CH 2 Motion along a Straight Line - USNA€¦ · CH 2 Motion along a Straight Line ... same direction at the same rate (no rotation or stretching) 5. ... –24 m/s2 2. Over a short

[SHIVOK SP211] August 26, 2015

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CH 2

Motion along a Straight Line

I. Motion

A. Wefindmovingobjectsallaroundus.

B. Thestudyofmotioniscalled_________________________;itistheclassificationandcomparisonofmotions.

1. Examples:

a) TheEarthorbitsaroundtheSun

b) AroadwaymoveswithEarth’srotation

2. Inthischapter,wewillstudymotionthattakesplaceinastraightline.

3. Forcescausemotion.Wewillfindout,asaresultofapplicationofforce,iftheobjectsspeedup,slowdown,ormaintainthesamerate.(Forthischapter,wediscussonlythemotionitself,nottheforcesthatcauseit.)

4. Themovingobjectherewillbeconsideredasaparticle.Ifwedealwithastiff,extendedobject,wewillassumethatallparticlesonthebodymoveinthesamefashion.Wewillstudythemotionofaparticle,whichwillrepresenttheentirebody.

a) Aparticleiseither:

(1) Apoint‐likeobject(suchasanelectron)

(2) Oranobjectthatmovessuchthateachparttravelsinthesamedirectionatthesamerate(norotationorstretching)

5. Exampleofmotioninstraightline


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II. Positionanddisplacement

A. Thelocationofanobjectisusuallygivenintermsofastandardreferencepoint,calledtheorigin.Thepositivedirectionistakentobethedirectionwherethecoordinatesareincreasing,andthenegativedirectionasthatwherethecoordinatesaredecreasing.

B. Achangeinthecoordinatesofthepositionofthebodydescribesthe_______________ofthebody:∆xisthechangeinx,(finalposition)–(initialposition)

C. Forexample,ifthex‐coordinateofabodychangesfromx1tox

2,thenthe

displacement,

D. Displacementisa________________________quantity.Thatis,aquantitythathasbothmagnitudeanddirectioninformation.

E. Anobject’sdisplacementisx=‐4mmeansthattheobjecthasmovedtowardsdecreasingx‐axisby4m.Thedirectionofmotion,here,istowarddecreasingx.

F. Theactualdistancecoveredisirrelevant

G. Exampleproblem:

1. Herearethreepairsofinitialandfinalpositions,respectively,alongthex‐axis.Whichpairsgiveanegativedisplacement?

(a) ‐3m, +5m ; (b) ‐3m, ‐7m ; (c) 7m, ‐3m


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III. AverageVelocityandAverageSpeed

A. Acommonwaytodescribethemotionofanobjectistoshowagraphofthepositionasafunctionoftime.

B. Averagevelocity,orvavg,isdefinedasthedisplacementoverthetime

duration.

1. Averagevelocityhasunitsof(distance)/(time)

2. Meterspersecond,m/s

C. Theaveragevelocityhasthesamesignasthedisplacement


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D. Themagnitudeoftheslopeofthex‐tgraphgivestheaveragevelocity

1. Here,theaveragevelocityis:

E. AverageSpeed

1. Averagespeedistheratioofthetotaldistancetraveledtothetotaltimeduration.Itisascalarquantity,anddoesnotcarryanysenseofdirection.

2. Averagespeedisalwayspositive(nodirection)

F. ExampleProblems:

1.


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2.

3.


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4.

Note: Notice savg is not always the same as the magnitude of vavg.

IV. InstantaneousVelocityandSpeed

A. Theinstantaneousvelocityofaparticleataparticularinstantisthevelocityoftheparticleatthatinstant.

B. Obtainedfromaveragevelocitybyshrinking∆t;heretapproachesalimitingvalue:

1. v,theinstantaneousvelocity,istheslopeofthetangentoftheposition‐timegraphatthatparticularinstantoftime.

C. Velocityisavectorquantity(units(distance)/(time))andhaswithitanassociatedsenseofdirection.Thesignofthevelocityrepresentsitsdirection.

D. Speedisthemagnitudeof(instantaneous)velocity


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E. Sampleproblem:

1.


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V. Averageandinstantaccelerations

A. Averageaccelerationisthechangeofvelocityoverthechangeoftime.

B. Assuch,

1. Herethevelocityisv1attimet

1,andthevelocityisv

2attimet

2.

