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1|Introduction
47
1.5CheckingSolutions
Thisis notakeytoalltheproblems inthebook,butitisakeytondingoutforyourselfifyou’ve
Ifyouhavesolvedfortheaccelerationofthemass
m
1
intheAtwoodsystemofFigure1.4and
you’vegottenaresultfortheaccelerationof
m
1
tobe
a
y
=
m
1
g=
(
m
1
m
2
)howcanyoutellifitis
likelytoberight?
0. Doesithavethecorrectdimensions? mass
.
g=
massisanacceleration,sothishastherightdimen-
sions.
1. If
m
1
m
2
youexpect
m
1
toacceleratedown,andthatispositive
a
y
. Thisexpressionsaysthat
if
m
1
islargeenough
a
y
m
1
g=m
1
=
g
,andthatisright. Sofar,sogood.
2. If
m
2
m
1
thisexpressiongoestozero. Thatdoesn’tseemtoolikely.
3. If
m
1
=
m
2
,thisexpressionhaszerointhedenominator. Itblowsup. Thatisveryunreasonable.
It’swrong,sogobackandndtheerror.
Thesesimpleproceduresaretheprimarytoolsyouwillusetoanalyzeyourresults,thoughother
techniqueswillbebuiltaroundthem. Takeanotherexample,onethatyoucansolvewithalittleeort,
butthat’snotwhatIwanttodonow.
x
~
F
0
Fig.1.5
Example
A blockofmass
m
isonarampinclinedatanangle
tothehorizontal,
andyouarepushingithorizontally,tryingtokeepitfromsliding.Yourapplied
forcehas magnitude
F
0
. Thebox x isinitiallymovingdownhillandthere isa
coecientofslidingfriction
k
betweentheboxandtheramp. Whatisthe
accelerationdowntheramp?
Examinesomeproposedsolutionsanddetermineifanyareplausible. Ileavethedimensionchecksto
you.
(1)
a
x
=
g
[sin
k
cos
F
0
m
[sin
+
k
cos
]
(a) If
F
0
and
k
bothequalzero,thisis
g
sin
.Thatisaresultyou’veseenbefore. Itmakessense,
goingfrom0to
g
as
variesfrom0to
=
2,soitpassesthistest.
(b) If
=0,thisis
a
x
k
g
F
0
m
k
.Ifthefrictionisnowzero,
k
=0,thiswholethingvanishes.
Thisiswrong,becausetheforce
F
0
shouldbeslowingitdown. Drawapicture.
(c) If
=
=
2,thisis
a
x
=
g
F
0
=m
. Theonlythingthat
F
0
shouldbedoingisincreasingthe
normalcomponentoftheforceon
m
,andsoincreasingtheeectoffriction.This
g
F
0
=m
is
independentof
k
,sothatmakesnosense. Again,drawapicture.
(2)
a
x
=
g
[sin
+
k
cos
]+
F
0
m
[cos
+
k
sin
]
(a) Thisis s always positive(aslongas
F
0
and
are positive),butifthefriction is bigenough,it
shoulddeceleratethemass,making
a
x
negative.
(b) Letthefrictioncoecientbeverylarge. . Thisexpressionfor
a
x
saysthattheaccelerationgets
bigger(andstayspositive).Frictionshouldbeslowingitdown.
(c) Thebigger
F
0
gets,themorepositive
a
x
gets. That’sthewrongdirection.
(3)
a
x
=
g
[cos
k
sin
]+
F
0
m
[cos
k
sin
]
(a) Thesignsinfrontofthe
’sarenegative,tendingtomaketheaccelerationmorenegative. Good.
That’sthewayitshouldbe.
(b) If
=0and
F
0
=0,thisis
a
x
=
g
,sayingthatonalevelsurfacewithnootherforces,itwill
acceleratehorizontallyatarate
g
.Notverylikely.
(c) If
=
=
2andif
F
0
=0,thisshoulddropwithanacceleration
g
a
x
k
g
,soitiswrong.
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1|Introduction
48
Exercises
Verifythechainrulewithanexampleconstructedsothatithasenoughstructuretoshowsomething,
butissimpleenoughtoworkwitheasily.
h
(
t
)=
f
g
(
t
)
then
dh
dt
=
df
dg
dg
dt
Take
g
(
t
)=
At
2
and
f
(
x
)=
Bx
3
andevaluate
dh=dt
twoways. Firstusethechainrule. Nextwrite
h
explicitlyintermsof
t
anddierentiatethat.
a
y
=
m
1
+
m
2
m
1
m
2
g
or
a
y
=
m
1
m
2
m
1
+
m
2
+1
g
explainwhyeachoftheseisobviously wrong.
