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Solution Manual for Engineering Mechanics Dynamics in SI Units, 14th Edition - Document preview page 1

Solution Manual for Engineering Mechanics Dynamics in SI Units, 14th Edition - Page 1

Document preview content for Solution Manual for Engineering Mechanics Dynamics in SI Units, 14th Edition

Solution Manual for Engineering Mechanics Dynamics in SI Units, 14th Edition

Solution Manual for Engineering Mechanics Dynamics in SI Units, 14th Edition provides the perfect textbook solutions, giving you the help you need to succeed in your studies.

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Solution Manual for Engineering Mechanics Dynamics in SI Units, 14th Edition - Page 1 preview imageSolutiona=2t-6dv=adtLv0dv=Lt0(2t-6) dtv=t2-6tds=vdtLs0ds=Lt0(t2-6t) dts=t33-3t2Whent=6s,v=0Ans.Whent=11s,s=80.7 mAns.Starting from rest, a particle moving in a straight line has anacceleration ofa=(2t-6) m>s2, wheretis in seconds. Whatis the particle’s velocity whent=6 s, and what is its positionwhent=11 s?Ans:v=0s=80.7 m112–1.
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Solution Manual for Engineering Mechanics Dynamics in SI Units, 14th Edition - Page 2 preview image
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Solution Manual for Engineering Mechanics Dynamics in SI Units, 14th Edition - Page 3 preview image12–2.SOLUTIONWhen,Ans.Ans.SincethenAns.d=6-=6mvZ0s=639.5 mv=>st=s=1t-12t2+t+Lsds=Lt0(t-t+dtv=t-t+Lvdv=Lt0(4t-1)dtThe acceleration of a particle as it moves along a straightline is given bywheretis in seconds. Ifandwhendeterminetheparticle’svelocityandpositionwhenAlso,determine the total distance the particle travels during thistime period.t=t=0,v=>ss=a=1t-12m>s ,255 m2 m5 s.3345245)55525 s625 m39.5237.54v=625 m>ss=639.5 md=637.5 mAns:2
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Solution Manual for Engineering Mechanics Dynamics in SI Units, 14th Edition - Page 4 preview image12–3.The velocity of a particle traveling in a straight line is givenbyv=(6t-3t2) m>s, wheretis in seconds. Ifs=0 whent=0, determine the particle’s deceleration and positionwhent=3 s. How far has the particle traveled during the3-s time interval, and what is its average speed?Solutionv=6t-3t2a=dvdt=6-6tAtt=3sa=-12 m>s2Ans.ds=vdtLs0ds=Lt0(6t-3t2)dts=3t2-t3Att=3ss=0Ans.Sincev=0=6t-3t2,whent=0 andt=2 s.whent=2 s,s=3(2)2-(2)3=4 msT=4+4=8 mAns.(vsp)avg=sTt=83=2.67 m>sAns.a=-12 m>s2s=0sT=(vsp)avg=2.67 m>s=8 mAns:3
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Solution Manual for Engineering Mechanics Dynamics in SI Units, 14th Edition - Page 5 preview image*12–4.A particle is moving along a straight line such that itsposition is defined by, wheretis inseconds. Determine (a) the displacement of the particleduring the time interval fromto, (b) theaverage velocity of the particle during this time interval,and (c) the acceleration when.t=1 st=5 st=1 ss=(10t2+20) mmSOLUTION(a)Ans.(b)Ans.(c)Ans.a=d2sdt2=20 mm s2(for allt)vavg= ¢s¢t=2404=60 mm>s¢t=5-1=4 s¢s=270-30=240 mms|5s=10(5)2+20=270 mms|1s=10(1)2+20=30 mms=10t2+20Ans:∆s=240 mmvavg=60 mm>sa=20 mm>s24
