kinematika 2003
TRANSCRIPT
![Page 1: Kinematika 2003](https://reader034.vdocuments.mx/reader034/viewer/2022052121/55999dda1a28ab9e658b459a/html5/thumbnails/1.jpg)
Oleh :Egi Nur Purnama Ramadhan (10)/XIA6Rizky Oktavian (27)/XIA6
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PERPIND AHAN
k
jir
kjir
rrr
)(
)()(
0
00
0
zz
yyxx
zyx
−+−+−=∆
∆+∆+∆=∆−=∆
Perpindahan
Posisi akhir :
Posisi awal : kjir 0000 zyx ++=
kjir zyx ++=
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kji
rrr 0
t
z
t
y
t
xv
tttv
∆∆+
∆∆+
∆∆=
∆∆=
−−=
0
Vektor kecepatan rata2
t
lv
∆∆==
waktuselang
lintasan panjang
Laju rata-rata
kjiv
kjir
v
rv
zyx
t
vvvdt
dz
dt
dy
dt
dx
dt
dt
Lim
++=
++==
∆∆=
→∆ 0
Vektor kecepatan sesaat
KEC EPATANA
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t
tt
∆∆=
−−=
va
vva 0
0
kjia
kjia
vva
zyx
zyx
t
aaadt
dv
dt
dv
dt
dv
dt
d
tLim
++=
++=
=∆∆=
→∆ 0
PERC EPATAN
Vektor percepatan rata-rata
Vektor percepatan sesaat
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Animasi
![Page 6: Kinematika 2003](https://reader034.vdocuments.mx/reader034/viewer/2022052121/55999dda1a28ab9e658b459a/html5/thumbnails/6.jpg)
Animasi
![Page 7: Kinematika 2003](https://reader034.vdocuments.mx/reader034/viewer/2022052121/55999dda1a28ab9e658b459a/html5/thumbnails/7.jpg)
Animasi
![Page 8: Kinematika 2003](https://reader034.vdocuments.mx/reader034/viewer/2022052121/55999dda1a28ab9e658b459a/html5/thumbnails/8.jpg)
Contoh Soal
![Page 9: Kinematika 2003](https://reader034.vdocuments.mx/reader034/viewer/2022052121/55999dda1a28ab9e658b459a/html5/thumbnails/9.jpg)
GERAK TRANSLASI 1- DIMENSI
2
2
0
0
0
0
0
:sesaat Percepatan
:rata-rata Percepatan
:sesaatKecepatan
ditempuh yang waktu selang
ditempuh yglintasan panjang:rata-rataLaju
:rata-rataKecepatan
-atau :arah :nPerpindaha
dt
xd
dt
dva
t
v
tt
vva
dt
dxv
t
lv
t
x
tt
xxv
xxx
==
∆∆=
−−
=
=
∆∆==
∆∆=
−−
=
+−=∆
![Page 10: Kinematika 2003](https://reader034.vdocuments.mx/reader034/viewer/2022052121/55999dda1a28ab9e658b459a/html5/thumbnails/10.jpg)
Gerak KhususGERAK DENGAN PERCEPATAN TETAP (1 D)
( ) tvvx
xxavv
attvxx
dtatvxx
ttavv
adtvv
t
tt
t
t
t
t
t
t
)4
)(2 )3
)( )2
)(
)1
021
020
2
221
00
0
00
00
0
0
+=−+=
++=
+=−
−+=
=−
∫
∫
Persamaan Kinematika
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GERAK JATUH BEBAS
( ) tvvy
yyavv
tatvyy
dttavyy
tavv
dtavv
yy
yyy
yy
t
yy
yy
t
yy
).4
)(2 ).3
)( ).2
).1
021
020
2
221
00
0
00
0
0
0
+=
−+=
++=
+=−
+=
=−
∫
∫
ja gy −=
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ANALISA GRAFIK
x
t
a
t
v
t
-Kemiringan-Luas-Rata-rata
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Gerak KhususGERAK DENGAN PERCEPATAN TETAP (2D)
Arah x
( ) tvvx
xxavv
tatvxx
dttavxx
tavv
dtavv
xx
xx
xx
t
t
xx
xx
t
t
xx
)(2
)(
021
020
2
221
00
00
0
0
0
0
+=−+=
++=
+=−
+=
=−
∫
∫
( ) tvvy
yyavv
tatvyy
dttavyy
tavv
dtavv
yy
yyy
yy
t
t
yy
yy
t
t
yy
)(2
)(
021
020
2
221
00
00
0
0
0
0
+=
−+=
++=
+=−
+=
=−
∫
∫
Arah y
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Gerak KhususGERAK PELURU (2 D)
),0(00
0
tetapva
tvxx
vv
xx
x
xx
==+=
=
)(
220
2
221
00
0
tetapga
gyvv
gttvyy
gtvv
y
yy
y
yy
=−=
−=
−+=
−=
Persamaan Gerak Dalam Arah Horisontal
Persamaan Gerak Dalam Arah Vertikal
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vPG = vPT + vTG
vPG: Kecepatan Penumpang relatif thd Tanah
vPT: Kecepatan Penumpang relatif thd Kereta
vTG: Kecepatan Kereta relatif thd Tanah
KECEPATAN RELATIF
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GERAK MELINGKAR(UMUM)
Posisi sudut θ dinyatakan dalam radian (rad)
Vektor perpindahan sudut: ∆θ = θ2 − θ1
Vektor kecepatan sudut rata2: <ω> = (θ2 − θ1)/(t2-t1)
Vektor kecepatan sudut sesaat: ω = dθ/dt
Vektor percepatan sudut rata2: <α> = (ω2 − ω1)/(t2-t1)
Vektor percepatan sudut sesaat: α = dω/dt
![Page 17: Kinematika 2003](https://reader034.vdocuments.mx/reader034/viewer/2022052121/55999dda1a28ab9e658b459a/html5/thumbnails/17.jpg)
RR
a
Ra
Rv
Rs
s2
2
tan
v ω
αωθ
==
===
Gerak KhususGERAK MELINGKAR BERATURAN
Gerak melingkar dengan laju tetap
R
vas
2
=
Gerak melingkar dengan percepatan tetap
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SUWUN JEH !!!!