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1|Introduction
49
Problems
1.1 DeriveEq.(1.12)again,butstartingfrom
~
F
=
d~p=dt
,somassmayormaynotbeconstant.
1.2 WriteoutEq.(1.11)forthecaseofthreemasses.
1.3 Inthedeepoceans,thespeedofawaveisexpectedtobeafunctionofatmostthegravitational
eld,
g
,thedensityofthewater,
,andthewavelengthofthewaves,
. Assumethatthespeed d is
givenbyanexpression
v
=
g
a
b
c
where
a
,
b
,and
c
areunknownnumbers. Youknowtheunitsof
alltheitems,
v
,
g
,
,
intheequation;determinethevaluesof
a
,
b
,and
c
sothattheequationis
dimensionallyconsistent.
1.4 Thespeedofsoundina uidsuchasairmaydependonthepressure,thedensityandthewavelength
ofthesound,as
v
=
p
a
b
c
where
a
,
b
,and
c
are unknownnumbers. . Solveforthevaluesof
a
,
b
,
and
c
sothatthe equation is dimensionally consistent. . In n this casethereisanotherparameterthat
canentertheequation:
=
c
p
=c
v
istheratioofthespecicheats,butitisdimensionless,sothere’s
nowaybylookingatthedimensionstotelliforhowitaectstheresults. Inairatlowfrequenciesit
entersas
1
=
2
,butatveryhighfrequencies(megaHertz)itapproachesone.
1.5 Solvefortheacceleration forFigure 1.5,thensubjectyoursolutiontothesamesortofanalysis
shownthere.
y
^
y
M
m
1
1.6 Amass
m
1
hangsfromastringthatiswrappedaround apulleyofmass
M
. Asthe
mass
m
1
fallswithacceleration
a
y
,thepulleyrotates.Someoneclaimsthattheacceleration
of
m
1
plausible,andremember: thatdoesnot meansolvingtheproblemand d comparingtheseto
(a)
a
y
=
(
M
+
m
1
)
g
m
1
(b)
a
y
=
m
1
g
m
1
M
(c)
a
y
=
m
1
g
M
+1
(d)
a
y
=
m
1
g
M
(e)
a
y
=
Mg
m
1
M
(f)
a
y
=
Mg
M
+
m
1
(g)
a
y
=
Mg
m
1
+1
(h)
a
y
=
Mg
m
1
1.7 Amassmovesinastraightlineanditsvelocitysatisestheequation
v
x
=
C=x
forsomeconstant
C
.Findtheforceactingonthismassasafunctionof
x>
0.Sketchagraphofthisforce. Ans:/1
=x
3
1.8 A mass
m
2
is atrest. . Anothermass
m
1
,having velocity
v
1
x
collideswithit. Assume e that all
themotionisalongastraightlineandthatbothkineticenergyand momentumareconserved. . Find
thevelocitiesof
m
1
and
m
2
afterthecollision. Examinespecialcasestoseeifyourresultisplausible.
Ans:
v
0
2
x
=2
m
1
v
1
x
=
(
m
1
+
m
2
)
1.9 Twoparticleshavethesamemass
m
. Onehaszerovelocityandtheotherhasvelocity
~v
0
. They
collideelastically, so thatkinetic energy y is conserved, , and they y move owith velocities
~v
1
and
~v
2
,
notalong astraightlinethis time. . Writedowntheconservationofkinetic c energyandconservation
ofmomentumandshowthatthenaltwovelocitiesareperpendiculartoeachother. Nocomponents
1.10(a) The sameas problem1.8,exceptthatmassisexchangedsothatthenalmasses are
m
3
and
m
4
. Take
m
1
=
m
2
=
m
to save e on the algebra. . Don’t t forget t the e conservation of mass
equation. Test t this in n detail, , looking at the eect when n the e nal l masses s are the same e and when
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1|Introduction
50
theyareverydierent. Also,arethetwoequationsfor
v
3
andfor
v
4
properlyrelatedtoeachother?
