Kinematics*in*One*Dimension

Chapter(2

My#lecture#slides#may#be#found#on#my#website#athttp://www.asc.ohio9state.edu/humanic.1/

and#this#link#is#also#on#Carmen

Acceleration

ttt o

o

Δ

Δ=

−

−=

vvva

DEFINITION(OF(AVERAGE(ACCELERATION

a = lim

Δt→0

ΔvΔt

DEFINITION(OF(INSTANTANEOUS(ACCELERATION

SI#unit:##m/s2

Acceleration

Example: Acceleration+and+Increasing+Velocity

Determine+the+average+acceleration+of+the+plane.

vo = 0 km h hkm260=v s 0=ot s 29=t

!a =!v− !vot − to

=260km h− 0km h

29 s− 0 s= +9.0 km h

s= +2.5 m

s2

Equations*of*Kinematics*for*Constant*Acceleration

When%working%in%1-dimension,%it%is%customary%to%dispense%with%the%use%of%boldface%symbols%overdrawn%with%arrows%for%the%displacement,%velocity,%and%acceleration%vectors.%We%will,%however,%continue%to%convey%the%directions%with%a%plus%or%minus%sign.

a =v− vot − too

o

tt −−

=xx

v

o

o

ttxx

v−

−=

o

o

ttvv

a−

−=

_ _

Equations*of*Kinematics*for*Constant*Acceleration

o

o

ttxx

v−

−=

0=ox 0=ot

( )tvvtvx o +== 21

Let$the$object$be$at$the$origin$when$the$clock$starts.

tx

v =True$forconstant$acceleration

Equations*of*Kinematics*for*Constant*Acceleration

a = a = v− vot − to t

vva o−=

ovvat −=

atvv o +=

True%for%constant%acceleration

Equations*of*Kinematics*for*Constant*Acceleration

Five%kinematic%variables:

1.%displacement,%x

2.%acceleration%(constant),%a

3.%final%velocity%(at%time%t),%v

4.%initial%velocity,%vo

5.%elapsed%time,%t

Equations*of*Kinematics*for*Constant*Acceleration

atvv o +=

x = 12 vo + v( ) t = 1

2 vo + vo + at( ) t

221 attvx o +=

Equations*of*Kinematics*for*Constant*Acceleration

x = vot + 12 at

2

= 6.0m s( ) 8.0 s( )+ 12 2.0m s2( ) 8.0 s( )2

= +110 m

Boat%moving%with%constant%acceleration%22>%find%x

Equations*of*Kinematics*for*Constant*Acceleration

Example:* Catapulting*a*Jet

Find*its*displacement.sm0=ov

??=x

2sm31+=asm62+=v

Equations*of*Kinematics*for*Constant*Acceleration

( ) ( )( )avv

vvtvvx ooo

−+=+= 2

121

tvv

a o−=

avv

t o−=

avv

x o

2

22 −=

Equations*of*Kinematics*for*Constant*Acceleration

x = v2 − vo

2

2a=

62m s( )2− 0m s( )2

2 31m s2( )= +62 m

Equations*of*Kinematics*for*Constant*Acceleration

Equations*of*Kinematics*for*Constant*Acceleration

( )tvvx o += 21

221 attvx o +=

atvv o +=

axvv o 222 +=

Applications+of+the+Equations+of+Kinematics

Reasoning)Strategy1.))Make)a)drawing.

2.)Decide)which)directions)are)to)be)called)positive)(+))and)negative)(?).

3.))Write)down)the)values)that)are)given)for)any)of)the)fivekinematic)variables.

4.))Verify)that)the)information)contains)values)for)at)least)threeof)the)five)kinematic)variables.))Select)the)appropriate)equation.

5.))When)the)motion)is)divided)into)segments,)remember)thatthe)final)velocity)of)one)segment)is)the)initial)velocity)for)the)next.

6.))Keep)in)mind)that)there)may)be)two)possible)answers)to)a)kinematics)problem.

Example:))An#Accelerating#Spacecraft

A#spacecraft#is#traveling#with#a#velocity#of#+3250#m/s.##Suddenlythe#retrorockets#are#fired,#and#the#spacecraft#begins#to#slow#downwith#an#acceleration#whose#magnitude#is#10.0#m/s2.##What#isthe#velocity#of#the#spacecraft#when#the#displacement#of#the#craftis#+215#km,#relative#to#the#point#where#the#retrorockets#began#firing?

