Analisis Gerak
7
3?43C6;D3?B;93633:9C38;=B7CB;63:3 D74F3:D7B763J3947C97C3=E7C:363BH3=EF
+7EF=3=757B3E3D7B763B363D33E 3
D7=A 4
D7=A63 5
D7=A 4
7B3399C38;=97C3=P97C3=D7B7636;439;633?E;9393C;DFCFDJ3;EF93C;D 93C;D6393C;D
3 363D33ED7=A9C38;=97C3=PD7B76347C363B36393C;DFCFD
7D3C=757B3E3D7B33993C;DE7CD74FE3633: E3
? D
?D 4
363D33ED7=A9C38;=97C3=PD7B76347C363B36393C;DFCFD 7D3C=757B3E3J33633:
E3
D
D7B763E;63=47C97C3= 5
363D33E
D7=AD7B76347C363B36393C;DFCFD757B3E3D7B763 ?7CFB3=3=7?;C;9393C;DJ3;EF
E3
? D
P
?D +363793E;8?7FF==3D7B76347C43;=3C3:
x m 6 0
3 3 0
A B
C
D 6
8 1 0
1 4 1 5 t s
c. Menghitung Posisi dari Kecepatan
+73: 63 =7E3:F; 43:H3 =757B3E3 ?7CFB3=3 EFCF3 B7CE3?3 63C; 8F9D; BAD;D; J3;EF 3
753C3 ?3E7?3E;D BAD;D; D74F3: B3CE;=7 63B3E 6;B7CA7: 63C; 8F9D; =757B3E3J3 ?73F; BCAD7D
;E79C3D; 7D3C =757B3E3 633? 3C3: DF?4F-
P P
74F3:B3CE;=747C97C3=6793B7CD3?33;E3D3 P
?6793
633??7E7C63633?D7=A+7EF=3=757B3E3B3CE;=7=7E;=3D7=A
4
757B3E36;B7CA7:63C;6;87C7D;3B7CD3?33BAD;D;793?7?3DF==3H3=EF 6;B7CA7:
P
?D ?D
Contoh
1.5
Contoh
1.6
Tantangan
untuk Anda
Pada saat balapan A1-GP, pembalap Indonesia, Ananda Mikola memantau
kecepatannya melalui speedometer. Menurut Anda, bagaimanakah cara
kerja speedometer? Gunakan bahasa Anda sendiri untuk menerangkan cara
kerja speedometer.
Mudah dan Aktif Belajar Fisika untuk Kelas XI
8
77=AC=7;5;47C336;3E3DCF?BFEB3634;639 7E3=3H3=7;5;B363 =AAC6;3E?A?BA7=757B3E3J33633:
63 ;=3
63 633??D63633?D7=AE7EF=33:
3 G7=EACBAD;D;=7;5;63
4 BAD;D;=7;5;B363D33ED7=A
4 3
AAC6;3E3H3?6347D3C=A?BA7=757B3E3J33633: 63
D7:;993
-7=EACBAD;D;=7;5;3633: 4
AAC6;3E=7;5;B363D33ED7=A3633: ?
?
?
?
??
? 36;G7=EACBAD;D;B363D33ED7=A3633:
?
?
d. Menghitung Perpindahan dan Jarak dari Grafik Kecepatan terhadap Waktu
C38;= =757B3E3 E7C:363B H3=EF 63C; 97C3= DF3EF 4763 63B3E 6;;:3EB363 + 753C39C38;=F3D637C3:J396;3CD;CJ3;EF
637C3: J39 6;43E3D; 9C38;= 47D3C =757B3E3 D74393; 8F9D; H3=EF 6793 DF?4F :AC;KAE3 3633: BAD;D; 63C; 4763
;=3D74F3:476347C97C3=?77?BF:93C;DFCFDE3B347C43;=3C3: 47D3C B7CB;63:3 D73F D3?3 6793 3C3= J39 6;E7?BF: 4763 =3
E7E3B; FEF= 4763 J39 47C97C3= FCFD 63 D7D33E =7?F6;3 47C43;= 3C3: 3=3 ?7?;;=; 47D3C 3C3= J39 47C4763 6793 47D3C B7CB;63:3
7C:3E;=3 93?43C + F3D 637C3: J39 6;3CD;C ?7FF==3 47D3CJ3 3C3= 793 ?79;E79C3=3 B7CD3?33
93C;DJ3 6;B7CA7: 47D3C 3C3= D74393; 47C;=FE
+363?FE3=6;9F3=3FEF=?7?3DE;=343:H347D3C3C3=D73F 47CE363 BAD;E;8 63BF FEF= ?79:;EF9 47D3C B7CB;63:3J3
6;9F3=3 B7CD3?33 47C;=FE
Gambar 1.9
a Grafik fungsi kecepatan v terhadap waktu t.
b Grafik v–t untuk gerak benda yang berbalik arah.
Contoh
1.7
7D3C =757B3E3 633? 3C3: DF?4F-
P P
t
1
t
2
t
3
v t
4
v t
t
2 1
t t
vtdt t
1
t
2
a t
3
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