Ans:
v
3
=
1
2
v
1
1
m
4
=m
3
1
=
2
1.11 Iftwomassescollideandintheprocesstheybreakintothree,mustmassstillbeconserved?
1.12 Twoparticleshave momenta
~p
1
and
~p
2
,with coordinates
~r
1
and
~r
2
respectively. Write e their
totalangularmomentum,andshowthatifthetotallinearmomentumis zerothenthetotalangular
momentumisindependentoftheoriginchosenforthetwo
~r
’s. Recalltheidentity: : (
~
A
+
~
B
)
~
C
=
~
A
~
C
+
~
B
~
C
(anddrawpictures).
1.13 Readsection1.4again. Youcansometimesnd d anamusementparkridecalledaRotor. . You
walkintoacylindricalroomandstandagainsttheverticalwall. Theroomstartstorotate,andafter
ithascometoasucientrotationrate,the oordropsbyafewfeet. Ifthereisenoughfrictionwith
R
,
!
,andthecoecientof
staticfriction
s
(
F
fr
<
s
F
N
)sothatyoustayup?Checkthedimensionsofyourresultandseeifthis
minimum
!
behavesinaplausiblewayas
s
ischanged.Recallthatthemagnitudeoftheacceleration
ofanobjectmovingalongacircleatconstantspeedis
v
2
=r
=
r!
2
. Ans:
!
2
>g=
s
R
1.14_AvariationonaRotorisaGravitron. Thedierenceisthatthecylindricalwallisnotvertical;
ittiltsoutbyanangle
(whichwaszerofortheRotorintheprecedingproblem.). Whatdetermines
whetheryouwillstayagainstthewallandnotslideeitherupordownwhenyoucannolongerusethe
oor?
!
istheangularspeedoftheapparatus,
R
ismeasuredfromtheaxis,andtherequirementis
claimedtobe
~!
g
R
.
s
tan
tan
+
s
<!
2
<
g
R
.
1+
s
tan
tan
s
The second inequality applies only if tan
>
s
. If f that’snottrue thenthereisnoupperlimit on
!
. Isthisplausible? ? That t is: : checkunits; ; checkwhathappensif
s
is large or r small; checkwhat
happensif
isincreasedordecreased;checkwhathappensif
isnegative(theydon’tnormally do
that|probablytoomanycustomercomplaints);whatif
R
islargeorsmall;youcanevenvary
g
. See
1.15 Usingthesamestatedresultasintheprecedingproblem,whathappensifyoucoatthewallwith
te onandwearsilk?I.e.makefrictionverysmall.Howfarupthewallwillyoucometoanequilibrium?
Let
R
0
bethedistanceupthetiltedwall. ThetypicalGravitron
m
1
m
2
m
3
1.16 In the e complicated-looking pulley system sketched, the masses and moments of
inertiaofthepulleysarenegligiblysmall. Theaxleofthetoppulleyisattachedtowalland
everythingisreleasedtorotateandfall. Herearefourproposedsolutionsforthetension
inthecordthatisattachedtothemass
m
1
,thatis,themagnitudeoftheforcethecord
exertson
m
1
.Examineeachandexplainwhyitcouldnotpossiblybecorrect.
(
a
)
T
=
m
2
m
3
g
m
1
+
m
2
+
m
3
(
b
)
T
=
2
m
1
(
m
2
m
3
+
m
3
m
1
)
g
m
1
m
2
+
m
2
m
3
+
m
3
m
1
(
c
)
T
=
8
m
1
m
2
m
3
g
m
1
m
2
+
m
1
m
3
4
m
2
m
3
(
d
)
T
=
m
1
(2
m
2
+2
m
3
m
1
)
g
m
1
+
m
2
+
m
3
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1|Introduction
51
m
1
m
2
^
x
1.17 Inthesystemasdrawn,thereisnofriction,andthemassofthe
pulley isnegligible. . Theaccelerationofmass
m
2
isclaimedtobeone
ofthefollowing.Theycan’tallbecorrect,soshowwhatiswrongwith
most(orall)ofthem. Thatis,analyzealltheseproposedsolutionsand
determinewhethertheyarereasonable.
(a)
a
x
=
m
2
sin
m
1
m
2
+
m
1
cos
g
(b)
a
x
=
m
2
cos
+
m
1
m
2
sin
+
m
1
g
(c)
a
x
=
m
2
sin
m
1
cos
m
2
+
m
1
g
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OneDimensionalMotion
.