C. Theinstantaneousaccelerationisdefinedas:

1. Slopeoftangentlineofthevelocityvs.timegraph

D. Intermsofthepositionfunction,theaccelerationcanbedefinedas:

E. TheSIunitsforaccelerationare________________.

F. Ifaparticlehasthesamesignforvelocityandacceleration,thenthatparticleisspeedingup.

G. Conversely,ifaparticlehasoppositesignsforthevelocityandacceleration,thentheparticleisslowingdown.

H. Ourbodiesoftenreacttoaccelerationsbutnottovelocities.Afastcaroftendoesnotbothertherider,butasuddenbrakeisfeltstronglybytherider.Thisiscommoninamusementcarrides,wheretherideschangevelocitiesquicklytothrilltheriders.


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I. ThemagnitudeofaccelerationfallingneartheEarth’ssurfaceis

9.8m/s2,andisoftenreferredtoasg.

1. Onarollercoaster,youmightexperiencebriefaccelerationsofupto3g,whichis(3)*(9.8m/s2),orabout29m/s2,whichisathrill.Anun‐trainedindividualcanblackoutbetween4and6g,particularlyifthisispulledsuddenly.

J. Sampleproblems:

1. Overashortintervalneartimet=0thecoordinateofanautomobileinmetersisgivenbyx(t)=27t–4.0t3,wheretisinseconds.Attheendof1.0stheaccelerationoftheautois:A)27m/s2B)4.0m/s2C)–4.0m/s2D)–12m/s2E)–24m/s2

2. Overashortinterval,startingattimet=0,thecoordinateofanautomobileinmetersisgivenbyx(t)=27t–4.0t3,wheretisinseconds.Themagnitudesoftheinitial(att=0)velocityandaccelerationoftheautorespectivelyare:A) 0; 12 m/s2

B) 0; 24 m/s2

C) 27 m/s; 0 D) 27 m/s; 12 m/s2


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VI. Constantacceleration

A. Whentheaccelerationisconstant,itsaverageandinstantaneousvaluesarethesame…so:

means that Eqn (2‐11) 1. Here,velocityatt=0isv

o.

B. Similarly,,whichmeansthat

C. Alsovavg=½(vo+v)=>vavg=½vo+½vvavg=vo+½at

D. ThusX=Xo+(vo+½at)tEqn(2‐15)

E. EliminatingtfromtheaboveboxedEquationsgivesus

Eqn(2‐16)

F. Eliminatingafromequations2‐11and2‐15givesusoreliminatingvogivesusEqn’s2‐17and2‐18listedinthetablebelow.Proofscanbeshowninmyofficeifyouareinterested.

Can only be used for CONSTANT

Acceleration!


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G. Thefirsttwoequationscanbeobtainedbyintegratingaconstantaccelerationsoneverneedtobememorized.

1. Theotheradditionalusefulformsallcomefromalgebrarearrangementsofthefirsttwoequationsdependingontheknownandmissingvalues.

a) NowunderVF=VO+at.Squarebothsides.YougetVF2=(VO+at)(VO+at)whichusingthefoilmethod=

b) VO2+2aVOt+a2t2.NowifIfactoroutthe2anoticeIgetVF2=VO2+2a(VOt+(1/2)at2).Solookaboutisn’tthepartinparenthesisjustDX.Sodon’tyouhaveVF2=VO2+2a(DX).Hmmmnomemorizationrequiredhereeither.Equation2‐16learned.Threeof5done.

c) Nowlookatthefirstequationifwearenotgivena,wecanrearrangethefirstequationforathensubstitutethatequationintothesecondequationlikethis.

d) a=(Vf‐Vo)/tXf‐Xo=VOt+(1/2)t2(Vf‐Vo)/t.Thenoneofthet’scancelsoIhaveVot+(1/2)Vft‐(1/2)Vot.SoifIfactorout½IgetXf‐Xo=½(Vo+Vf)t.Fourthequationlearned.Nothingmemorized.

e) FINALLYifIgobacktothefirstequationandassumetheydidnotgivemeVo,IrearrangeforVo=Vf‐at.

f) SubstitutethatintothefourthequationandIgetXf‐Xo=½(Vf‐at+Vf)t,thusfactorout…Xf‐Xo=Vft‐(1/2)at2.Fifthequationlearned.Nothingmemorized.


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H. SampleProblems:

1. Howfardoesacartravelin6sifitsinitialvelocityis2m/sanditsaccelerationis2m/s2intheforwarddirection?

A) 12 m B) 14 m C) 24 m D) 36 m E) 48 m

2. Adragracingcarstartsfromrestatt=0andmovesalongastraightlinewithvelocitygivenbyv=bt2,wherebisaconstant.Theexpressionforthedistancetraveledbythiscarfromitspositionatt=0is?