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Solution Manual for Engineering Mechanics Dynamics in SI Units, 14th Edition - Page 6 preview image12–5.A particle moves along a straight line such that its positionis defined bys=(t2-6t+5) m. Determine the averagevelocity, the average speed, and the acceleration of theparticle whent=6 s.Solutions=t2-6t+5v=dsdt=2t-6a=dvdt=2v=0 whent=3st=0=5st=3=-4st=6=5vavg=∆s∆t=06=0Ans.(vsp)avg=sT∆t=9+96=3 m>sAns.at=6=2 m>s2Ans.Ans:vavg=0(vsp)avg=3 m>sat=6 s=2 m>s25
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Solution Manual for Engineering Mechanics Dynamics in SI Units, 14th Edition - Page 7 preview image12–6.stoneAis dropped from rest down a well, and in 1 sanotherstoneBisdroppedfromrest. Determinethedistance between the stones another second later.SOLUTIONAns.¢s=-. 1=.sB=sA=0+0+12(9.81)(1)2sA=sA=0+0+12( .)(2)2+ Ts=s1+v1t+12act2A9 8119.62 m4.91 m19.624 914 71 ms=14.71 mAns:6
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Solution Manual for Engineering Mechanics Dynamics in SI Units, 14th Edition - Page 8 preview image12–7.A bus starts from rest with a constant acceleration of 1.Determine the time required for it to attain a speed of 25and the distance traveled.m>sm>s2SOLUTIONKinematics:and.Ans.Ans.s=312.5 m252=0+2(1)(s-0)v2=v20+2ac(s-s0)A+:Bt=25 s25=0+(1)tv=v0+actA+:Bac=1 m>s2s0=0,v=25 m>s,v0=0,t=25 ss=312.5 mAns:7
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Solution Manual for Engineering Mechanics Dynamics in SI Units, 14th Edition - Page 9 preview image*12–8.A particle travels along a straight line with a velocityv=(12-3t2) m>s, wheretis in seconds. Whent=1 s, theparticle is located 10 m to the left of the origin. Determinetheaccelerationwhent=4 s, thedisplacementfromt=0 tot=10 s, and the distance the particle travels duringthis time period.SOLUTIONv=12-3t2(1)a=dvdt=-6tt=4=-24 m>s2Ans.Ls-10ds=Lt1vdt=Lt1(12-3t2)dts+10=12t-t3-11s=12t-t3-21st=0=-21st=10=-901∆s=-901-(-21)=-880 mAns.From Eq. (1):v=0 whent=2sst=2=12(2)-(2)3-21=-5sT=(21-5)+(901-5)=912 mAns.Ans:a=-24 m>s2∆s=-880 msT=912 m8
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Solution Manual for Engineering Mechanics Dynamics in SI Units, 14th Edition - Page 10 preview image12–9.SOLUTIONMotion of carA:Motion of carB:The distance between carsAandBisAns.sBA=|sB-sA|=`vAvBaA-v2A2aA`=`2vAvB-v2A2aA`sB=vBt=vBavAaAb=vAvBaAsA=v2A2aA0=vA2+2(-aA)(sA-0)v2=v02+2ac(s-s0)0=vA-aAtt=vAaAv=v0+actWhen two carsAandBare next to one another, they aretraveling in the same direction with speedsandrespectively. IfBmaintains its constant speed, whileAbeginstodecelerateatdeterminethedistancedbetween the cars at the instantAstops.aA,vB,vAABdSBA=`2vAvB-v2A2aA`Ans:9
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Solution Manual for Engineering Mechanics Dynamics in SI Units, 14th Edition - Page 11 preview image12–10.A particle travels along a straight-line path such that in 4 sit moves from an initial positionsA=-8 m to a positionsB=+3 m. T heninanother5sitmovesfromsBtosC=-6 m. Determine the particle’s a verage velocity andaverage speed during the 9-s time interval.SOLUTIONAverage Velocity:The displacement fromAtoCis∆s=sC-SA=-6-(-8)=2 m.vavg=∆s∆t=24+5=0.222 m>sAns.Average Speed:The distances traveled fromAtoBandBtoCaresASB=8+3=11.0 m andsBSC=3+6=9.00 m, respectively. Then, the total distance traveledissTot=sASB+sBSC=11.0+9.00=20.0 m.(vsp)avg=sTot∆t=20.04+5=2.22 m>sAns.Ans:vavg=0.222 m>s(vsp)avg=2.22 m>s10
