vdbn

1. Bab
1
Meriam dengan peluru manusia ditembakkan dengan sudut kemiringan dan kecepatan
awal tertentu agar peluru jatuh tepat pada sasaran.
1
Analisis Gerak
A. Persamaan Gerak
Lurus
B. Gerak Parabola
C. Gerak Melingkar
Hasil yang harus Anda capai:
menganalisis gejala alam dan keteraturan dalam cakupan mekanika benda titik.
Setelah mempelajari bab ini, Anda harus mampu:
(7C@3:=3: @63 ?7;:3E 3EC3=D; ?7C;3? 67@93@ B7FCF ?3@FD;3
*33: D3EF =7A?BA= D;C=FD J3@9 E7C=7@3 67@93@
3EC3=D;
@J3 E3:F@

2. 3@ 3633: =7A?BA= D;C=FD =7F3C93
035:;@; (C7DE3D; ?7C7=3 J3@9 E7C53E3E D74393; C7=AC 6;3=F=3@ A7:
?3@F7 035:;@; E3:F@
*33E ;EF ;3 47C:3D; E7C43@9 ?77H3E; E;93
4F3: =;@5;C 63@ ?7@63C3E 6; 3C;@9 D73F: ? 63C; E;E;= B7@7?43=3@
(C7DE3D;87@A?7@3B363=736;3@E7CD74FE?7CFB3=3@D33:D3EF4F=E;
B7@7C3B3@ =3;63:=3;63: ;D;=3 633? :3 ;@; 3633: 97C3= B3C34A3
7@93@ =757B3E3@ 3H3 J3@9 DF63: 6;=7E3:F; 63@ DF6FE 77G3D; ?7C;3?
E7C:363B :AC;KA@E3 DF63: 6;3EFC ?3=3 E;E;= ?3=D;?F? =7E;@99;3@ 63@
E;E;=E7C3F:63B3E6;=7E3:F;'7:=3C7@3;EF?7C7=363B3E?7@7@EF=3@
D7E;@99; 3B3 C;@E3@93@ J3@9 6;9F@3=3@ 63@ 6; ?3@3 :3CFD ?77E3==3@
3C;@9@J3
+3:F=3: @63 43:H3 97C3= B3C34A3 63B3E 6;3@3;D;D ?73F;
B7CB36F3@ 6F3 97C3= FCFD J3@9 E73: @63 B733C; 6; #73D / ,@EF=
?7?B7C?F63: 3@3;D;D 97C3= B7CD3?33@ 97C3= 6;F43: 633? 47@EF=
G7=EAC ,@EF= 74;: ?7?3:3?;@J3 B733C;3: 434 ;@; 67@93@ 43;=
Sumber: Fundamentals of Physics, 2001
menganalisis gerak lurus, gerak melingkar, dan gerak parabola dengan
menggunakan vektor.

3. Tes Kompetensi Awal
$!$*2++$+.$* ( /')-,0$., *'0'0$/ ))$/( ) ,* 0- *0- *!$/')21# * +!2)2* 1' ,
A. Persamaan Gerak Lurus
1. Posisi dan Arah Partikel Berdasarkan Vektor
33?;D;=3BAD;D;DF3EFB3CE;=7D73F6;EF;D=3@633?DF3EFDF?4F
=AAC6;@3E *F?4F =AAC6;@3E J3@9 D7C;@9 6;9F@3=3@ J3;EF =AAC6;@3E
53CE7D;FD=AAC6;@3ED;;@67C63@=AAC6;@3E4A3,@EF=B7?43:3D3@B363
434 ;@; DF?4F =AAC6;@3E J3@9 6;9F@3=3@ 3633: =AAC6;@3E 53CE7D;FD
67@93@ BAD;D; B3CE;=7 6;@J3E3=3@ 633? DF?4F- DF?4F- 63@ DF?4F-!
,@EF= ;EF =;E3 3B;=3D;=3@ G7=EAC =3; B7CE3?3 B363 97C3= FCFD
a. Vektor Satuan dan Vektor Posisi
*F3EFG7=EAC633?=AAC6;@3E53CE7D;FD?7?;;=;=A?BA@7@6;DF?4F-
DF?4F- 63@DF?4F-!(363D7E;3BDF?4FE7CD74FEE7C63B3EG7=EACD3EF3@
J3@947D3C@J3D3EF63@?7?;;=;3C3:D3?367@93@3C3:DF?4F@J3(363
DF?4F- G7=EAC D3EF3@@J3 3633: ' (363 DF?4F- G7=EAC D3EF3@@J3
3633: ( 63@ DF?4F-! G7=EAC D3EF3@@J3 3633: ) *74393; 5A@EA:
B7C:3E;=3@ +! /
-7=EACJ3@9E7C7E3=B3634;63@9 63B3E
6;EF;D=3@ 633? 47@EF= G7=EAC D74393; 47C;=FE
y
P = Pxi + Pyj
x
0 x
Dr
2 Mudah dan Aktif Belajar Fisika untuk Kelas XI
' (
63BF@ F@EF= G7=EAC J3@9 E7C7E3= 633? CF3@9 +! /
!
G7=EAC 63B3E 6;EF;D=3@ 633? @AE3D; G7=EAC D74393; 47C;=FE
'J(K)
P
67@93@ 63@ ! ?7CFB3=3@ 47D3C3@ D=33C
b. Vektor Perpindahan
*74F3:B3CE;=7J3@947C97C3=B363DF3EF4;63@963E3CJ3;EFE7C:363B
DF?4F- 63@ DF?4F- G7=EAC BAD;D;@J3 B363 DF3EF E;E;= 6;EF@F==3@
D7B7CE;B363 +! / ;=3B3CE;=7?73=F=3@B7CB;@63:3@63C;BAD;D;
/
=7 BAD;D; / B7CB;@63:3@ BAD;D; E7CD74FE 6;@J3E3=3@ 67@93@ /
(7CB;@63:3@ D74F3: B3CE;=7 / 63B3E 6;B7CA7: 67@93@ ?7@99F@3=3@
3EFC3@ G7=EAC B7CB;@63:3@ J3;EF D74393; 47C;=FE
//P/
P
;=3 G7=EAC BAD;D;/
'
( 63@ G7=EAC BAD;D; / ' (
B7CD3?33@ B7CB;@63:3@ / ?7@36;
/' (P
'
(
/P
'P P
(
;=3
63@
?3=3 3=3@ 6;B7CA7:
3
4
Gambar 1.2
(a) Posisi partikel pada
bidang xy.
(b) Vektor P dalam ruang.
Gambar 1.3
Perpindahan vektor
posisi r .
y
Py
j
i Px
Py j
Px i
Pzk
P
y
x
z
y
x
A B
0
r1
r2
5

