f104 gerak 2d_1
TRANSCRIPT
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GERAK DALAM DUA DIMENSI
TIU
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A
Dimanakah A berada ?
OKerangka acuan
Pusat acuan
Vektor posisirjarak
arah
Y
X
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PENGURAIAN VEKTOR ATAS KOMPONEN-KOMPONENNYA
X
Y
O
ay a
ax
a
a
ay = a sin ax = a cosa2 = ax
2 + ay2
a a ax y 2 2
tan aa
y
x
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X
Y
O
aya
ax
a
a
VEKTOR SATUAN
i
j
jia ˆˆyx aa
- Menunjukkan satu arah tertentu- Panjangnya satu satuan- Tak berdimensi- Saling tegak lurus (ortogonal)
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PENJUMLAHAN VEKTOR
a
a
b
+ b
R
= R
b
= b
a
+ aPenjumlahan vektor adalah komutatif
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PENJUMLAHAN VEKTOR MENGGUNAKAN KOMPONEN-KOMPONENNYA
a
bR
X
Y
o ax
ay
bx
by
Rx
Ry
R a bx x x
R a by y y
R R Rx y 2 2
tan RR
y
x
jiR ˆˆyx RR
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PENGURANGAN VEKTOR
a
b
-ba - b
a b a ( b)
Apakah pengurangan vektor komutatif ?
-a
b - aa b b a
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PENJUMLAHAN BEBERAPA VEKTOR
a
b
c
d
R
R = a + b + c + d
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P,ti
O
ri
Posisi awal
Q,t2rPergeseran
rfPosisi akhir
rir = rf
VEKTOR PERGESERANY
X
r = rf
ri
C
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O
ri
r
rf
Y
Xxxi
yi
y
yf
xf
xxx if yyy if
jir ˆˆfff yx
jir ˆ)(ˆ)( yyxx iif
jir ˆˆ yx
jir ˆˆiii yx
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KECEPATAN rata-rata
O
ri
r
rf
Y
X
t r
if
ifav tt
rrv
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KECEPATAN SESAAT
O
r1
Y
X
tav rv
if
if
tt
rr
r
r2r2
r2
v rr tt
rv0
limdtdr
dtyxd )ˆˆ( ji
ji ˆˆdtdy
dtdx
ji ˆˆyx vv
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PERCEPATAN
O
r1 r2
Y
X
t vv1
v2
v1
dtdv
ji ˆˆyx aa
12
12
ttav vva
tt
va0
lim
vaav
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Gerak dalam Dua Dimensi dengan Percepatan Tetap
jiv ˆˆyx vv vxo + axt vxo + axt
ji ˆ)(ˆ)( tavtav yyoxxo
)ˆˆ()ˆˆ( jiji tatavv yxyoxo
to avv
A. Kecepatan
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jir ˆˆ yx 221 tatvx xxoo 2
21 tatvx xxoo
jir ˆ)(ˆ)( 2212
21 tatvytatvx yyooxxoo
)ˆˆ()ˆˆ()ˆˆ( 2212
21 jijiji tatatvtvyx yxyoxooo
221 ttoo avrr
B. Posisi
Contoh Soal :
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GERAK PELURU Asumsi-asumsi :
Selama bergerak percepatan gravitasi, g, adalah konstan dan arahnya ke bawah
Pengaruh gesekan udara dapat diabaikan
Benda tidak mengalami rotasi
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0 X
Y
vxo
vyovo
vxo
vy vvxo
vy = 0
vxo
vy v
vyo
vxo
vo
g
o
konstan xox vv
gtvv yoy
tvx xooov cos tv oo )cos(
gtv oo sin
221 gttvy yo
221sin gttv oo Problem :
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TUGAS Dalam gerak parabola tunjukkan bahwa
lintasan partikel dapat dinyatakan seperti berikut ini : 2
22 )cos2
()(tan xv
gxyoo
o