Analisis Gerak
9 757B3E3B3CE;=7J3947C97C3=FCFD?7?7F:;B7CD3?333
PP 679347D3C633??D63633?D7=A+7EF=3B7CB;63:3633C3=J39
6;E7?BF:B3CE;=7FEF=D739H3=EF3E3C3D7=A63
D7=A 4
3?43C=39C38;= PPE7C74;:63:FF6793B7C:;EF9347C;=FE
O +;E;=BAEA9E7C:363BDF?4F
-
6;B7CA7:63C; PP
PP P
63P O
+;E;=BF53=9C38;= D7:;993=757B3E36;
D3633:
P
PP
?D ,EF=D739H3=EFD7=A:;993
D7=A6;B7CA7:
O 7CB;63:3
PP P
P
P P
P? O
3C3= P P
P P
P ?
Tugas Anda 1.2
Diskusikan dengan teman sebangku Anda, apa perbedaan
antara jarak dan perpindahan? Bagaimana Anda menerangkan
konsep jarak dan perpindahan ini pada kasus mobil F1 yang sedang
balapan di sirkuit? Pada balapan F1, garis start dan finish berada di
tempat yang sama, dan mobil hanya bergerak mengelilingi
sirkuit.
4. Percepatan
7E;3B 4763 J39 ?763B3E 93J3
3=3 ?7933?; B7CF43:3 =757B3E3 D7:;993 4763 E7CD74FE ?7?;;=; B7C57B3E3 3?3 :3J3
6793 =757B3E3 B363 B7C57B3E3 6;=73 F93 ;DE;3: B7C57B3E3 C3E3 C3E363B7C57B3E3D7D33E7:=3C73=757B3E3E7C?3DF=47D3C3G7=EAC
?3=3B7C57B3E3F93?7CFB3=347D3C3G7=EACJ39;3;J3?7CFB3=3 EFCF3 B7CE3?3 63C; =757B3E3 ,EF= 4763 J39 47C97C3= G7CE;=3 =7
3E3D B7C57B3E3 J39 6;?;;=; 3633: B7C57B3E3 9C3G;E3D; 63 47C;3; 793E;8 D7639=3 FEF= 97C3= 3EF: B7C57B3E3J3 47C;3; BAD;E;8
a. Percepatan Rata-Rata
7C57B3E3C3E3C3E3B36397C3=6F36;?7D;?7?;;=;B797CE;3D3?3 6793 B7C57B3E3 C3E3C3E3 B363 97C3= D3EF 6;?7D; J3;EF :3D; 439;
B7CF43:3 =757B3E3 E7C:363B ;E7CG3 H3=EF + ?7?B7C;:3E=3 9C38;= :F4F93 =757B3E3 E7C:363B
H3=EF 363 D33E 4763 47C363 6; E;E;= 6793 =757B3E3 J39
6;?;;=; 363 D33E
4763 47C363 6; E;E;= 6793 =757B3E3 J39 6;?;;=;
7C57B3E3 C3E3C3E3 4763 63C; D3?B3; 3633: P
7E7C393 B7C57B3E3 C3E3C3E3 ?D
B7CF43:3 =757B3E3 ?D D739 H3=EF D
Gambar 1.10
Percepatan rata-rata. v
2
v
1
A B
t
1
t
2
t v
a =
v t
v
t
Contoh
1.8
v ms
t s – 1
2 3
4 5
6 7
8
– 1 – 2
– 3 – 4
– 5 – 6
– 7 – 8
– 9 –10
–11 –12
– 2 – 3
– 4 4
3 2
1 5
1 v = 3t
2
– 6t – 9
Mudah dan Aktif Belajar Fisika untuk Kelas XI
10
-7=EAC B7C57B3E3 D74F3: 4763 J39 47C97C3= B363 4;639 J3;EF B363DF?4F:AC;KAE3DF?4F-63DF?4FG7CE;=3DF?4F- ;3;J3
3633:
a
a P
63BF 47D3C B7C57B3E3 C3E3C3E3 ?7?7F:; B7CD3?33
2 2
x y
a a
P 7E7C393
B7C57B3E3 C3E3C3E3 ?D
a 47D3CB7C57B3E3B363DF?4F-?D
a 47D3CB7C57B3E3B363DF?4F- ?D
b. Percepatan Sesaat
7B7CE;B363=757B3E3D7D33EB7C57B3E3D7D33E97C3=D74F3:B3CE; =7 ?7?4FEF:=3 D739 H3=EF
J39 D393E D;9=3E J3;EF ?767=3E; A 36; B7C57B3E3 D7D33E ?7CFB3=3 EFCF3 B7CE3?3 63C;
B7CD3?33 =757B3E3 FEF= D739 H3=EF ?767=3E; A ;?
;?
3
t
3
d dt
P 7C57B3E3D7D33E?7CFB3=3EFCF3=76F363C;8F9D;BAD;D;=3C73
3 7:=3C73;EF
3
P +3:F=3: 63 53C3 ?7FF==3 BCAD7D ;?;E 633? ?77EF=3
B7C57B3E3D7D33E47C63D3C=39C38;=,EF=?797E3:F;J3B7C:3E;=3 +
363 + 63 + G7=EAC 3 63 3
?7CFB3=3 G7=EAC =757B3E3 B363 D33E 63
D7639=3
3
?7CFB3=3 B7CF43:3 =757B3E3
6;4F3E E7E3B D7639=3 6;4F3E ?767=3E;
D7:;993 7C63D3C=3 678;;D;
3
?7?;;=; 3C3: J39 D3?3
6793
3
7: =3C73 ;EF ?7?;;=; 3C3: J39 D3?3 6793
3
=7E;=3 ?767=3E; A 363 D33E
6;53B3; D7B7CE; 6;EFF==3 + 633
6;EFF==36793633D7:;993 ?736; -7=EAC B7C57B3E3 D7D33E D73F ?736; D;D; 79=F9 ;E3D3 E;E;= ?3E7C;
D7639=3 G7=EAC =757B3E3 3 E7E3B ?7J;99F9 ;E3D3 6; ;=3B3CE;=747C97C3=B3634;639 6;63B3E=A?BA7=A?BA7
B7C57B3E3 47C;=FE ;;
3
d d
dt dt
y x
dv dv
dt dt
63 E7EF E73: ?797E3:F; 43:H3 63
793 67?;=;3 63
D7:;993
2 2
2 2
d y d x
dt dt
P
Gambar 1.11
Percepataan sesaat a =
lim
t
t
v
merupakan percepatan pada saat t
2
– t
1
menuju nol atau t menuju
nol.
3
4
5
y
x
v
2
v
1
v
1
v
1
t
1
t
2
v
2
t
v a
v
2
v
1
t
v a
y
x
a
t
1
v
2
t
2
y
x t
v v
1
