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

Preview (16 of 1273 Pages)

100%

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

Loading page ...

Solutiona=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.

Solution Manual for Engineering Mechanics Dynamics in SI Units, 14th Edition - Page 2 preview image

Loading page ...

Solution Manual for Engineering Mechanics Dynamics in SI Units, 14th Edition - Page 3 preview image

Loading page ...

12–2.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

Solution Manual for Engineering Mechanics Dynamics in SI Units, 14th Edition - Page 4 preview image

Loading page ...

12–3.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

Solution Manual for Engineering Mechanics Dynamics in SI Units, 14th Edition - Page 5 preview image

Loading page ...

*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

Solution Manual for Engineering Mechanics Dynamics in SI Units, 14th Edition - Page 6 preview image

Loading page ...

12–5.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

Solution Manual for Engineering Mechanics Dynamics in SI Units, 14th Edition - Page 7 preview image

Loading page ...

12–6.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

Solution Manual for Engineering Mechanics Dynamics in SI Units, 14th Edition - Page 8 preview image

Loading page ...

12–7.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

Solution Manual for Engineering Mechanics Dynamics in SI Units, 14th Edition - Page 9 preview image

Loading page ...

*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

Solution Manual for Engineering Mechanics Dynamics in SI Units, 14th Edition - Page 10 preview image

Loading page ...

12–9.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

Solution Manual for Engineering Mechanics Dynamics in SI Units, 14th Edition - Page 11 preview image

Loading page ...

12–10.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

Solution Manual for Engineering Mechanics Dynamics in SI Units, 14th Edition - Page 12 preview image

Loading page ...

12–11.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

Solution Manual for Engineering Mechanics Dynamics in SI Units, 14th Edition - Page 13 preview image

Loading page ...

*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

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

Loading page ...

12–13.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

Solution Manual for Engineering Mechanics Dynamics in SI Units, 14th Edition - Page 15 preview image

Loading page ...

12–14.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

Solution Manual for Engineering Mechanics Dynamics in SI Units, 14th Edition - Page 16 preview image

Loading page ...

12–15.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

Preview Mode

This document has 1273 pages. Sign in to access the full document!

Report

Study Now!

Document Details

Related Documents

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

Engineering Mechanics: Dynamics 14th Edition Solution Manual

Solution Manual for Engineering Mechanics Statics, 13th Edition

A system is a group of parts that work together to

Solution Manual For Thermodynamics: An Interactive Approach, 1st Edition

Solution Manual For Thermal Radiation Heat Transfer, 6th Edition

Solution Manual for Introduction to Robotics: Mechanics and Control, 4th Edition

Mechanical Vibrations: Theory and Applications 1st Edition Solution Manual

Solution Manual For Design Of Machine Elements, 8th Edition

Solution Manual For Theory Of Vibrations With Applications, 5th Edition

Company

Explore

Study Tools