3. Thefollowingequationsgivethepositionofx(t)ofaparticleinfoursituations.Forwhichofthesescenariosdothe“big5equations”apply?

a) ( ) 3 4x t t

b) 3 2( ) 5 4 6x t t t

c) 2

2 4( )x t

t t

d) 2( ) 5 3x t t


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VII. Free‐FallAcceleration

A. InthiscaseobjectsclosetotheEarth’ssurfacefalltowardstheEarth’ssurfacewithnoexternalforcesactingonthemexceptfortheirweight.

B. Usetheconstantaccelerationmodelwith“a”replacedby“‐g”,whereg=9.8m/s2formotionclosetotheEarth’ssurface.NOTICEthatgdoesnotequal‐9.8m/s2BUTa=‐g!

1. Interestingwebsitewhichwillgiveyoutheactualgvalueforyourgeographiclocationis:http://www.physicsclassroom.com/class/circles/u6l3e.cfm.

a) ThevalueofgforAnnapolis,MDis9.80171m/s2.

b) Interestingly,thevalueofgforHonnolulu,HIis9.78452m/s2.

C. Inaddition,itisimportanttonotetheaccelerationisnotalwaysaconstant=9.80m/s2(thisisonlytrueinFree‐fallacceleration).

D. Invacuum,afeatherandanapplewillfallatthesamerate.Youcanseethisforyourselfat:

http://www.youtube.com/watch?v=_XJcZ‐KoL9o

E. FreefallQuantitative:

1. ChoosepositiveYasupwards(a=‐g.AgainNotegisnon‐negative!)

Vx =

X =

V2x =

Vy =

Y =

V2y =

Monday’s Lesson Review


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VIII. Exampleproblems:

A. Boyontopofbuildingreleasesfromrestarock,whichstrikestheground3.5seclatter.Howtallisthebuilding?Whatisthevelocityatthebottompicosecondsbeforeitstruck?

Solution: Y =

h =

V=

B. Abatterhitstheball40mverticallyupward.Whatisaccelerationbefore,at,andafterballreachesthemaximumheight?Whatisvelocityatthemaximumheight?Whatistheinitialandfinalvelocity?Howlongistheballinflight?

Solution:


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C. Astoneisthrownverticallyupwardwithaninitialspeedof19.5m/s.Itwillrisetoamaximumheightof:A) 4.9 m B) 9.8 m C) 19.4 m D) 38.8 m E) none of these

Solution:

D. Aprojectileisshotverticallyupwardwithagiveninitialvelocity.Itreachesamaximumheightof100m.If,onasecondshot,theinitialvelocityisdoubledthentheprojectilewillreachamaximumheightof:A) 70.7 m B) 141.4 m C) 200 m D) 241 m E) 400 m

Solution:


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E. Anelevatorismovingupwardwithconstantacceleration.Thedashedcurveshowsthepositionyoftheceilingoftheelevatorasafunctionofthetimet.Attheinstantindicatedbythedot,aboltbreakslooseanddropsfromtheceiling.Whichcurvebestrepresentsthepositionoftheboltasafunctionoftime?

A) A B) B C) C D) D E) E


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IX. GraphicalIntegrationinMotionAnalysis

A. Integratingacceleration:

1. Givenagraphofanobject'saccelerationaversustimet,wecanintegratetofindvelocity.

a) TheFundamentalTheoremofCalculusgives:

2. FromyourknowledgeoftheendofCalcIandfromCalcII,youshouldrememberthatiftheaxesofthegrapharetheitemsbehindtheintegralsymbol,thatanothergroupofwordsforintegralis________________________________.

a) Thusthedefiniteintegralontherightcanbeevaluatedfromagraph:

B. Integratingvelocity:

1. Givenagraphofanobject'svelocityvversustimet,wecanintegratetofindposition.

a) TheFundamentalTheoremofCalculusgives:

b) Similarlyasabove,thedefiniteintegralontherightcanbeevaluatedfromagraph:


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C. Example:

1. AparticularsalamanderofthegenusHydromantescapturespreybylaunchingitstongueasaprojectile:Theskeletalpartofthetongueisshotforward,unfoldingtherestofthetongue,untiltheouterportionlandsontheprey,stickingtoit.TheFigurebelowshowstheaccelerationmagnitudeaversustimetfortheaccelerationphaseofthelaunchinatypicalsituation.

Theindicatedaccelerationsare 22 400 /a m s and 2

1 100 /a m s .Whatisthe

outwardspeedofthetongueattheendoftheaccelerationphase?

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CH 2 Motion along a Straight Line - USNA€¦ · CH 2 Motion along a Straight Line ... same direction at the same rate (no rotation or stretching) 5. ... –24 m/s2 2. Over a short