Thesimplestproblemsthatyousawinarstintroductiontomechanicsstarttobecomeharder
as soon as youadd some reality. . Remembering g someofthe formulas forconstant acceleration will
getyouonly sofarandthenyoucomeupagainst various formsoffrictionandagainst position-or
time-dependentforcesandlifebecomesdicult.
When you u toss s an n object straight up, the simplest model representing the force acting on n it
assumesthatonlygravityactsandthatthegravitationalforceisaconstantintimeandspace. Inthis
one-dimensionalcase,
~
F
=
m~a
is(
y
ispositiveupasusual.)
F
y
mg
=
ma
y
=
m
d
2
y
dt
2
Themassconvenientlycancelsandthenyouneedtodoacoupleofintegrals.
v
y
(
t
)=
Z
a
y
dt
gt
+
C;
y
(
t
)=
Z
v
y
dt
1
2
gt
2
+
Ct
+
D
Thetwoarbitraryconstant
C
and
D
comefromtheintegrationandyouneedtwomoreequationsto
determinethem. Boundaryconditionsareequationssuchas
y
(0)=0
v
y
(0)=
v
0
or
y
(0)=0
y
(
T
)=0
or
y
(0)=0
v
y
(
T
)=
v
1
(2
:
1)
Intherstpairyouspecify theinitialpositionandvelocity. . Inthesecondpairyouspecify y theinitial
positionandthetimeyouwantittoreturn. Inthethirdpairyouspecifytheinitialpositionandthe
velocityyoudesireatalatertime. Thesegivetherespectiveresults
v
y
(
t
)=
v
0
gt
y
(
t
)=
v
0
t
1
2
gt
2
y
(
t
)=
1
2
gt
(
t
T
)
v
y
(
t
)=
gt
+
1
2
gT
 
y
(
t
)=
1
2
gt
2
+(
v
1
+
gT
)
t
v
y
(
t
)=
g
(
t
T
)+
v
1
Youshouldcheckrstiftheseequationsarereasonable. Takesomecasestoseeiftheydowhatthey
aresupposedtodo. Doproblem2.1.
Whatifyounowrememberthatthere’sanatmosphereonEarth.Whateectwillthathaveon
themotion? Usuallylots. Iftheobjectismassiveenoughandmovingslowlyenoughyoucangetaway
withignoringsuchairresistance,butmostofthetimeyoucan’t.Airresistanceiscomplicated,andyou
willhavetomakesomesimplifyingmodeltodescribeit.Thereareafewcommonlyusedmathematical
modelsforairresistance,withvaryingsimplicityandaccuracy. Moreoftheformertypicallymeansless
ofthelatter.
typicalequationissomethinglike
F
friction
=
k
F
N
. Thisfrictionalforcedependsonthevelocity,but
onlythroughitsdirectionandnotitsmagnitude.Itisnotevencloseasawaytodescribeairresistance.
Wetfriction is amodel along the lines
F
x

bv
x
,so thatitisexplicitly dependenton the
velocity,bothinmagnitudeanddirection.Thissortofbehaviorischaracteristicofobjectsslidingalong
lubricatedsurfacesorofobjectsmovingat(very)lowspeedthroughair.
52
2|OneDimensionalMotion
53
Arespectablemodelforairresistanceis
F
=
cv
2
longasthespeedisn’ttoohigh.
Thentherearetheexperimentally determinedfunctionsthatyouneed forexactwork. . Golfis
Iwouldtry toreferyoutotheexperimentallydeterminedairresistancefunctionsforcommercialgolf
balls,butyoudon’tthinkthatTitleistisgoingtosharetheirdatawithCallawaydoyou?
Don’t forgetalso thatthe gravitational eld of the Earthdecreaseswith altitude. . Does s that
Yes,andthiswillshowupinproblem4.46.
2.1SolvingF=ma: F(t)
Timetogetawayfromthegeneralities andtostartsolvingsomeproblems. . Todothat,stayinone
dimensionforawhilesothatthebasicequationtosolveforasinglemassis
F
x
(
t;x;v
x
)=
m
dv
x
dt
=
m
d
2
x
dt
2
Whetherthisiseasy,hard,ornightmarishdependsonthefunction
F
x
. Startwiththesimplestcase,
forwhich
F
x
dependson
t
alone,youthensimplyhavetodoacoupleofintegrals.
