches the ground?hSolution: The equations that govern the motion are:a = g = 9.81 m/s2v = gts = 12gt2+h(a) v = gt = (9.81 m/s2)(1 s) = 9.81 m/s.The downward velocity is 9.81 m/s.(b) We need to rst determine the time at which the vehicle hits thegrounds = 0 = 12gt2+h t =

2hg=

2(6 m)9.81 m/s2 = 1.106 sNow we can solve for the velocityv = gt = (9.81 m/s2)(1.106 s) = 10.8 m/s.The downward velocity is 10.8 m/s.Problem 13.2 The milling machine is programmed sothat duringtheinterval oftimefromt = 0tot = 2 s,the position of its head (in inches) is given as a functionof time bys = 4t 2t3. What are the velocity (in in/s)and acceleration (in in/s2) of the head att = 1 s?sSolution: The motion is governed by the equationss = (4 in/s)t (2 in/s2)t2,v = (4 in/s) 2(2 in/s2)t ,a = 2(2 in/s2).Att = 1 s, we havev = 0, a = 4 in/s2.9c 2008PearsonEducation South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from thepublisherprior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Problem 13.3 In an experiment to estimate the accel-eration due to gravity, a student drops a ball at a distanceof 1 m above the oor. His lab partner measures the timeit takes to fall and obtains an estimate of 0.46 s.(a) What do they estimate the acceleration due to grav-ity to be?(b) Let s betheballspositionrelativetotheoor.Using the value of the acceleration due to gravitythat theyobtained, andassumingthat theball isreleased att = 0, determines(in m) as a functionof time.ss 0Solution: The governing equations area = gv = gts = 12gt2+h(a) When the ball hits the oor we have0 = 12gt2+h g =2ht2=2(1 m)(0.46 s)2 = 9.45 m/s2g = 9.45 m/s2(b) The distancesis then given bys = 12(9.45 m/s2) +1 m.s = (4.73 m/s2)t2+1.0 m.Problem 13.4 Theboatspositionduringtheintervaloftimefromt = 2 stot = 10 sisgivenbys = 4t +1.6t20.08t3m.(a) Determinetheboatsvelocityandaccelerationatt = 4 s.(b) Whatistheboatsmaximumvelocityduringthisinterval of time, and when does it occur?Solution:s = 4t +1.6t20.08t3v =dsdt= 4 +3.2t 0.24t2a)v(4s) = 12.96 m/s2a(4s) = 1.28 m/s2b)a = 3.2 0.48t = 0 t = 6.67sv(6.67s) = 14.67 m/sa =dvdt= 3.2 0.48t10c 2008PearsonEducation South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from thepublisherprior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Problem 13.5 The rocket starts from rest att = 0 andtravels straight up. Its height above the groundas afunction of time can be approximated bys = bt2+ct3,wherebandcareconstants. At t = 10 s, therocketsvelocity and acceleration arev = 229 m/s anda = 28.2m/s2. Determinethetimeat whichtherocket reachessupersonicspeed(325 m/s). What isitsaltitudewhenthat occurs?sSolution: The governing equations ares = bt2+ct3,v = 2bt +3ct2,a = 2b +6ct .Using the information that we have allows us to solve for the constantsb andc.(229 m/s) = 2b(10 s) +3c(10 s)2,(28.2 m/s2) = 2b +6c(10 s).Solving these two equations, we ndb = 8.80 m/s2, c = 0.177 m/s3.When the rocket hits supersonic speed we have(325 m/s) = 2(8.80 m/s2)t +3(0.177 m/s3)t2 t = 13.2 s.The altitude at this time iss = (8.80 m/s2)(13.2 s)2+(0.177 m/s3)(13.2 s)3s = 1940 m.Problem 13.6 The position of a point during the inter-val of time from t = 0 to t = 6 s is given by s = 12t3+6t2+4t m.(a) What is the maximum velocity during this intervalof time, and at what time does it occur?(b) What is the accelerationwhenthe velocityis amaximum?Solution:s = 12t3+6t2+4t mv = 32t2+12t +4 m/sa = 3t +12 m/s2Maximum velocity occurs where a =dvdt= 0 (it could be a minimum)This occurs at t = 4 s. At this pointdadt= 3 so we have a maximum.(a) Max velocity is att = 4 s. wherev = 28 m/s and(b) a = 0 m/s211c 2008PearsonEducation South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from thepublisherprior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Problem 13.7 The position of a point during the inter-val of timefromt = 0tot = 3secondsis s = 12 +5t2t3(a) What is the maximum velocity during this intervalof time, and at what time does it occur?(b) What is the accelerationwhenthe velocityis amaximum?Solution:(a) The velocity isdsdt= 10t 3t2. The maximumoccurs whendvdt= 10 6t = 0, from whicht =106= 1.667 seconds.This is indeed a maximum, sinced2vdt2 = 6 < 0. The maximumvelocity isv =10t 3t2t =1.667 = 8.33 /s(b) The acceleration isdvdt= 0 when the velocity is a maximum.Problem 13.8 Therotatingcrankcausesthepositionof point P as a function of time to be s = 0.4 sin(2t ) m.(a) Determinethevelocityandaccelerationof P att = 0.375 s.(b) What isthemaximummagnitudeofthevelocityofP?(c) Whenthe magnitude of the velocityof P is amaximum, what is the acceleration ofP?PsSolution:s = 0.4 sin(2t )v =dsdt= 0.8 cos(2t ) a)v(0.375s) = 1.777 m/sa(0.375) = 11.2 m/s2b)vmax = 0.8 = 2.513 m/s2c)vmax t = 0, n a = 0a =dvdt= 1.62sin(2t )Problem 13.9 For the mechanismin Problem 13.8,drawgraphs of the position s, velocity v, andacce-leration a of point P as functions of time for 0 t 2 s.Using yourgraphs, conrmthatthe slope ofthe graphofsis zero at times for whichvis zero, and the slopeof the graph ofv is zero at times for whichais zero.Solution:12m.mc 2008PearsonEducation South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from thepublisherprior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Problem 13.10 A seismograph measures the horizon-tal motion of the ground during an earthquake. An engi-neer analyzing the data determines that for a 10-s intervalof time beginning att = 0, the position is approximatedby s = 100 cos(2t ) mm. What are(a) themaximumvelocityand(b) maximumaccelerationof thegroundduring the 10-s interval?Solution:(a) The velocity isdsdt= (2)100 sin(2t ) mm/s = 0.2 sin(2t ) m/s.The velocity maxima occur atdvdt= 0.42cos(2t ) = 0,from which2t =(2n 1)2, ort =(2n 1)4,n = 1, 2, 3, . . . M, where(2M 1)4 10 seconds.These velocity maxima have the absolute value

dsdt

t =(2n1)4= [0.2] = 0.628 m/s.(b) The acceleration isd2sdt2 = 0.42cos(2t ).The acceleration maxima occur atd3sdt3 =d2vdt2 = 0.83sin(2t ) = 0,from which 2t = n, ort =n2, n = 0, 1, 2, . . . K, whereK2 10 seconds.These acceleration maxima have the absolute value

dvdt

t =n2= 0.42= 3.95 m/s2.Problem 13.11 Inanassemblyoperation, therobotsarmmovesalongastraight horizontal line. Duringaninterval of time from t = 0 to t = 1 s, the position of thearm is given bys = 30t220t3mm. (a) Determine themaximum velocity during this interval of time. (b) Whatare the position and acceleration when the velocity is amaximum?sSolution:s = 30t220t3mmv = 60t 60t2mm/sa = 60 120t mm/s2dadt= 120 mm/s3(a) Maximumvelocity occurs whendvdt= a = 0. This occurs at0 = 60 120t or t = 1/2second. (since da/dt < 0, wehavea maximum). The velocity at this time isv = (60)

12

60

14

mm/sv = 15 mm/s(b) The position and acceleration at this time ares = 7.5 2.5 mms = 5 mma = 0 mm/s213c 2008PearsonEducation South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from thepublisherprior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Problem 13.12 In Active Example 13.1, the accelera-tion (in m/s2) of pointPrelative to pointOis given asa function of time bya = 3t2. Suppose that att = 0 theposition and velocity ofPares = 5 m andv = 2 m/s.Determine the position and velocity ofPatt = 4 s.O Pss Solution: The governing equations area = (3 m/s4)t2v =13(3 m/s4)t3+(2 m/s)s =112(3 m/s4)t4+(2 m/s)t +(5 m)Att = 4 s, we have s = 77 m, v = 66 m/s.Problem 13.13 ThePorschestarts fromrest at timet = 0. Duringthe rst 10seconds of its motion, itsvelocity in km/h is given as a function of time byv =22.8t 0.88t2, wheret isinseconds. (a) What isthecarsmaximumaccelerationinm/s2,andwhendoesitoccur? (b) What distance in km does the car travel dur-ing the 10 seconds?Solution: First convert the numbers into meters and seconds22.8kmhr

1000 m1 km

1hr3600 s

= 6.33 m/s0.88kmhr

1000 m1 km

1 hr3600 s

= 0.244 m/sThe governing equations are thens =12(6.33 m/s)(t2/s) 13(0.88 m/s)(t3/s2),v = (6.33 m/s)(t /s) (0.88 m/s)(t2/s2),a = (6.33 m/s2) 2(0.88 m/s)(t /s2),ddta = 2(0.88 m/s)(1/s2) = 1.76 m/s3.Themaximumaccelerationoccurs at t = 0(anddecreases linearlyfrom its initial value).amax = 6.33 m/s2@t = 0In the rst 10 seconds the car travels a distances =12

