Solution Manual - Dynamics 11th Engineering Mechanics - Statics and Dynamics by Russell C Hibbeler edition.pdf

TableofContentsChapter121Chapter13145Chapter14242Chapter15302Chapter16396Chapter17504Chapter18591Chapter19632Chapter20666Chapter21714Chapter22786EngineeringMechanics-DynamicsChapter12Problem12-1Atrucktravelingalongastraightroadatspeedv1,,:kmkmv1=20v2=120t=15shrhrSolution:v2−v1ma=a==1t+atd=-:v=80d=500ftsSolution:22vftv=2ada=a==t=t=-:ftfth=50ftg==182ssSolution:2ftvv=0+2ghv=-DynamicsChapter12vv−0t=t=*Problem12–4Startingfromrest,elerationofa=(bt+c).Whatistheparticle’svelocityatt1andwhatisitspositionatt2?mmGiven:b=2c=−6t1=6st2=11s32ssSolution:⌠t⌠tat()=bt+cvt()=⎮at()dtdt()=⎮vt()dt⌡0⌡0mvt()1=0dt()2=-?Also,throughwhatdistancedoesthecartravelduringthistime?kmkmkmGiven:v0=70a=6000vf=120hr2hrhrSolution:vf−v0vf=v0+att=t=30sa2222vf−v0vf=v0+2ass=s=792m2aProblem12-6−btAfreighttraintravelsatvv=0()1−,-DynamicsChapter12Given:ftv0=60s1b=st1=3sSolution:t−btd⌠vt()=v0()1−eat()=vt()d()t=⎮vt()dtdt⌡0ftd()t1=()1=-732Thepositionofaparticlealongastraightlineisgivenbysp=at+bt+≤t≤:a=1b=−9c=15t0=0stf=10s32sssSolution:32sp=at+bt+ctd2vp=sp=3at+2bt+cdtdd2a=v=s=6at+=max()6at0+b,6atf+2bamax=uratthebeginning,attheend,.−btcr=tcr=3s3a3EngineeringMechanics-DynamicsChapter12222ftvmax=max()3at0+2bt0+c,3atf+2btf+c,3atcr+2btcr+cvmax=135s*Problem12-8Fromapproximatel