F
x
(
t
)=
m
d
2
x
dt
2
=)
v
x
(
t
)=
Z
dt
1
m
F
x
(
t
)
;
=)
x
(
t
)=
Z
dt
Z
dt
1
m
F
x
(
t
)
(2
:
2)
Each integral has aconstantof integration thatmust be determined, so this means that you need
twomoreequationsinordertondthesetwoconstantsand tosolvetheproblem. . Commonlythese
equationswillcomefromspecifyinganinitialpositionandaninitialvelocity,thoughnotalways,asin
theexamplesofEqs.(2.1).
Example
Take
F
x
(
t
)=
F
0
sin
!t
(
t>
0).Assumethatthemassstartsfromtheoriginandwasatrestbefore
time zero. . Beforedoingthemathematics,whatdoyouexpectthemass s todo? ? You’reapplyingan
oscillatingforce,asmuchpositiveasitisnegative,sowill
m
wanderaroundtheoriginorwillitmove
away?Well,whichdoyouexpect?Testyourintuition;pre-judgetheproblemandsaywhatyouexpect,
thepossibilitythatyourintuitionisrightandyourcalculationiswrong.Thatcanhappentoo.
v
x
(
t
)=
1
m
Z
dtF
0
sin
!t
F
0
m!
cos
!t
+
C;
then
x
(
t
)=
Z
dtv
x
(
t
)=
F
0
m!
2
sin
!t
+
Ct
+
D
Applytheinitialconditions.
v
x
(0)=0=
F
0
m!
+
C;
so
C
=
F
0
m!
x
(0)=0=
F
0
m!
2
.
0+
C
.
0+
D;
so
D
=0
Putthistogetherandyouhave
x
(
t
)=
F
0
m!
t
F
0
m!
2
sin
!t
(2
:
3)
2|OneDimensionalMotion
54
Doesthismakesense? Firstcheckthedimensions.
F
0
is(dimensionally)[
F
0
]=ML
=
T
2
. This
usesthefairlycommonnotationthatthesquarebracketsdenote\thedimensionsof",andtheindividual
dimensionsare
M=mass
;
L=length
;
T=time
Theparameter
!
hastobesuchthat
!t
isanangle,so
[!]
=1
=
T. Thismeansthatthetwo
termsinEq.(2.3)havethesamedimensions,becausedimensionally,
t
isthesameas1
=!
. Now,are
theythecorrectdimensions?
[
F
0
m!
2
]=
ML
=
T
2

M
=
T
2
=
ML
=
T
2

T
2
=
M
=L
andthedimensionspasstheirtest.
Whatisthebehaviorofthesolutionforsmalltime?Notzerotime! Forsmalltimeusethepower
seriesexpansionofthesine. Remembertheequations(0.1),(b)inparticular.
x
(
t
)=
F
0
m!
t
F
0
m!
2
!t
1
6
(
!t
)
3
+
=
F
0
6
m
!t
3
+
Checkthedimensionsagain. Youcannotchangeany y dimensionsby makinganapproximationtothe
result,butyoucanalwaysmakeamistake.Thisstartsoas
t
3
. CanIunderstandwhy?Forcomparison
Iknowthatifitstartedas
a
0
t
2
=
2thenthiswouldimplythattherewasaforce
ma
0
.Thisforcefunction
F
0
sin
!t
howeverstartsatzeroforce. Thatimpliesthatithastostartmoreslowlythan
t
2
,andthat’s
t
3
.
x
t
Fig.2.1
What is the behavior forlarge time? ? The e sine term m oscillates and
goes nowhere, , but the
t
term means that it has an average e drift t velocity
of
F
0
=m!
thesinewithacosine,andthenpuzzleoutthedierencesbetweenthetwo
results. ThisgraphshowsEq.(2.3);besuretosketchagraphoftheresult
ofproblem2.3.Thedashedlineisthelineardrift,
F
0
t=m!
.
Whatifthetimedependenceoftheforcecan’tbewritteninasingle,simpleequation? What
if
F
x
(
t
) is asin
!t
on Mondayand acos
!t
on Tuesday and an
e
!t
on Wednesday? ? Answer: : you
noticethattheendofMonday isthebeginningofTuesday and theend ofTuesdayis the beginning
of Wednesday. . Thepositionandthe e velocity attheend ofone day determine the position and the
velocityatthestartofthenextday. Youbreaktheproblemintoseveralpieces. . Canthisgettobea
lotofwork?Ofcourse.