22.8kmhr

(10 s)2s 13

0.88kmhr

(10 s)3s2 1 hr3600 s

s = 0.235 km.14c 2008PearsonEducation South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from thepublisherprior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Problem 13.14 The acceleration of a point is a =20t m/s2. When t = 0, s = 40 m and v = 10 m/s.What are the position and velocity att = 3 s?Solution: The velocity isv =

a dt +C1,whereC1is the constant of integration. Thusv =

20t dt +C1 = 10t2+C1.At t = 0, v = 10 m/s, hence C1 = 10andthe velocityis v =10t210 m/s. The position iss =

v dt +C2,whereC2is the constant of integration.s =

(10t210) dt +C2 =

103

t310t +C2.Att = 0, s = 40 m, thusC2 = 40. The position iss =

103

t310t +40 m.Att = 3 seconds,s =103t310t +40t =3= 100 m.The velocity att = 3 seconds isv =10t210t =3 = 80 m/s.Problem 13.15 The acceleration of a point is a =60t 36t2 2t = 0, s = 0 and v = 20 m/s.What are position and velocity as a function of time?Solution: The velocity isv =

a dt +C1 =

(60t 36t2) +C1 = 30t212t3+C1.At t = 0, v = 20 m/s,henceC1 = 20,andthevelocityas a functionof time isv = 30t212t3+20 m/s.The position iss =

v dt +C2 =

(30t212t3+20) +C2= 10t33t4+20t +C2.Att = 0, s = 0, henceC2 = 0, and the position iss = 10t33t4+20t15m/s. Whenmc 2008PearsonEducation South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from thepublisherprior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Problem 13.16 As a rst approximation, a bioengineerstudyingthemechanicsofbirdight assumesthat thesnow petrel takes offwithconstant acceleration. Videomeasurements indicate that a bird requires a distance of4.3 m to take off and is moving at 6.1 m/s when it does.What is its acceleration?Solution: The governing equations area = constant, v = at, s =12at2.Using the information given, we have6.1 m/s = at, 4.3 m =12at2.Solving these two equations, we ndt = 1.41 s anda = 4.33 m/s2.Problem 13.17 Progressively developing a more real-istic model, the bioengineer next models the accelerationof the snowpetrel by an equation of the forma =C(1 +sin t ), where C and are constants. From videomeasurements of a birdtakingoff, he estimates that = 18/sanddeterminesthat thebirdrequires1.42 sto take off and is moving at 6.1 m/s when it does. Whatis the constantC?Solution: We nd an expression for the velocity by integrating theaccelerationa = C(1 +sin t ),v = Ct +C(1 cos t ) = C

t +11cos t

.Using the information given, we have6.1 m/s = C

1.42 s +s18 s18 cos[18(1.42)]

Solving this equation, we ndC = 4.28 m/s2.16c 2008PearsonEducation South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from thepublisherprior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Problem 13.18 Missilesdesignedfordefenseagainstballisticmissileshaveattainedaccelerationsinexcessof 100gs, or 100 times the acceleration due to gravity.Suppose that the missile shown lifts off from the groundandhasaconstant accelerationof 100gs. Howlongdoes it take to reach an altitude of 3000 m? How fast isit going when it ranches that altitude?Solution: The governing equations area = constant, v = at, s =12at2Using the given information we have3000 m =12100(9.81 m/s2)t2 t = 2.47 sThe velocity at that time isv = 100(9.81 m/s2)(2.47 s) = 243 m/s. v = 2430 m/s.Problem 13.19 Supposethat themissileshownliftsoff from the ground and, because it becomes lighter asits fuel is expended, its acceleration (ings) is given asa function of time in seconds bya =1001 0.2t.What is the missiles velocity inkilometres per hour 1safter liftoff?Solution: We nd an expression for the velocity by integrating theacceleration (valid only for 0 < t < 5s).v =

t0adt =

t0100g1 0.2(t /s)dt =100gs0.2ln

11 0.2[t /s]

At timet = 1 s, we havev =100( .2)(1 s)0.2ln

11 0.2

= .v = 0 m h.179 81 m/s1095m/s394 k /c 2008PearsonEducation South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from thepublisherprior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Problem 13.20 The airplane releases its drag para-chuteattimet = 0. Itsvelocityisgivenasafunctionof time byv =801 +0.32tm/s.What is the airplanes acceleration att = 3 s?Solution:v =801 +0.32t; a =dvdt=25.6(1 +0.32t )2 a(3 s) = 6.66 m/s2Problem 13.21 How far does the airplane in Problem13.20travel duringtheinterval oftimefromt = 0tot = 10 s?Solution:v =801 +0.32t; s =

10 s0801 +0.32tdt = 250 ln

1 +3.21

= 359 mProblem 13.22 The velocity of a bobsled is v =10t t = 2 s,thepositioniss = 25 m. Whatis its position att = 10 s?Solution: Theequationforstraightline displacementundercon-stant acceleration iss =a(t t0)22+v(t0)(t t0) +s(t0).Chooset0 = 0. Att = 2, the acceleration isa =dv(t )dtt =2= 10 m/s2,the velocity isv(t0) = 10(2) = 20 m/s, and the initial displacement iss(t0) = 25 m. Att = 10 seconds, the displacement iss =102(10 2)2+20(10 2) +25 = 505 m18m/s. Whenc 2008PearsonEducation South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from thepublisherprior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Problem 13.23 In September, 2003, Tony Schumacherstartedfromrestanddrove402 km in 4.498 seconds inaNational Hot RodAssociation race. Hisspeedashecrossed thenish line wa 5 8 /h. Assumethat thecarsacceleration can be expressed bya linear functionof timea = b +ct .(a) Determine the constantsb andc.(b) What was thecars speed2s after thestart ofthe race?Solution:a = b +ct , v = bt +ct22, s =bt22+ct36Both constants of integration are zero.(a) 528 km/h = b(4.498 s) +c2(4.498 s)2=b2(4.498 s)2+c6(4.498 s)3b =2c = 3(b) v = b(2 s) +c2(2 s)2= 9 /sProblem 13.24 The velocity of an object is v = 200 2t2m/s. Whent = 3 seconds, its position iss = 600 m.Whatarethepositionandaccelerationoftheobjectatt = 6 s?Solution: The acceleration isdv(t )dt= 4t m/s2.At t = 6seconds,theaccelerationisa = 24 m/s2.Choosetheini-tial conditionsat t0 = 3seconds. Thepositionisobtainedfromthevelocity:s(t t0) =

63v(t ) dt +s(t0) =200t 23t363+600 = 1070 m.Problem 13.25 An inertial navigation systemmea-sures the acceleration of a vehicle fromt = 0 tot = 6 sanddetermines it tobe a = 2 +0.1t m/s2. At t = 0,the vehicles position and velocityares = 240 m, v =42 m/s, respectively. What are the vehicles position andvelocity att = 6 s?Solution:a = 2 +0.1t m/s2v0 = 42 m/s s0 = 240 mIntegratingv = v0+2t +0.1t2/2s = v0t +t2+0.1t3/6 +s0Substituting the known values att = 6 s, we getv = 55.8 m/ss = 531.6 m1940254m/s9.5m/s8 ms 2 kmkmc 2008PearsonEducation South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from thepublisherprior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Problem 13.26 In Example 13.3, suppose that the chee-tahs acceleration is constant and it reaches its top speedof 120 km/h in 5 s. What distance can it cover in 10 s?Solution: The governing equations while accelerating area = constant, v = at, s =12at2.Using the information supplied, we have = a(5 s) a =2Thedistancethat hetravelsintherst10s(5secondsacceleratingand then the last 5 seconds traveling at top speed) iss =12(2) (5 s)2+ (10 s 5 s) = .s = .Problem 13.27 The graph shows the airplanes accel-eration during its takeoff. What is the airplanes velocitywhen it rotates (lifts off) att = 30 s?at9 m/ s23 m/s25 s 0030 sSolution: Velocity = Area under the curvev =12(3 m/s2+9 m/s2)(5 s) +(9 m/s2)(25 s) =255 m/sProblem 13.28 Determine the distance traveled duringits takeoff by the airplane in Problem 13.27.Solution: for 0 t 5 sa =

6 m/s25 s

t +(3 m/s2), v =

6 m/s25 s

t22 +(3 m/s2)ts =

6 m/s25 s

t36 +(3 m/s2)t22v(5 s) = 30 m/s, s(5 s) = 62.5 mfor 5 s t 30 sa = 9 m/s2, v = (9 m/s2)(t 5 s) +30 m/s,s =(9 m/s2)(t 5 s)22+(30 m/s)(t 5 s) +62.5 ms(30 s) = 362520120 1000 m3600 s6. 67 m/s6.67 m/s120 1000 m3600 sm 250m 250mc 2008PearsonEducation South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from thepublisherprior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Problem 13.29 Thecaris traveling at 48 km/hwhenthetrafclight 90 maheadturnsyellow. Thedrivertakes one second to react before he applies the brakes.(a) Afterheappliesthebrakes, whatconstantrateofdecelerationwill causethecartocometoastopjust as it reaches the light?(b) How long does it take the car to travel the 90 m?90 m /h 48 kmSolution: for 0 t 1 sa = 0, v = = , s = ( )ts(1 s) =fort > 1 sa = c(constant), v = ct + , s = ct22 +( )t +At the stop we have90 m = ct22 +( )t +a) c = . 7 /s2b) t = 11.41 s0 = ct +Problem 13.30 Thecaristravelingat 48 km/h whenthetrafclight 90 maheadturns yellow. The drivertakes1 storeact beforeheappliestheaccelerator. Ifthecar has aconstant accelerationof 2 m/s2andthelight remains yellow for 5 s, will the car reach the lightbefore it turns red? How fast is the car moving when itreaches the light?Solution: First, convert the initial speed into m/s.= .At the end of the 5 s, the car will have traveled a distanced = ( )(1 s) +