Applications)of)the)Equations)of)Kinematics

a = -10.0 m/s2

v =#?

x a v vo t

+215000&m (10.0&m/s2 ? +3250&m/s

Applications+of+the+Equations+of+Kinematics

a = -10.0 m/s2

v =&?

Applications+of+the+Equations+of+Kinematics

axvv o 222 +=

x a v vo t

+215000&m (10.0&m/s2 ? +3250&m/s

v = ± vo2 + 2ax

v = ± 3250m s( )2+ 2 −10.0m s2( ) 215000 m( )

= ±2500m s

Applications+of+the+Equations+of+Kinematics

Freely&Falling&Bodies

In#the#absence#of#air#resistance,#it#is#found#that#all#bodiesat#the#same#location#above#the#Earth#fall#vertically#with#the#same#acceleration.##If#the#distance#of#the#fall#is#smallcompared#to#the#radius#of#the#Earth,#then#the#accelerationremains#essentially#constant#throughout#the#descent.

This#idealized#motion#is#called#free$fall and#the#accelerationof#a#freely#falling#body#is#called#the#acceleration,due,to,gravity.

22 sft2.32or sm80.9=g

Freely&Falling&Bodies

2sm80.9=g

Freely&Falling&Bodies

Example:&&A"Falling"Stone

A"stone"is"dropped"from"the"top"of"a"tall"building.""After"3.00sof"free"fall,"what"is"the"displacement"y of"the"stone?

Freely&Falling&Bodies

y a v vo t? "9.80'm/s2 0'm/s 3.00's

Freely&Falling&Bodies

y a v vo t? "9.80'm/s2 0'm/s 3.00's

y = vot + 12 at

2

= 0m s( ) 3.00 s( )+ 12 −9.80m s2( ) 3.00 s( )2

= −44.1 m

Freely&Falling&Bodies

Example:&&How$High$Does$it$Go?

The$referee$tosses$the$coin$upwith$an$initial$speed$of$5.00m/s.In$the$absence$if$air$resistance,how$high$does$the$coin$go$aboveits$point$of$release?

Freely&Falling&Bodies

y a v vo t? "9.80'm/s2 0'm/s +5.00'

m/s

Freely&Falling&Bodies

y a v vo t? "9.80'm/s2 0'm/s +5.00'

m/s

ayvv o 222 +=avv

y o

2

22 −=

y = v2 − vo

2

2a=

0m s( )2− 5.00m s( )2

2 −9.80m s2( )=1.28 m

Freely&Falling&Bodies

Conceptual&Example:&&Acceleration+Versus+Velocity

There+are+three+parts+to+the+motion+of+the+coin.++On+the+wayup,+the+coin+has+a+vector+velocity+that+is+directed+upward+andhas+decreasing+magnitude.+At+the+top+of+its+path,+the+coin+momentarily+has+zero+velocity.++On+the+way+down,+the+coinhas+downward=pointing+velocity+with+an+increasing+magnitude.

In+the+absence+of+air+resistance,+does+the+acceleration+of+thecoin,+like+the+velocity,+change+from+one+part+to+another?

Graphical)Analysis)of)Velocity)and)Acceleration

sm4s 2m 8

Slope +=+

=Δ

Δ=tx

Object'moving'with'constant'velocity'(zero'acceleration)

= vo

x = vot + 12 at

2 = vot

x = vot

Graphical)Analysis)of)Velocity)and)Acceleration

v = 2"m/s

v = 0"m/s

v = '1"m/s

Changing"velocity"during"a"bike"trip

Graphical)Analysis)of)Velocity)and)Acceleration

Object'moving'with'changing'velocity

Slope'of'the'tangent'lineis'the'instantaneousvelocity'at'the't ='20'spoint:

Slope'='!x/!t='(26'm)/(5's)='5.2'm/s'='v

x = vot + 12 at

2

Graphical)Analysis)of)Velocity)and)Acceleration

2sm6s 2

sm 12 Slope +=

+=

Δ

Δ=tv

Object'moving'with'constant'acceleration

= a

atvv o +=