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Chapter 3Chapter 3
KINEMATICS IKINEMATICS IRahmat Rasyid M.SiRahmat Rasyid M.Si
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IntroductionIntroduction
The mechanics science is the sciencestudying about motion, force concept, andenergy.
They divided to the two form :• Kinematics
• Dynamics
Kinematics is the science studying the
motion without oo!ing the cause of motion. Dynamics is the science studying the force
cause of motion.
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Motion ?Motion ?
The ob"ect can referred motion if changeof the position and need time to do that.
Three variabe of the motion : Speed #eocity
$cceeration
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SpeedSpeed
The measurement to how fast the ob"ectmove.
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VelocityVelocity
%&pressing vaue and direction of speed orrate of change of position.
They divided to the two form :
• 'nstantaneous veocity
• $verage veocity
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The Difference Speed & VelocityThe Difference Speed & Velocity
Comparing both of the pictures
Aboe ! the car hae re"ain #peed
but the direction chan$e for eery
"o"ent% o' 'ith the picture
be#ide?
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Acceleration Acceleration
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Acceleration Acceleration
The e&pression that how fast the veocity ofchange.
They divided to the two form : 'nstantaneous acceeration
$verage acceeration
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Acceleration for Inclined (lane Acceleration for Inclined (lane
%ver greater theob"ect ange, henceever greater the
acceeration of ob"ect.
(ase :• )ow much vaue of the
acceeration if theob"ect in perpendicuarpane as shown picturebeside*
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)ree )all Motion)ree )all Motion
The ob"ect can freefa as conse+uenceof the earth
gravitation withacceeration ag-. m/s0
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Motion in * di"en#ionMotion in * di"en#ion
'n 12D, we usuay write position as x(t 1 ).
Since it3s in 12D, a we need to indicate directionis 4 or −.Dispacement in a time ∆t = t 2 - t 1 is
∆ x = x(t 2 ) - x(t 1 ) = x 2 - x 1
t
x
t 1 t 2
∆ x
∆ t
x 1
x 2 #o"e particle+# tra,ectory
in *-D
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*-D .ine"atic#*-D .ine"atic#
#eocity v is the 5rate of change of position6
$verage veocity V av in the time ∆ t = t 2 - t 1 is:
t
x
t t
)t ( x )t ( x v
12
12 av
∆
∆=
−
−≡
t
x
t 1 t 2
∆ x
x 1
x 2 tra,ectory
∆ t
V av = #lope of line connectin$ x 1 and x 2 %
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*-D .ine"atic#%%%*-D .ine"atic#%%%
(onsider imit t1 t0 'nstantaneous veocity v is defined as:
dt
t dxt v
)()( =
t
x
t 1 t 2
∆ x
x 1
x 2
∆ t
#o v(t 2 ) = #lope of line tan$ent to path at t 2 %
( )t
t xt v
t
∆
∆=
→∆
)(lim
0
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E/a"pleE/a"ple
7ind instantaneous veocity for every times, by7ind instantaneous veocity for every times, bye+uatione+uation x(t) = 5t 2
Answer : Answer : 7or the ne&t time7or the ne&t time , , t = t + ∆t , , so that:so that:
x(t +∆t) = 5(t + ∆t)2
then :then : ∆x = x(t +∆t) - x(t)
7inay :7inay :
∆t→0 then :then : vavg = 10t
( )t t t
t t t
t
xvavg ∆+=∆
∆+∆
=∆
∆
= 5105.10 2
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*-D .ine"atic#%%%*-D .ine"atic#%%%
• $cceeration a is the 5rate of change of veocity6
• $verage acceeration aav in the time ∆t = t 2 - t 1
is:
t
v
t t
)t ( v )t ( v
a 12
12
av ∆
∆
=−
−
≡
$nd then the
instantaneous acceeration a isdefined as:
2
2
dt
)t ( x d
dt
)t ( dv
)t ( a ==
dt
)t ( dx )t ( v =u#in$
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0ecap0ecap
'f the position x is !nown as a function of time,then we can find both veocity v and acceerationa as a function of time8
a dv dt
d x dt
= =
2
2
v dx
dt =
x x t = ( )
x
a
v t
t
t
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*-D Motion 'ith con#tant*-D Motion 'ith con#tant
accelerationacceleration 9y cacuus:
$so reca that
Since a is constant, we can integrate this usingthe above rue to find:
Simiary, since we can integrate againto get:
const t 1n
1dt t 1nn ++
= +∫
a dv
dt =
v dx
dt =
∫ ∫ +=== 0 v at dt adt av
0 0 2
0 x t v at 2
1dt )v at ( dt v x ++∫ ∫ =+==
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0ecap0ecap
at v v 0 +=
2 0 0 at 2
1t v x x ++=
a const =
x
a
v t
t
t
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Deriation!Deriation!
ugging in for t:
at v v 0 += 2
0 0 at 2
1t v x x ++=
Soving for t:
a
v v t 0
−=
2
0 0 0 0
a
v v a
2
1
a
v v v x x
−+
−+=
) x x ( a2 v v 0 2
0 2 −=−
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(roble" !(roble" !
%&press your comment for under the picture by%&press your comment for under the picture by
using average and instantaneous veocity andusing average and instantaneous veocity and
acceeration in point a, b c, d, e, f and g :acceeration in point a, b c, d, e, f and g :
t ;sec<
= ; m
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