12(2)(5 s 1 s)2+( )(5 s 1 s)

= .When the light turns red, the driver will still be 7.35 mfrom the light.No.To nd the time at which the car does reach the light, we solve90m=( )(1 s) +

12(2)(t 1 s)2+ ( )(t 1 s)

t = 5. .The speed at this time isv = +(2) (5. 1 s) =v = . m h.2148 km/h 13.33m/s 13.33m/s13. 33m13.33 m/s 13.33 m13. 33m/s1 1 m48 km/h 13.33m/s13. 33m/s 2 m/s 13.33 m/s m 82.6513.33 m/s 2m/s 13.33 m/s34 s13.33 m/s 2 m/s 34 s 22.01 m/s.79 2k /13.33 m/s 13.33m/s13.33mc 2008PearsonEducation South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from thepublisherprior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Problem 13.31 Ahigh-speedrail transportationsys-tem has a top speed of 100 m/s. For the comfort of thepassengers, the magnitude of the acceleration and decel-eration is limited to 2 m/s2. Determine the time requiredfor a trip of 100 km.Strategy: A graphical approach can help you solve thisproblem. Recall that the change in the position from aninitial timet0toatimet isequal totheareadenedby the graph of the velocity as a function of time fromt0tot .Solution: Divide the time of travel into three intervals: The timerequiredtoreachatopspeedof 100 m/s, thetimetravelingat topspeed, and the time required to decelerate from top speed to zero. Fromsymmetry, the rst and last time intervals are equal, and the distancestraveled duringthese intervals are equal.Theinitial timeis obtainedfromv(t1) = at1, fromwhicht1 = 100/2 = 50 s. Thedistancetrav-eled during this time iss(t1) = at21/2 from whichs(t1) = 2(50)2/2 =2500 m. The third time interval is given byv(t3) = at3+100 = 0,from which t3 = 100/2 = 50 s. Check. The distance traveled is s(t3) =a2t23 +100t3, from which s(t3) = 2500 m. Check. The distance trav-eled at top speed is s(t2) = 100000 2500 2500 = 95000 m= 95 km. The time of travel is obtained from the distance traveled atzeroacceleration: s(t2) = 95000 = 100t2, fromwhicht2 = 950. Thetotal time of travel isttotal = t1+t2+t3 = 50 +950 +50 = 1050 s= 17.5 minutes.Aplot ofvelocityversustimecanbemadeandtheareaunder thecurve will be the distance traveled.The length of the constant speedsection of the trip can be adjusted to force the length of the trip to bethe required 100 km.Problem 13.32 The nearest star, Proxima Centauri, is4.22 light years from the Earth. Ignoring relative motionbetweenthesolar systemandProximaCentauri, sup-posethat aspacecraft acceleratesfromthevicinityoftheEarthat0.01 g(0.01timestheaccelerationduetogravity at sea level) until it reaches one-tenth the speedof light, coasts until it is time to decelerate, then decel-erates at 0.01 g until it comes to rest in the vicinity ofProxima Centauri. How long does the trip take? (Lighttravels at 3 108m/s.)Solution: The distance to Proxima Centauri isd = (4.22 light - year)(3 108m/s)(365.2422 day)

86400 s1 day

= 3.995 1016m.Divide the time of ight into the three intervals. The time required toreach 0.1 times the speed of light ist1 =va=3 107m/s0.0981 m/s2 = 3.0581 108seconds.The distance traveled iss(t1) =a2t21 +v(0)t +s(0),wherev(0) = 0ands(0) = 0(fromtheconditionsintheproblem),fromwhich s(t1) = 4.587 1015m. Fromsymmetry, t3 = t1, ands(t1) = s(t3). The length of the middle interval iss(t2) = d s(t1) s(t3) = 3.0777 1016m. The time of ight at constant velocity ist2 =3.0777 1016m3 107= 1.026 109seconds.Thetotal timeof ight is ttotal = t1+t2+t3 = 1.63751 109sec-onds. In solar years:ttotal =

1.63751 109sec

1 solar years365.2422 days

1 days86400 sec

= 51.9 solar years22c 2008PearsonEducation South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from thepublisherprior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Problem 13.33 A race car starts fromrest and accel-erates at a = 5 +2t2for 10 seconds. The brakesarethenapplied, andthecar has aconstant acceler-ation a = 30 m/s2until it comes torest. Determine(a) themaximumvelocity, (b) thetotal distancetrav-eled; (c) the total time of travel.Solution:(a) For the rst interval, the velocity isv(t ) =

(5 +2t ) dt +v(0) = 5t +t2sincev(0) = 0. The velocity is an increasing monotone function;hencethe maximumoccursat theend ofthe interval, t = 10 s,from whichvmax = 150 m/s .(b) The distance traveled in the rst interval iss(10) =

100(5t +t2) dt =52t2+ 13t3100= 583.33 .The time of travel in the second interval isv(t210) = 0 = a(t210) +v(10), t2 10 s,from which(t210) = 15030 = 5, and(c) the total time of travel ist2 = 15. The total distance traveled iss(t210) =a2(t210)2+v(10)(t210) +s(10),from which (b)s(5) = 30252+150(5) +583.33 = 958.33 mProblem 13.34 Whent = 0, the position of a point iss = 6 m and its velocity is v = 2 m/s. From t = 0 to t =6 s, theaccelerationof thepoint is a = 2 +2t2m/s2.Fromt = 6 suntil it comestorest, itsaccelerationisa = 4 m/s2.(a) What is the total time of travel?(b) What total distance does the point move?Solution: For the rst interval the velocity isv(t ) =

(2 +2t2) dt +v(0) =2t + 23t3+2 m/s.Thevelocityat theendof theinterval is v(6) = 158 m/s. Thedis-placement in the rst interval iss(t ) = 2t + 23t3+2

dt +6 =t2+ 16t4+2t+6.The displacement at the end of the interval iss(6) = 270 m. For thesecond interval, the velocity isv(t 6) = a(t 6) +v(6) = 0, t 6,from which(t 6) = v(6)a= 1584 = 39.5.The total time of travel is(a) ttotal = 39.5 +6 = 45.5 seconds.(b) The distance traveled iss(t 6) = 42(t 6)2+v(6)(t 6) +s(6)= 2(39.5)2+158(39.5) +270,from which the total distance is stotal = 3390 m23m/smc 2008PearsonEducation South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from thepublisherprior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Problem 13.35 Zoologists studying the ecology of theSerengetiPlainestimatethattheaverageadultcheetahcanrun100 km/handthattheaveragespringbuckcanrun 65 km/h. If the animals run along the same straightline, start at thesametime, andareeachassumedtohaveconstant accelerationandreachtopspeedin4 s,how close must the a cheetah be when the chase beginsto catch a springbuck in 15 s?Solution: The top speeds areVc = 100 km/h = 27.78 m/s for thecheetah, and Vs = 65 km/h = 18.06 m/s. The acceleration is ac =Vc4= 6.94 m/s2forthecheetah, andas =Vs4= 4.513 m/s2forthespringbuck. Dividetheintervalsintotheaccelerationphaseandthechase phase. For the cheetah, the distance traveled inthe rst issc(t ) =6.942(4)2= 55.56 m. Thetotal distancetraveledat theendof the second phase is stotal = Vc(11) +55.56 = 361.1 m. For thespringbuck, the distance traveled during the acceleration phase isss(t ) =4.5132(4)2= 36.11 m. The distance traveled at the end of thesecond phase is ss(t ) = 18.06(11) +36.1 = 234.7 m. The permissibleseparation between the two at the beginning for a successful chase isd = sc(15) ss(15) = 361.1 234.7 = 126.4 mProblem 13.36 Suppose that a person unwisely drivescers beginconstant accelerationat theinstant they are passed andincrease their speed to 129 km/h in 4 s , how long does ittake them to be even with the pursued car?Solution: The conversion from mi/h to m/s isThe acceleration of the police car isa =( )( . )4 s=2.The distance traveled during acceleration iss(t1) =.2(4)2+ ( . )(4) = .The distance traveled by the pursued car during this acceleration issc(t1) = ( . ) t1= (4) =The separation between the two cars at 4 seconds isd = = .This distance is traversed in the timet2 =( )( . )= .The total time is ttotal = + 4 = seconds.Problem 13.37 If = 1 radandddt= 1 rad/s, whatis the velocity ofPrelative toO?Strategy: You can write the position ofPrelative toOas s = (2 m) cos +(2 m) cos and then take thederivative of this expressionwith respect to time todetermine the velocity.s2 mOP2 mSolution: The distancesfrom point Oiss = (2 m) cos +(2 m) cos .The derivative isdsdt= 4 sin ddt.For = 1 radian andddt= 1 radian/second,dsdt= v(t ) = 4(sin(1 rad)) = 4(0.841) = 3.37 m/s24120 km/h in a 88 km/h zone and passes a police car going88 km/h in the same direction. If the police ofkmh=1000 m3600s= 0.278m/s129 88 0 278 m/s2.85 m/s2 85880 278 m 121120 0 278 33. 36 . 133. 4 m133.4 121 12.4m12.4129 88 0 2781. 091. 09 5. 09c 2008PearsonEducation South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from thepublisherprior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Problem 13.38 In Problem 13.37, if = 1 rad,d/dt= 2 rad/s andd2/dt2= 0, what are the velocity andacceleration ofPrelative toO?Solution: The velocity isdsdt= 4 sin ddt= 4(sin(1 rad))(2) = 6.73 m/s .The acceleration isd2sdt2 = 4 cos

ddt

24 sin

d2dt2

,from whichd2sdt2 = a = 4 cos(1 rad)(4) = 8.64 m/s2Problem 13.39 If = 1 radandddt= 1 rad/s, whatis the velocity ofPrelative toO?sOP200 mm400 mmSolution: Theacuteangleformedbythe400 mmarmwiththehorizontal is given by the sine law:200sin =400sin ,from whichsin =