Example
Take aconcrete example ofa constant forceforthetimefrom zeroto
T
1
andanother constant
from
T
1
to
T
2
>T
1
. Start t from rest atthe origin. . I’lldothistwodierent t ways; the rst oneis
straight-forward,butratherclumsy.Thesecondoneusesatechniquethatseemsalmostobviousafter
you’vedoneitafewtimes,butyou’renotlikelytoinventit.Itwillsavealotofeortandconfusion.
F
x
(
t
)=
F
1
(0
<t<T
1
)
F
2
(
T
1
<t<T
2
)
F
x
x
(2
:
4)
TheFirstWay:
Intherstinterval
a
x
=
dv
x
dt
=
F
1
m
!
v
x
(
t
)=
F
1
m
t
+
C
!
x
(
t
)=
Z
v
x
dt
=
F
1
2
m
t
2
+
Ct
+
D
(2
:
5)
2|OneDimensionalMotion
55
Applytheinitialconditions:
v
x
(0)=0=
C; x
(0)=0=
D
giving
v
x
(
t
)=
F
1
m
t
and
x
(
t
)=
F
1
2
m
t
2
(2
:
6)
allveryfamiliar,theold
1
2
a
x
t
2
+
v
0
t
+
x
0
.
Inthesecondinterval
a
x
=
dv
x
dt
=
F
2
m
!
v
x
(
t
)=
F
2
m
t
+
C
0
!
x
(
t
)=
Z
v
x
dt
=
F
2
2
m
t
2
+
C
0
t
+
D
0
(2
:
7)
Now theconditionsatthestartofthis
F
2
-forcearevaluesof
x
and
v
x
attheendoftherstforce.
Thenewinitialconditionsaretheoldterminalconditions.
v
x
(
T
1
)=
F
1
m
T
1
and
x
(
T
1
)=
F
1
2
m
T
2
1
Applytheseequationstodeterminethevaluesof
C
0
and
D
0
.
v
x
(
T
1
)=
F
2
m
T
1
+
C
0
=
F
1
m
T
1
!
C
0
=
F
1
m
T
1
F
2
m
T
1
x
(
T
1
)=
F
2
2
m
T
2
1
+
C
0
T
1
+
D
0
=
F
1
2
m
T
2
1
!
D
0
=
F
1
2
m
T
2
1
F
2
2
m
T
2
1
C
0
T
1
D
0
=
F
1
2
m
T
2
1
F
2
2
m
T
2
1
F
1
m
T
1
F
2
m
T
1
T
1
F
1
2
m
T
2
1
+
F
2
2
m
T
2
1
!
Surelytheremustbeabettermethod!
TheSecondWay:
Thereis. InEq.(2.7)youarestartinganewparttotheproblem,butdoesthatmeanthatyoumust
keepthesamecoordinatesasintherstpart? Ofcoursenot.Choicesofcoordinatesareuptoyouand
inthiscasethecoordinateinquestionis
t
.Ijustautomaticallyusedthesame
t
inthesecondinterval
T
1
Callit
t
0
.
Nowtheequations(2.7)appearthesame,buttheirmeaningisdierentwhenyouapplytheterminal
conditionsfromtherstinterval.
a
x
=
dv
x
dt
0
!
v
x
(
t
0
)=
F
2
m
t
0
+
C
0
and
x
(
t
0
)=
F
2
2
m
t
02
+
C
0
t
0
+
D
0
(2
:
8)
v
x
(
t
0
=0)=
v
x
(
t
=
T
1
)  !
F
2
m
.
0+
C
0
=
F
1
m
T
1
x
(
t
0
=0)=
x
(
t
=
T
1
)  !
F
2
2
m
.
0
2
+
C
0.
0+
D
0
=
F
1
2
m
T
2
1
Theequations(2.8)nowbecome
v
x
(
t
0
)=
F
2
m
t
0
+
F
1
m
T
1
x
(
t
0
)=
F
2
2
m
t
02
+
F
1
m
T
1
t
0
+
F
1
2
m
T
2
1
(2
:
9)
You can nowcombineallof theseresults inafairlycompactsetof equations. . The e newtime
coordinateisjustashiftedversionoftheold:
t
0
=
t
T
1
. Theequations(2.6)and(2.9)become