Engineering Mechanics: Dynamics 14th Edition Solution Manual

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1Solutiona=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.12–1.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:s=80.7 m

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212–2.SOLUTION1S+2s=s0+v0t+12act2=0+12(10)+12(-2)(10)2=20 ftAns.If a particle has an initial velocity ofv0=12 ft>s to theright, ats0=0, determine its position whent=10 s, ifa=2 ft>s2to the left.Ans:s=20 ft

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312–3.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=4s, thedisplacement fromt=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 m

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4*12–4.SOLUTIONVelocity:To determine the constant acceleration, set,,andand apply Eq. 12–6.Using the result, the velocity function can be obtained by applyingEq. 12–6.Ans.v=A29.17s-27.7ft>sv2=32+2(4.583) (s-4)(:+)v2=v20+2ac(s-s0)ac=4.583 ft>s2ac=4.583 ft>s282=32+2ac(10-4)(:+)v2=v20+2ac(s-s0)v=8 ft>ss=10 ftv0=3 ft>ss0=4 ftacA particle travels along a straightline witha constantacceleration. When,and when,. Determine the velocity as a function of position.v=8 ft>ss=10 ftv=3 ft>ss=4 ftAAns:v=(19.17s-27.7)ft>s

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512–5.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.Ans:sT=8 mvavg=2.67 m>s

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612–6.SOLUTIONPosition:The position of the particle whenisAns.Total DistanceTraveled:The velocity of the particle can be determined by applyingEq. 12–1.The times when the particle stops areThe position of the particle at, 1 s and 5 s areFrom the particle’s path, the total distance isAns.stot=10.5+48.0+10.5=69.0 ftst=5 s=1.5(53)-13.5(52)+22.5(5)= -37.5 ftst=1 s=1.5(13)-13.5(12)+22.5(1)=10.5 ftst=0 s=1.5(03)-13.5(02)+22.5(0)=0t=0 st=1 sandt=5 s4.50t2-27.0t+22.5=0v=dsdt=4.50t2-27.0t+22.5s|t=6s=1.5(63)-13.5(62)+22.5(6)= -27.0 ftt=6 sThe position of a particle along a straight line is given by,wheretisinseconds.Determine the position of the particle whenand thetotal distance it travels during the 6-s time interval.Hint:Plot the path to determine the total distance traveled.t=6 ss=(1.5t3-13.5t2+22.5t) ftAns:st=6s=-27.0ftstot=69.0ft

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712–7.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>s2

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8*12–8.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>s2

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912–9.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 m

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1012–10.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>s

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1112–11.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>s

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12*12–12.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 m

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1312–13.SOLUTIONStoppingDistance:Fornormaldriver,thecarmovesadistanceofbefore he or she reacts and decelerates the car. Thestopping distance can be obtained using Eq. 12–6 withand.Ans.For a drunk driver, the car moves a distance ofbefore heor she reacts and decelerates the car. The stopping distance can be obtained usingEq. 12–6 withand.Ans.d=616 ft02=442+2(-2)(d-132)A:+Bv2=v20+2ac(s-s0)v=0s0=d¿ =132 ftd¿ =vt=44(3)=132 ftd=517 ft02=442+2(-2)(d-33.0)A:+Bv2=v20+2ac(s-s0)v=0s0=d¿ =33.0 ftd¿ =vt=44(0.75)=33.0 ftTests reveal that a normal driver takes aboutbeforehe or she canreactto a situation to avoid a collision. It takesabout 3 s for a driver having 0.1% alcohol in his system todo the same. If such drivers are traveling on a straight roadat 30 mph (44) and their cars can decelerate at,determine the shortest stopping distancedfor each fromthe moment they see the pedestrians.Moral: If you mustdrink, please don’t drive!2 ft>s2ft>s0.75 sdv144 ft/sAns:Normal:d=517ftdrunk:d=616ft

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1412–14.The position of a particle along a straight-line path isdefined bys=(t3-6t2-15t+7) ft, wheretis in seconds.Determine the total distance traveled whent=10 s. Whatare the particle’s average velocity, average speed, and theinstantaneous velocity and acceleration at this time?Solutions=t3-6t2-15t+7v=dsdt=3t2-12t-15Whent=10 s,v=165 ft>sAns.a=dvdt=6t-12Whent=10 s,a=48 ft>s2Ans.Whenv=0,0=3t2-12t-15The positive root ist=5 sWhent=0,s=7 ftWhent=5 s,s=-93 ftWhent=10 s,s=257 ftTotal distance traveledsT=7+93+93+257=450 ftAns.vavg=∆s∆t=257-710-0=25.0 ft>sAns.(vsp)avg=sT∆t=45010=45.0 ft>sAns.Ans:v=165 ft>sa=48 ft>s2sT=450 ftvavg=25.0 ft>s(vsp)avg=45.0 ft>s

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1512–15.SOLUTIONAns.Ans.s=1kB ¢2kt+¢1n20≤ ≤12-1n0Rs=2a2kt+a1n20b b122k3t0Ls0ds=Lt0dta2kt+a1v20b b12ds=ndtn=a2kt+a1n20b b-12-121n-2-n-202=-ktLnn0n-3dn=Lt0-k dta=dndt=-kn3A particleis moving witha velocity ofwhenandIf it is subjected to a deceleration ofwherekis a constant, determine its velocity and position asfunctions of time.a= -kv3,t=0.s=0v0Ans:v=a2kt+1v20b-1>2s=1kc a2kt+1v20b1>2-1v0d

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