200400

sin .For = 1 radian, = 0.4343 radians. The position relative toOis.s = 200 cos +400 cos .The velocity isdsdt= v(t ) = 200 sin

ddt

400 sin

ddt

.From the expression for the angle, cos

ddt

= 0.5 cos

ddt

,from which the velocity isv(t ) = (200 sin 200 tan cos )

ddt

.Substitute: v(t ) = 218.4 mm/s.25c 2008PearsonEducation South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from thepublisherprior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Problem 13.40 In Active Example 13.4, determine thetimerequiredfortheplanes velocitytodecrease from50 m/s to 10 m/s.Solution: From Active Example 13.4 we know that the accelera-tion is given bya = (0.004/m)v2.Wecanndanexpressionforthevelocityasafunctionoftimebyintegratinga =dvdt= (0.004/m)v2dvv2 = (0.004/m)t

10m/s50m/sdvv2 =

1v

10m/s50m/s=

1 s10 m +1 s50 m

=

4 s50 m

= (0.004/m)tt =

4 s50 m

1 m0.004

= 20 s.t = 20 s.Problem 13.41 An engineer designing a system to con-trol arouter for amachiningprocessmodelsthesys-tem so that the routers acceleration (in cm/s ) during an2interval of time is given bya = 0.4v, wherevis thevelocity of the router in cm/s. When t = 0, the position iss =0 and the velocity is v =2 cm/s. What is the positionatt = 3 s?sSolution: We will rst nd the velocity at t = 3.a =dvdt=

0.4s

v,

v22cm/sdvv=

3s0

0.4s

dt,ln

v22 cm/s

=

0.4s

(3 s) = 1.2v2 = (2 cm/s)e1 . 2= 0.602 cm/s.Now we can nd the positiona =vdvds=

0.4s

v,

0.602cm/s2cm/svdvv=

0.4s

s20ds(0.602 cm/s) (2 cm/s)=

0.4s

s2s2 = 1.398 cm/s0.4/ s= 3.49 cms2 = 3.49 cm26c 2008PearsonEducation South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from thepublisherprior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Problem 13.42 Theboat is movingat 10 m/s whenits engine is shut down. Due to hydrodynamic drag, itssubsequent accelerationis a = 0.05v2m/s2, wherevisthevelocityof theboat inm/s. What istheboatsvelocity 4 s after the engine is shut down?Solution:a =dvdt= (0.05 m1)v2

v10 m/sdvv2 = (0.05 m1)

t0dt 1v

v10 m/s= (0.05 m1)tv =10 m/s1 +(0.5 s1)tv(4 s) = 3.33 m/sProblem 13.43 In Problem 13.42, what distance doestheboat moveinthe4s followingtheshutdownofits engine?Solution: From Problem 13.42 we knowv =dsdt=10 m/s1 +(0.5 s1)ts(4 s) =

4s010 m/s1 +(0.5 s1)tdts(4 s) = (20 m) ln2 +(1 s1)(4 s)2

= 21.97 mProblem 13.44 Asteel ball isreleasedfrom rest in acontainer of oil. Itsdownward accelerationis a = 2. 4 0.6v2v is the balls velocity in cm/s. What isthe balls downward velocity 2 s after it is released?Solution:a =dvdt= (2.4 cm/s) (0.6 s1)v

v0dv(2.4 cm /s) (0.6 s1)v=

t0dt 53 ln

v +4 cm/s4 cm/s

= t v = (4 cm/s)

1 e(0.6s1)t

v(2 s) = 2.795 cm/sProblem 13.45 In Problem 13.44, what distance doesthe ball fall in the rst 2 s after its release?Solution: From 13.44 we knowv =dsdt= (4 cm/s)

1 e(0.6s1)t

s(2 s) =

t0(4 cm/s)

1 e(0.6s1)t

dt=20 cm3

e(0.6s1)t1

+(4 cm/s )ts(2 s) = 3.34 cm.27cm/s, wherec 2008PearsonEducation South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from thepublisherprior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Problem 13.46 Thegreatest oceandepthyet discov-ered is the Marianas Trench in the western Pacic Ocean.A steel ball released at the surface requires 64 minutestoreachthebottom. Theballsdownwardaccelerationis a = 0.9g cv, where g = 9.81 m/s2and the constantc = 3.02 s1. What is the depth of the Marianas Trenchin kilometers?Solution:a =dvdt= 0.9g cv.Separating variables and integrating,

v0dv0.9g cv=

t0dt = t .Integrating and solving forv,v =dsdt=0.9gc(1 ect).Integrating,

s0ds =

t00.9gc(1 ect) dt.We obtains =0.9gc

t +ectc 1c

.Att = (64)(60) = 3840 s, we obtains = 11,225 m.Problem 13.47 The acceleration of a regional airlinerduring its takeoff run isa = 14 0.0003v2 2v is its velocity in /s. How long does it take the airlinerto reach its takeoff speed of 200 m/s?Solution:a =dvdt= (14 /s2) (0.0003 m1)v2

200 m/s0dv(14 m/s2) (0.0003 m1)v2 =

t0dtt = 25.1 sProblem 13.48 In Problem 13.47, what distance doesthe airliner require to take off?Solution:a = vdvds= (14 m/s2) (0.0003 m1)v2

200m/s0vdv(14 m/s2) (0.0003 m1)v2 =

s0dss = 324328m/s , wheremmmc 2008PearsonEducation South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from thepublisherprior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Problem 13.49 Askydiverjumpsfromahelicopterand is falling straight down at 30 m/s when her parachuteopens. Fromthen on, her downward acceleration isapproximately a = g cv2, where g = 9.81 m/s2andcisaconstant. After aninitial transientperiodshedescends at a nearly constant velocity of 5 m/s.(a) What is the value ofc, and what are its SI units?(b) What maximum deceleration is the sky diversubjected to?(c) What is her downward velocity when she has fallen2meters fromthepoint at whichher parachuteopens?Solution: Assumec> 0.(a) After the initial transient, she falls at a constant velocity, so thatthe acceleration is zero andcv2= g, from whichc =gv2 =9.81 m/s2(5)2m2/s2 = 0.3924 m1(b) Themaximumacceleration(inabsolute value)occurswhen theparachute rst opens, when the velocity is highest:amax = |g cv2| = |g c(30)2| = 343.4 m/s2(c) Choose coordinates such that distance is measured positive down-ward. The velocity is related to position by the chain rule:dvdt=dvdsdsdt= vdvds= a,from whichv dvg cv2 = ds.Integrate:

12c

ln |g cv2| = s +C.When the parachute openss = 0 andv = 30 m/s, from whichC =

12c

ln |g 900c| = 7.4398.The velocity as a function of distance is ln |g cv2| =2c(s +C). Fors = 2 m,v = 14.4 m/sProblem 13.50 Therocket sledstarts fromrest andaccelerates at a = 30 +2t m/s2until its velocity is400 m/s. Itthenhits a waterbrake and itsaccelerationis a = 0.003v2m/s2until its velocity decreases to100 m/s. What total distance does the sled travel?Solution: Acceleration Phasea = 30 +2t m/s2v = 30t +t2m/ss = 15t2+t3/3 mWhenv = 400 m/s, acceleration ends. At this point,t = 10 s ands =1833 m. Deceleration Phase starts ats1 = 1833 m,v1 = 400 m/s. Letus start a new clock for the deceleration phase. vf = 100 m/sa = vdvds= 0.003v2

sfs1ds = 1(0.003)

vfv1v dvv2sf 1833 m = 1(0.003)[ln(100) ln(400)]sf = 2300 m29c 2008PearsonEducation South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from thepublisherprior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Problem 13.51 InProblem13.50, what isthesledstotal time of travel?Solution: FromthesolutiontoProblem13.50, theaccelerationtakes 10 s. At t = 10 s, the velocity is 400 m/s. We need to nd outhow long it takes to decelerate from 400 m/s to 100 m/s and add thisto the 10 s required for acceleration. The deceleration is given asa =dvdt= 0.003v2m/s20.003

td0dt =

100400dvv20.003td = 1v

100400=

1100 1400

0.003td =3400td = 2.5 st = 10 +td = 12.5 sProblem 13.52 Acars accelerationis relatedtoitspositionbya = 0.01sm/s2. Whens = 100 m, thecaris moving at 12 m/s. How fast is the car moving whens = 420 m?Solution:a = vdvds= 0.01sm/s2

vf12v dv = 0.01

420100s dsv22vf12m/s= 0.01s22420m100mv2f2=1222+0.01(42021002)2vf = 42.5 m/sProblem 13.53 Engineers analyzingthemotionof alinkage determine that the velocity of an attachmentpoint isgivenbyv = A +4s2Aisacon-stant. Whens = 2 m, itsaccelerationismeasuredanddeterminedtobea = 320 m/s.Whatisitsvelocityof2the point whens = 2 m?Solution: The velocity as a function of the distance isvdvds= a.Solve fora and carry out the differentiation.a = vdvds= (A+4s2)(8s).Whens = 2 m, a = 320 m/s, from which2A = 4.The velocity ats = 2 m isv = 4 +4(22) = 20 /s30m/s, wheremc 2008PearsonEducation South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from thepublisherprior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Problem 13.54 The acceleration of an object is givenasafunctionofitspositioninfeet bya = 2 s2(2).When s = 0, its velocity is v = 1 m/s. What is the veloc-ity of the object whens = 2 m?Solution: We are givena =vdvds=