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Solution Manual for Engineering Mechanics Dynamics in SI Units, 14th Edition - Page 12 preview image12–11.SOLUTIONAns.Ans.s=0.792 km=792 m(120)2=702+2(6000)(s-0)v2=v12+2ac(s-s1)t=8.33(10-3) hr=30 s120=70+6000(t)v=v1+actTraveling with an initial speed ofa car acceleratesatalong a straight road. How long will it take toreach a speed ofAlso, through what distancedoes the car travel during this time?120 km>h?6000 km>h270 km>h,Ans:t=30 ss=792 m11
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Solution Manual for Engineering Mechanics Dynamics in SI Units, 14th Edition - Page 13 preview image*12–12.SOLUTIONAns.v=1.29 m>s0.8351=12v2L215dsA3s13+s52B=Lv0vda ds=vda=5A3s13+s52BA particle moves along a straight line with an accelerationof,wheresisinmeters.Determine the particle’s velocity when, if it startsfrom rest when. Use a numerical method to evaluatethe integral.s=1 ms=2 ma=5>(3s1>3+s5>2) m>s2vvAns:v=1.29 m>s12
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Solution Manual for Engineering Mechanics Dynamics in SI Units, 14th Edition - Page 14 preview image12–13.The acceleration of a particle as it moves along a straightline is given bya=(2t-1) m>s2, wheretis in seconds. Ifs=1 mandv=2 m>swhent=0,determinetheparticle’svelocityandpositionwhent=6 s.Also,determine the total distance the particle travels during thistime period.Solutiona=2t-1dv=adtLv2dv=Lt0(2t-1)dtv=t2-t+2dx=vdtLstds=Lt0(t2-t+2)dts=13t3-12t2+2t+1Whent=6 sv=32 m>sAns.s=67 mAns.Sincev0 for 0…t…6 s, thend=67-1=66 mAns.Ans:v=32 m>ss=67 md=66 m13
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Solution Manual for Engineering Mechanics Dynamics in SI Units, 14th Edition - Page 15 preview image12–14.A trainstartsfrom rest atstationAand acceleratesatfor 60s. Afterwardsit travelswith a constantvelocity for 15 min. It then deceleratesat 1until it isbroughttorestatstationB.Determinethedistancebetween thestations.m>s20.5 m>s2SOLUTIONKinematics:Forstage (1) motion,andThus,Forstage (2) motion,andThus,Forstage (3) motion,andThus,Ans.=28 350 m=28.4 kms3=27 900+30(30)+12(-1)(302)s=s0+v0t+12act2+:t=30s0=30+(-1)tv=v0+actA+:Bac= -1 m>s2.v0=30 m>s,v=0,s0=27 900 ms2=900+30(900)+0=27 900 ms=s0+v0t+12act2A+:Bt=15(60)=900s.ac=0s0=900 m,v0=30 m>s,v1=0+0.5(60)=30 m>sv=v0+actA+:Bs1=0+0+12(0.5)(602)=900 ms=s0+v0t+12act2A+:Bac=0.5 m>s2.t=60s,s0=0,v0=0,Ans:s=28.4km14
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Solution Manual for Engineering Mechanics Dynamics in SI Units, 14th Edition - Page 16 preview image12–15.A particle is moving along a straight line such that itsvelocity is defined as, wheresis in meters.Ifwhen,determinethevelocityandacceleration as functions of time.t=0s=2 mv=(-4s2) m>sSOLUTIONAns.Ans.a=dvdt=16(2)(8t+1)(8)(8t+1)4=256(8t+1)3m>s2v= -4a28t+1b2= -16(8t+1)2m>ss=28t+1t=14 (s-1-0.5)-s-1|s2= -4t|0tLs2s-2ds=Lt0-4dtdsdt= -4s2v= -4s2v=16(8t+1)2m>sa=256(8t+1)3m>s2Ans:15
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Mechanical Engineering
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Engineering Mechanics: Statics Dyna...

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