Post on 25-Jul-2020
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 +=