22

s2,

v1m/svdv =

22

2m0s2dsv22(1 m/s)22=

22

(2 m)33v = 3.42 /s.Problem 13.55 Gas guns are used to investigatethe properties of materials subjectedtohigh-velocityimpacts. Aprojectileisacceleratedthroughthebarrelof thegunbygas at highpressure. Assumethat theacceleration of the projectile is given bya = c/s, wheresis the position of the projectile in the barrel in metersand c is a constant that depends on the initial gas pressurebehindtheprojectile. Theprojectilestartsfromrest ats = 1.5 m and accelerates until it reaches the end of thebarrelat s = 3 m.Determinethevalueoftheconstantcnecessary for the projectile to leave the barrel with avelocity of 200 m/s.sSolution:a = vdvds=cs,

200 m/s0vdv =

3 m1.5 mcsds(200 m/s)22= c ln

3 m1.5 m

c = 28.85 103m2/s2Problem 13.56 If thepropellinggas inthegas gundescribed in Problem13.55 is air, a more accuratemodeling of the acceleration of the projectile is obtainedbyassumingthat theaccelerationof theprojectileisgiven by a = c/s, where = 1.4 is the ratio of specicheatsforair. (Thismeansthatanisentropicexpansionprocess is assumedinsteadof the isothermal processassumed in Problem 13.55.) Determine the value of theconstant c necessary for the projectile to leave the barrelwith a velocity of 200 m/s.Solution:a = vdvds=cs1.4,

200 m/s0vdv =

3 m1.5 mcs1.4ds(200 m/s)22= 2.5c

(3m)0.4(1.5m)0.4

c = 38.86 103m2.4/s231m/sm-sm s -m-smc 2008PearsonEducation South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from thepublisherprior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Problem 13.57 A spring-mass oscillator consists of amass and a spring connected as shown. The coordinatesmeasures the displacement of the mass relative to itspositionwhenthespringis unstretched. If thespringislinear, themassissubjectedtoadecelerationpro-portional tos. Suppose thata = 4sm/s2, and that yougive the mass a velocity v = 1 m/s in the position s = 0.(a) How far will the mass move to the right before thespring brings it to a stop?(b) What will be the velocity of the mass when it hasreturned to the positions = 0?sSolution: The velocity of the mass as a function of its posi-tion is given byv dv/ds = a. Substitute the given acceleration, sepa-rate variables, and integrate: v dv = 4s ds, fromwhich v2/2 =2s2+C. Theinitial velocity v(0) = 1 m/s at s = 0, fromwhichC = 1/2. The velocity isv2/2 = 2s2+1/2.(a) The velocity is zero at the position given by0 = 2(s1)2+ 12,from which s1 =

14 = 12m.Since the displacement has the same sign as the velocity,s1 = +1/2 mis the distance traveled before the spring brings it to a stop.(b) At the return to s = 0, the velocity is v =

22 = 1 m/s. Fromthe physical situation, the velocity on the rst return is negative(opposite the sign of the initial displacement),v = 1 m/s.Problem 13.58 In Problem 13.57, suppose that att = 0youreleasethemass fromrest inthepositions = 1 m. Determine the velocity of the mass as a func-tion ofsas it moves from the initial position tos = 0.Solution: FromthesolutiontoProblem 13.57, thevelocityasafunction of position is given byv22= 2s2+C.At t = 0, v = 0 and s = 1 m, from which C = 2(1)2= 2. The velocityis given byv(s) = (4s2+4)12 = 21 s2m/s.From the physical situation, the velocity is negative (opposite the signof the initial displacement):v = 21 s2m/s.[Note: From the initial conditions, s2 1 always.]32c 2008PearsonEducation South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from thepublisherprior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Problem 13.59 A spring-mass oscillator consists of amass and a spring connected as shown. The coordinatesmeasures the displacement of the mass relative to itspositionwhenthespringis unstretched. Supposethatthe nonlinear spring subjects the mass to an accelerationa = 4s2s3m/s2and that you give the mass a velocityv = 1 m/s in the positions = 0.(a) How far will the mass move to the right before thesprings brings it to a stop?(b) What will be the velocity of the mass when it hasreturned to the positions = 0?sSolution:(a) Find the distance when the velocity is zero.a =vdvds=

4s2

s +

2m2s2

s3

01m/svdv =

d04s2

s +

2m2s2

s3ds(1 m/s)22=

4s2

d22

2m2s2

d44Solving fordwe nd d = 0.486m.(b) When the cart returns to the positions = 0, we havea =vdvds=

4s2

s +

2m2s2

s3

v1m/svdv =

004s2

s +

2m2s2

s3ds = 0v22(1 m/s)22= 0, v = 1 m/s.The cart will be moving to the left, so we choose v = 1 m/s.Problem 13.60 Themass is releasedfromrest withthe springs unstretched. Its downwardaccelerationisa = 32.2 50s2s is the position of the massmeasured fromthe position in which it is released.(a) How far does the mass fall? (b) What is themaximum velocity of the mass as it falls?sSolution: The acceleration is given bya =dvdt=dvdsdsdt= vdvds= 32.2 50s2.Integrating, we get

v0v dv =

s0(32.2 50s) ds orv22= 32.2s 25s2.(a) Themassfallsuntil v = 0. Settingv = 0, weget 0 = (32.2 25s)s. Wend v = 0at s = 0andat s = 1.288 m. Thus, themass falls 1.288 m before coming to rest.(b) Fromtheintegrationoftheequationofmotion, wehavev2=2(32.2s 25s2). The maximum velocity occurs wheredvds= 0.From the original equation for acceleration, we havea = vdvds=(32.2 50s)2assume that v = 0 at this point. Thus, 0 = (32.2 50s), ors =(32.2/50) v =vMAX. Substitutingthisvaluefor s intothe equation forv, we getv2MAX = 2

(32.2)250(25)(32.2)2502

,or vMAX = 4.55 /s33m/s, wherem/sm/s. Sincewewant maximumvelocity, wecanmwhenmc 2008PearsonEducation South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from thepublisherprior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Problem 13.61 Suppose that the mass in Problem13.60 is in the positions = 0 and is given a downwardvelocity of 10 m/s.(a) How far does the mass fall?(b) What is the maximumvelocity of the mass asit falls?Solution:a = vdvds= (32.2 m/s2) (50 s2)s

v10m/svdv =

s0[(32.2 m/s2) (50 s2)s]dsv22(10 m/s)22= (32.2 m/s2) s (50 s2)s22v2= (10 m/s)2+(64.4 m/s2)s (50 s2)s2(a) The mass falls untilv = 00 = (10 m/s)2+(64.4 m/s2)s (50 s2)s2s = 2.20 m(b) The maximum velocity occurs whena = 00 = (32.2 m/s2) (50 s2)s s = 0.644 mv2= (10 m/s)2+(64.4 m/s2)(0.644 m) (50 s2)(0.644 m)2v = 10.99 /sProblem 13.62 If a spacecraft is 161 km above the sur-face of the earth, what initial velocityv0straight awayfromthe earthwould be required for the vehicle toreachthemoonsorbit , m fromthecenteroftheearth?Theradiusoftheearth is 6372 km. Neglectthe effect of the moons gravity. (See Example 13.5.)0161 km382,942 kmSolution: For computational convenience, convert the accelerationdue to Earths gravity into the units given in the problem, namely milesand hours:g =

.1s2

1 km 36002s21 h2

= .2.The velocity as a function of position is given byvdvds= a = gR2Es2.Separate variables,v dv = gR2Edss2.Integrate:v2= 2gR2E

12

+C.Suppose that the velocity at the distance of the Moons orbit is zero.Then0 = 2( . )

2 +C,fromwhich C = 2/h. At the161 kmaltitude, the2equation for the velocity isv20 = 2g

R2ERE+161

+C.From whichv0 ==Converting:v0 =

1 h

1 km

1 h3600 s

= ,Check: Use the result of Example 13.5v0 =

2gR2E

1s01H

,(where H >s0always), and H = , ,from which v0 = ,4 m/h. check.34m382 942 k9 81 m1000 m127137 6km/h127137 6637238294226960164 km1553351991 39,413 km/h39413 km 1000 m10 948 m/s .382 94239 13 kc 2008PearsonEducation South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from thepublisherprior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Problem 13.63 Themoons radius is 1738 km. Themagnitude of the acceleration due to gravity of the moonat a distancesfrom the center of the moon is4.89 1012s2m/s2.Supposethataspacecraftislaunchedstraight upfromthe moons surface with a velocity of 2000 m/s.(a) What will the magnitude of its velocity be when itis 1000 km above the surface of the moon?(b) Whatmaximumheight abovethemoonssurfacewill it reach?Solution: SetG = 4.89 1012m3/s2, r0 = 1.738 106m,v0 = 2000 m/sa = vdvds= Gs2

vv0vdv =

rr0Gs2ds v22v022= G

1r1r0

v2= v02+2G

r0rrr0

(a) v(r0+1.0 106m) = 1395 m/s(b) The maximum velocity occurs whenv = 0r =2G2Gr0v02 = 6010 km h = r r0 = 4272 kmProblem 13.64* Thevelocityof anobject subjectedonly to the earths gravitational eld isv =v20 +2gR2E

1s1s01/2,wheresis the objects position relative to the center ofthe earth,v0 is the velocity at positions0, andRE is theearths radius. Using this equation, show that the objectsacceleration is given as a function of s by a = gR2E/s2.Solution:v = v20 +2gR2E

1s1s0

1/2a =dvdt= vdvdsRewrite the equation given asv2= v20 + 2gR2Es 2gR2Es0Take the derivative with respect tos.2vdvds= 2gR2Es2Thusa = vdvdsgR2Es2Problem 13.65 Suppose that a tunnel could be drilledstraight throughtheearthfromtheNorthPoletotheSouth Pole and the air was evacuated. An object droppedfromthesurfacewouldfall withtheaccelerationa =gs/RE, wheregistheaccelerationofgravityat sealevel, RE is radius of the earth, and s is the distance of theobject from the center of the earth. (The acceleration dueto gravitation is equal to zero at the center of the earthand increases linearly with the distance from the center.)What isthemagnitudeof thevelocityofthedroppedobject when it reaches the center of the earth?SNTunnelREsSolution: The velocity as a function of position is given byvdvds= gsRE.Separate variables and integrate:v2=

gRE

s2+C.Ats = RE, v = 0, from whichC = gRE. Combine and reduce:v2= gRE

1 s2R2E

At the center of the earths = 0, and the velocity is v =gRE35c 2008PearsonEducation South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from thepublisherprior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Problem 13.66 Determine the time in seconds requiredfor the object in Problem 13.65 to fall from the surfaceof the earth to the center. The earths radius is 6370 km.Solution: From Problem 13.65, the acceleration isa = vdvds= gREs

v0v du =

sRE

gRE

s dsv2=

gRE

(R2Es2)Recall thatv = ds/dtv =dsdt=

gRE

R2Es2

0REds

R2Es2=

gRE

tf0dt

gREtf = sin1

sRE

0RE= sin1(1)tf =

REg2 = 1266 s = 21.1minProblem 13.67 In a second test, the coordinates of theposition (in m) of the helicopter in Active Example 13.6are given as functions of time byx = 4 +2t ,y = 4 +4t +t2.(a) What is the magnitude of the helicopters velocityatt = 3 s?(b) What is the magnitude of the helicopters acceler-ation att = 3 s?yxSolution: We havex = (4 m) +(2 m/s)t , y = (4 m) +(4 m/s)t +(1 m/s2)t2,vx = (2 m/s), vy = (4 m/s) +2(1 m/s2)t ,ax = 0, ay = 2(1 m/s2).Att = 3 s, we havex = 10 m, vx = 2 m/s, ax = 0,y = 25 m, vy = 10 m/s, ay = 2 m/s2.Thusv =

v2x +v2y = 10.2 m/s v = 10.2 m/s.a =

02+(2 m/s2)2= 2 m/s2a = 2 m/s2.36c 2008PearsonEducation South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from thepublisherprior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Problem 13.68 In terms of a particular referenceframe, thepositionof thecenter of massof theF-14at thetimeshown (t = 0) is r = 10i +6j +22k (m).The velocity fromt = 0 tot = 4 s is v = (52 +6t )i +(12 +t2)j (4 +2t2)k (m/s). What is thepositionofthe center of mass of the plane att = 4 s?Solution:r0 = 10i +6j +22k mv = (52 +6t )i +(12 +t2)j (4 +2t2)k m/sx4 =

40vx dt = 52t +3t2+x0x4 = (52)(4) +3(4)2+10 m = 266.0 my4 =

40vy dt = 12t +t3/3 +y0y4 = 12(4) +(4)3/3 +6 m = 75.3 mz4 =

40vz dt = (4t +2t3/3) +z0z4 = 4(4) 2(4)3/3 +22 = 36.7 mr|t =4s = 266i +75.3j 36.7k(m)Problem 13.69 In Example 13.7, suppose that the anglebetween the horizontal and the slope on which the skierlands is 30 instead of 45. Determine the distancedtothe point where he lands.453 md20Solution: The skier leaves the 20surface at 10 m/s.The equations areax = 0, ay = 9.81 m/s2,vx = (10 m/s) cos 20, vy = (9.81 m/s2)t (10 m/s) sin 20sx = (10 m/s) cos 20t , sy = 12(9.81 m/s2)t2(10 m/s) sin 20tWhen he hits the slope, we havesx = d cos 30 = (10 m/s) cos 20t,sy = (3 m) d sin 30 = 12(9.81 m/s2)t2(10 m/s) sin 20tSolving these two equations together, we nd t = 1.01 s, d = 11.0 m.37c 2008PearsonEducation South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from thepublisherprior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Problem 13.70 Aprojectileislaunchedfromgroundlevel withinitial velocity v0 = 20 m/s. Determine itsrangeR if (a)0 = 30; (b)0 = 45(c)0 = 60.R0u0yxSolution: Setg = 9.81 m/s2, v0 = 20 m/say = g, vy = gt +v0 sin 0, sy = 12gt2+v0 sin 0tax = 0, vx = v0 cos 0, sx = v0 cos 0tWhen it hits the ground, we have0 = 12gt2+v0 sin 0t t =2v0 sin 0gR = v0 cos 0t R =v02sin 20ga) 0 = 30 R = 35.3 mb) 0 = 45 R = 40.8 mc) 0 = 60 R = 35.3 mProblem 13.71 Immediately after the bouncinggolfball leaves the oor, its components of velocity are vx =0.662 m/s andvy = 3.66 m/s.(a) Determinethehorizontal distancefromthepointwheretheballlefttheoortothepoint whereithits the oor again.(b) Theball leavestheoor at x = 0, y = 0. Deter-minetheballs ycoordinateasafunctionof x.(The parabolic function you obtain is shown super-imposed on the photograph of the ball.)xySolution: The governing equations areax = 0, ay = g,vx = vx0, vy = gt +vy0,x = vx0t , y = 12gt2+vy0t(a) When it hits the ground again0 = 12gt2+vy0t t =2vy0gx = vx0t = vx0

2y0g

=2vx0vy0gx =2(0.662 m/s)(3.66 m/s)9.81 m/s2x = 0.494 m.(b) At any point of the ight we havet =xvx0, y = 12g

xvx0

2+vy0

xvx0

y = 12

9.81 m/s2[0.662 m/s]2

x2+3.66 m/s0.662 m/sxy =

11.2m

x2+5.53x.38c 2008PearsonEducation South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from thepublisherprior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Problem 13.72 Suppose that you are designing amortartolauncharescuelinefromcoastguardvesselto ships in distress. The light line is attached to a weightredbythemortar. Neglectaerodynamicdragandtheweight of the line for your preliminary analysis. If youwant the line tobe ableto reachaship91 mawaywhenthemortar isredat 45abovethehorizontal,what muzzle velocity is required?y45xSolution: From 13.70 we know thatR =v02sin 20gv0 =

Rgsin 20v0 =

( )(2)sin(90)=Problem 13.73 In Problem 13.72, what maximumheight above the point where it was red is reached bythe weight?Solution: From Problem 13.70 we havevy = gt +v0 sin 0, sy = 12gt2+v0 sin 0tWhen we reach the maximum height,0 = gt +v0 sin 0 t =v0 sin 0gh = 12gt2+v0 sin 0t h = 12g

v0 sin 0g

2+v0 sin 0

v0 sin 0g

h =v02sin202gPutting in the numbers we haveh =( )2sin2(45)2(2)=3991 9.81 m/s29.9 m/s29.9 m/s9.81 m/s22.78mc 2008PearsonEducation South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from thepublisherprior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Problem 13.74 Whentheathletereleasestheshot, itis 1.82 m above the ground and its initial velocity is v0 =13.6 m/s. Determine the horizontal distance the shot trav-elsfromthepointofreleasetothepointwhereit hitsthe ground.30v0Solution: The governing equations areax = 0,vx = v0 cos 30,sx = v0 cos 30t ,ay = gvy = gt +v0 sin 30sy = 12gt2+v0 sin 30t +hWhen it hits the ground, we havesx = (13.6 m/s) cos 30t,sy = 0 = 12(9.81 m/s2)t2+(13.6 m/s) sin 30t +1.82 m.Solving these two equations, we ndt = 1.62 s, sx = 19.0 m.Problem 13.75 A pilot wants to drop survey markersat remote locations in the Australian outback. If he iesat a constant velocityv0 = 40 m/s at altitudeh = 30 mand the marker is released with zero velocity relative tothe plane, at what horizontal dfrom the desired impactpoint should the marker be released?dh0Solution: We want to nd the horizontal distance traveled by themarker before it strikes the ground (ygoes to zero fort > 0.)ax = 0 ay = gvx = vx0vy = vy0 gtx = x0+vx0t y = y0+vy0t gt2/2From the problem statement,x0 = 0, vy0 = 0, vx0 = 40 m/s, andy0 =30 m The equation for y becomesy = 30 (9.81)t2/2Solving withy = 0, we gettf = 2.47 s.Substituting this into the equation forx, we getxf = 40tf = 98.9 m40c 2008PearsonEducation, Inc., Upper SaddleRiver, NJ. All rightsreserved. Thismaterial isprotectedunder all copyright lawsastheycurrently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.Problem 13.76 If the pitching wedge the golfer isusing gives the ball an initial angle 0 = 50, what rangeof velocitiesv0 will cause the ball to land within 3 m ofthe hole? (Assume the hole lies in the plane of the ballstrajectory).30 m3 mv00Solution: Setthecoordinateoriginatthepointwherethegolferstrikes the ball. The motion in the horizontal (x) direction is given byax = 0, Vx = V0 cos 0, x = (V0 cos 0)t . Themotioninthevertical(y) direction is given byay = g, Vy = V0 sin 0gt, y = (V0 sin 0)t gt22.From thexequation, we can ndthe time at whichthe ball reachesthe required value ofx(27 or 33 metres). This time istf = xf/(V0 cos 0).Wecansubstitute thisinformation theequationforYwithYf =3 mand solve for V0. The results are: For hitting (27,3) , V0 = 31.2 m/s.For hitting (33,3) metre,V0 = 34.2 /s.Problem 13.77 A batter strikes a baseball 0.9 m abovehome plate and pops it up. The second baseman catchesit1.8 mabovesecond base3.68 s after it was hit.What was the balls initial velocity, and what wastheanglebetweentheballsinitialvelocityvectorandthe horizontal? SecondbaseHome plate27 m27mSolution: The equations of motiong =2ax = 0 ay = gvx = v0 cos 0vy = gt +v0 sin 0sx = v0 cos 0t sy = 12gt2+v0 sin 0t +When the second baseman catches the ball we have=v0 cos 0(3.68 s)= 12(2) (3.68 s)2+v0 sin 0(3.68 s)Solving simultaneously we ndv0 = , 0 = 60.441m9.81m/s0. 9 m38.2m1.8 m 9.81 m/s21 m/smetrec 2008PearsonEducation South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from thepublisherprior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Problem 13.78A baseball pitcher releases a fastball withaninitial velocity v0= be the initialangleof theballsvelocityvectorabove the horizontal.When it is released, the ball is 1.83 mabovethegroundand 17.7 mfromthebatters plate. The batters strikezoneextends from0.56 mabovethegroundto 1.37 mabove the ground. Neglecting aerodynamic effects,determinewhether the ball will hit the strike zone (a) if = 1; (b) if = 2. Solution: The initial velocity is v0 = =velocity equations are(1)dvxdt= 0, from whichvx = v0 cos .(2)dvydt= g, from whichvy = gt +v0 sin .(3)dxdt= v0 cos , fromwhich x(t ) = v0 cos t , since the initialposition is zero.(4)dydt= gt +v0 sin , from whichy(t ) = g2t2+v0 sin t + ,since the initial position is y(0) = d =the height is h. The time of passage across the home plate isx(tp) = d = v0 cos tp, from whichtp =dv0 cos .Substitute:y(tp) = h = g2

dv0 cos

2+d tan + .For = 1, h = . , Yes, the pitcher hits the strike zone.For = 2, h = . No, the pitcher misses the strike zone.Problem 13.79 In Problem 13.78, assume that thepitcherreleasestheball at anangle = 1abovethehorizontal, and determine the range of velocitiesv0(instrike zone.Solution: From the solution to Problem 13.78,h = g2

dv0 cos

2+d tan + ,where d = h Solve for the initial velocity:v0 =

gd22 cos2(d tan + h).For h = . v0 = h = . v0 =pitcher will hit the strike zone for velocities of release of. v0 . ,and areleaseangle of = 1. Check: The range of velocities in milesper hour is 112. v0161.1 km/h, which is within the range ofmajor league pitchers, although the 160.9kmupper value is achievableonlybyatalentedfew(NolanRyan, while with the Houston Astros,would occasionally in a game throw a 168.9km fast ball, as measuredby hand held radar from behind the plate).42144.8km/h. Let144.8km/h 40.3m/s.The1.831.83 m. At a distance 17.7 m,1.831 2m1 5mm/s) withinwhichhemust releasetheball tohit the1.8317.7 m, and 1.37 0.56 m.1.831 37, 44.74 m/s. For 0 56 m, 31.15 m/s. The31 15 44 74m/s1km/ h1.37 m0.5617.7mmc 2008PearsonEducation South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from thepublisherprior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Problem 13.80 Azoologystudentisprovidedwithabowandanarrowtippedwitha syringe of sedativeandisassignedtomeasurethetemperatureofablackrhinoceros (Diceros bicornis). The range of his bowwhen it is fully drawn and aimed 45 above the horizon-talis100 m.Atruculentrhinochargesstraighttowardhim at 30 km/h. If he fully draws his bow and aims 20above the horizontal, how far away should the rhino bewhen the student releases the arrow?20Solution: The strategy is (a) to determine the range and ight timeof the arrow when aimed 20 above the horizontal, (b) to determine thedistance traveled by the rhino duringthis ight time, andthen (c) toadd this distance to the range of the arrow. Neglect aerodynamic dragon the arrow. The equations for the trajectory are: Denote the constantsof integration byVx, Vy, Cx, Cy, and the velocity of the arrow byVA.(1)dvxdt= 0, from whichvx = Vx. Att = 0, Vx = VA cos .(2)dvydt= g, from which vy = gt +Vy. At t = 0, Vy =VA sin .(3)dxdt= vx = VA cos , fromwhich x(t ) = VA cos t +Cx. Att = 0, x(0) = 0, from whichCx = 0.(4)dydt= vy = gt +VA sin , from whichy = g2t2+VA sin t +Cy.At t = 0, y = 0, fromwhich Cy = 0. The time of ight isgiven byy(tight) = 0 =

g2tight+VA sin

tight,from whichtight =2VA sin g.The range is given byx(tight) = R = VA cos tight =2V2A cos sin g.The maximumrange (100 meters) occurs when the arrowisaimed 45 above the horizon. Solve for the arrow velocity:VA =gRmax = 31.3 m/s. The time of ight when the angle is 20istight =2VA sin g= 2.18 s,and the range is R = VA cos tight = 64.3 m. The speed oftherhinois30 km/h = 8.33 m/s. Therhinotravelsadistanced = 8.33(2.18) = 18.2 m. The required range when the arrow isreleased isd +R = 82.5 m.43c 2008PearsonEducation South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from thepublisherprior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Problem 13.81 The crossbar of the goalposts inAmericanfootball is yc=kick a eld goal, the ball must make the ball go betweenthetwouprightssupportingthecrossbarandbeabovethecrossbarwhenit doesso. Supposethat thekickerattemptsa36.58 meldgoal , and kickstheball withaninitial velocityv0 = 0=40.Bywhatvertical distancedoestheball clear thecrossbar?ycxc0v0Solution: Set thecoordinateoriginat thepoint wheretheballiskicked. Thex(horizontal)motionoftheballisgivenbyax = 0,Vx = V0 cos 0,x = (V0 cos 0)t . Theymotion is given byay = g,Vy =V0 sin 0gt , y =(V0 sin 0)t gt 22. Setx =xc=nd the timetcat which the ball crossed the plane of the goal posts.SubstitutethistimeintotheyequationtondtheycoordinateYBof the ball as it passes over the crossbar. Substituting in the numbers(g =2), we get tc=2.24 s andyB = .clears the crossbar by .0 .Problem 13.82 An American football quarterbackstandsat A. At theinstant thequarterbackthrowsthefootball, thereceiverisat Brunningat 6.1m/s towardC,wherehecatchestheball. Theballisthrownatanangle of 45above the horizontal, and it is thrown andcaught at the same height above the ground. Determinethe magnitude of the balls initial velocity and the lengthof time it is in the air.90CBReceiverspathPath of the ballASolution: Set xas the horizontal motion of the football, yas theverticalmotionofthefootballandzas thehorizontalmotionofthereceiver. Setg =2,0 = 45. We haveaz = 0, vz = , sz = ( )tay = g, vy = gt +v0 sin 0, sy = 12gt2+v0 sin 0tax = 0, vx = v0 cos 0, sx = v0 cos 0tWhen the ball is caught we havesz = ( )t0 = 12gt2+v0 sin 0tsx = v0 cos 0tsx2= sz2+( )2We can solve these four equations for the four unknownssx, sz, v0, tWe nd t = 1.67 s, v0 =443.05 m abovethe ground. To21.3 m/s and36.58 m and9. 81 m/s 6 11m. Thus, the ball3 7 m9.81 m/s6.1m/s 6.1 m/s6.1 m/s9.1 m11.6m/s9.1 mc 2008PearsonEducation South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from thepublisherprior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Problem 13.83 ThecliffdiversofAcapulco,Mexicomust timetheir divesthat theyenter thewater at thecrest (highpoint) of awave. Thecrestsof thewavesare 1 mabove the mean water depth h = 4 m. Thehorizontalvelocityofthewavesisequalto gh. Thedivers aiming point is 2 m out from the base of the cliff.Assumethat hisvelocityishorizontal whenhebeginsthe dive.(a) What is the magnitude of the drivers velocity whenhe enters the water?(b) How far from his aiming point must a wave crestbe when he dives in order for him to enter the waterat the crest?2 m 6.4 m1 m26 mhSolution:t = 0, vy = 0, y = 27 m, x0 = 0ay = g = 9.81 m/sVy = V0y0 gty = y0gt2/2y = 1 m attIMPACTfor an ideal dive to hit the crest of the wavet1 = tIMPACT = 2.30 sVy(t1) = 22.59 m/sax = 0Vx = Vx0XI = Vx0t1+X0At impactXI = 8.4 m.For impact to occur as planned, thenVx = 8.4/t1 = 3.65 m/s = constantThe velocity at impact is(a) |V| =

(Vx)2+[Vy(t1)]2= 22.9 m/sThe wave moves at gh = 6.26 m/s.Thewavecrest travels2.30secondswhilethediverisintheir s =ght1 = 14.4 m.27 m8.4 m45c 2008PearsonEducation South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from thepublisherprior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Problem 13.84 A projectile is launched at 10 m/s fromaslopingsurface. The angle = 80. Determine therangeR.a10 m/sR30Solution: Setg = 9.81 m/s2, v0 = 10 m/s.The equations of motion areax = 0, vx = v0 cos(8030), sx = v0 cos 50tay = g, vy = gt +v0 sin(8030)t , sy = 12gt2+v0 sin 50tWhen the projectile hits we haveRcos 30 = v0 cos 50tt = 2.32 s, R = 17.21 mRsin 30 = 12gt2+v0 sin 50tProblem 13.85 A projectile is launched at 100 m/s at60 above the horizontal. The surface on which it landsis described by the equation shown. Determine the pointof impact.yy = 0.001x260100 m/sxSolution: The motion in thex direction is ax = 0,vx = V0 cos 0,x = (V0 cos 0)t , and the motion in the y direction is given byay = g, vy = (V0 sin 0) gt , y = (V0 sin 0)t gt2/2. Weknowthat V0 = 100 m/sand 0 = 60. Theequationof thesurfaceuponwhich the projectile impacts is y = 0.001x2. Thus, the time ofimpact, tI, canbedeterminedbysubstitutingthevalues of x andy fromthe motion equations into the equation for the surface.Hence, we get (V0 sin 0)tI gt2I2 = 0.001(V0 cos 0)2t2I . Evaluatingwiththe knownvalues, we get tI = 6.37 s Substitutingthis valueinto the motion equations reveals that impact occurs at (x, y) =(318.4, 101.4) .46mc 2008PearsonEducation South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from thepublisherprior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Problem 13.86 At t = 0,asteelballinatankofoilis given a horizontal velocity v = 2i (m/s). The compo-nents of the balls acceleration in m/s2are ax = 1.2vx,ay = 8 1.2vy, az = 1.2vz. What is the velocity ofthe ball att = 1 s?xySolution: Assume that the effect of gravity is included in the givenaccelerations. The equations for the path are obtained from:(1)dvxdt= ax = 1.2vx. Separate variables and integrate:dvxvx= 1.2dt,fromwhich ln(vx) = 1.2t +Vx. At t = 0, vx(0) = 2, fromwhichln

vx2

= 1.2t .Inverting: vx(t ) = 2e1.2t.(2)dvydt= ay = 8 1.2vy. Separate variables and integrate:dvy81.2 +vy= 1.2 dt,from whichln

81.2 +vy

= 1.2t +Vy.Att = 0, vy(0) = 0, fromln

1 + 1.28vy

= 1.2t .Inverting: vy(t ) =81.2(e1.2t1).(3)dvzdt= az = 1.2vz, from which ln(vz) = 1.2t +Vz. Invert toobtainvz(t ) = Vze1.2t. At t = 0, vz(0) = 0, henceVz = 0 andvz(t ) = 0. Att = 1 second,vx(1) = 2e1.2= 0.6024 m/s, andvy(1) =

812

(1 e1.2) = 4.66 m/s, orv = 0.602i 4.66j (m/s).47c 2008PearsonEducation South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from thepublisherprior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Problem 13.87 In Problem 13.86, what is the positionof the ball att = 1 s relative to its position att = 0?Solution: Use the solution for the velocity components from Prob-lem 13.86. The equations for the coordinates:(1)dxdt= vx = 2e1.2t, from whichx(t ) =

21.2

e1.2t+Cx.Att = 0, x(0) = 0, from whichx(t ) =

21.2

(1 e1.2t).(2)dydt=

81.2

(e1.2t1), from whichy(t ) =

81.2

e1.2t1.2+t

+Cy.Att = 0, y(0) = 0, from whichy(t ) =

81.2

e1.2t1.2+t 11.2

.(3) Sincevz(0) = 0 andz(0) = 0, thenz(t ) = 0. Att = 1,x(1) =

21.2

(1 e1.2) = 1.165 m.y(1) =

81.2

e1.21.2+1 11.2

= 2.784 m, orr = 1.165i 2.784j (m).Problem 13.88 Thepoint P moves alongacircularpath with radius R. Show that the magnitude of its veloc-ity is |v| = R|d/dt |.Strategy: Use Eqs. (13.23).PyxSolution:x = Rcos y = Rsin vx = Rsin

ddt

vy = Rcos

ddt

|V| =

V2x +V2y|V| =

R2sin2

ddt

2+R2cos2

ddt

2|V| =

R2

ddt

2(sin2 +cos2)|V| = R

ddt

48c 2008PearsonEducation South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from thepublisherprior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Problem 13.89 Ify = 150 mm,dydt= 300 mm/s, andd2ydt2 = 0, whatarethemagnitudesofthevelocityandacceleration of pointP?300 mmPySolution: Theequationforthelocationofthepoint Pis R2=x2+y2, from whichx = (R2y2)12 = 0.2598 m, anddxdt=

yx

dydt

= 0.1732 m/s,d2xdt2= 1x

dydt

2+yx2

dxdt

dydt

yx

d2ydt2

= 0.4619 m/s2.The magnitudes are:|vP| =

dxdt

2+

dydt

2= 0.3464 m/s|ap| =

d2xdt2

2+

d2ydt2

2= 0.4619 m/s2Problem 13.90 Acar travels at aconstant speedof100 km/h on a straight road of increasing grade whosevertical prole canbe approximatedbythe equationshown. When the cars horizontal coordinate is x =400 m, what is the cars acceleration?y = 0.0003x2yxSolution: Denote C = 0.0003 and V = 100 km/h = 27.78 m/s. Themagnitude of the constant velocity isV =

dydt

2+

dxdt

2The equation for the road isy = Cx2from whichdydt= 2Cx

dxdt

.Substitute and solve:

dxdt

=V

(2Cx)2+1= 27.01 m/s.dxdtis positive (car is moving to right in sketch). The acceleration isd2xdt2=ddt

V

(2Cx)2+1

=4C2Vx((2Cx)2+1)32

dxdt

= 0.0993 m/s2.d2ydt2= 2C

dxdt

2+2Cx

d2xdt2

= 0.4139 m/s2, ora = 0.099i +0.414j(m/s2)49c 2008PearsonEducation South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from thepublisherprior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Problem 13.91 Suppose that a projectile has the initialconditionsshowninFig. 13.12. Showthat intermsofthex

y

coordinatesystemwithitsoriginat thehigh-est point ofthetrajectory, theequationdescribingthetrajectory isy

= g2v20 cos20(x

)2.xyyxSolution: The initial conditions are t = 0, x(0) = 0, y(0) = 0,vx(0) = v0 cos 0, andvy(0) = v0 sin 0. Theaccelerationsareax(t )= 0, ay(t ) = g. The pathof the projectile inthe x, y systemisobtainedbysolvingthedifferential equations subject totheinitialconditions:x(t ) = (v0 cos 0)t , y(t ) = g2t2+(v0 sin 0)tEliminatet from the equations by substitutingt =xv0 cos 0to obtainy(x) = gx22v20 cos20+x tan 0.At the peak,

dydx

peak= 0,from whichxp =v20 cos 0 sin 0g,and yp =v20 sin202g.The primedcoordinates: y

= y ypx

= x xp. Substitute andreduce:y

= g(x

+xp)22v20 cos20+(x

+xp) tan 0yp.y

= g2v20 cos20((x

)2+x2p +2x

xp) +(x

+xp) tan 0v20 sin202g.Substitutexp =v20 cos 0 sin 0g,y

=g(x

)22v20 cos20v20 sin202gx

tan 0+x

tan 0+v20 sin20gv20 sin202g.y

= g2v20 cos20(x

)250c 2008PearsonEducation South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from thepublisherprior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.Problem 13.92 The acceleration components of apoint are ax = 4 cos 2t , ay = 4 sin 2t , az = 0. Att = 0, its position and velocity are r = i, v = 2j. Showthat (a) the magnitude of the velocity is constant; (b) thevelocityandaccelerationvectorsareperpendicular;(c)the magnitude of the acceleration is constant and pointstoward the origin; (d) the trajectory of a point is a circlewith its cente