4. 6
(36397C3=B3C34A33DF?D;97C3=3B3J3@96;9F@3=3@
633?3C3:DF?4F63@DF?4F

5. B3=3:B7C47633@=757B3E3@;@73C63@=757B3E3@
DF6FE
(363DF4434(7CD3?33@7C3=$FCFD@633=3@
43@J3=?7@99F@3=3@EFCF@3@63@;@E79C3
;EF@93:@;3;EFCF@3@3E3F;@E79C347C;=FE
3

6. 4
!

7. Gambar 1.1
Vektor satuan dalam sumbu-x,
sumbu-y, dan sumbu-z.
z
k
i
j

8. Ingatlah
Analisis Gerak 3
/ ' (
P

9. ,@EF= ?7@7@EF=3@ 47D3C G7=EAC B7CB;@63:3@ 63B3E 6;EF;D
/
P
63BF@ 3C3: B7CB;@63:3@ B3CE;=7 63B3E 6;B7CA7: ?73F; 47D3C DF6FE
J3@9 6;47@EF= E7C:363B DF?4F- J3;EF
E3@
y
x
3E3F 3C5E3@
y
x
(AD;D;3H3D74F3:B3CE;=73633:/
'P(?=7?F6;3@B3CE;=7E7CD74FE47CB;@63:
=7BAD;D;/

10. '(?+7@EF=3@3:G7=EACB7CB;@63:3@47D3CG7=EACB7CB;@63:3@
63@3C3:B7CB;@63:3@B3CE;=7;EF
4 !
;=7E3:F; /
'P(?
/

11. '(?
-7=EACB7CB;@63:3@
/ ' (
/

12. P'PP(
P'
(?
7D3CG7=EACB7CB;@63:3@
/

13. ?
C3:B7CB;@63:3@ 3C5E3@
N
Dr = r2 – r1
c. Menentukan Komponen-Komponen Vektor Jika Arah dan
Besarnya Diketahui
(7C:3E;=3@ +! /

14. (363 93?43C E7CD74FE G7=EAC 6;FC3;=3@
E7C:363B DF?4F- 63@ DF?4F- 3633: =A?BA@7@ G7=EAC B363
DF?4F- D763@9=3@ 3633: =A?BA@7@ G7=EAC B363 DF?4F- ;=3
3633:DF6FEJ3@96;47@EF=A7:G7=EACE7C:363BDF?4F-47D3C
63@ 63B3E 6;:;EF@9 67@93@ ?7@99F@3=3@ B7CD3?33@ 47C;=FE
5AD 63@ D;@
P
7@93@?7@97@63C3;D7B763+3C;93@?7@9;=FE;J3@96;D77@993C3=3@D7=A3:@J3
67@93@?7@7?BF:CFE7D7B7CE;6;EF@F==3@B36393?43C(7CE3?3D7B76347C97C3=
?7@FF=AE3D73F: =?633?3C3:

15. N=743C3E3FE=7?F6;3@47C97C3==7=AE3
D73F:

16. =?=7FE3C3+7@EF=3@BAD;D;=AE363@3C3:@J3E7C:363BD7=A3:+3C;93@
4 !
y (km)
-7=EACB7CB;@63:3@D7B7636;@J3E3=3@D74393;G7=EAC
63@#A?BA@7@G7=EAC
5AD
N =?P P=?
D;@
N =?

17. =?
#A?BA@7@G7=EAC
5AD N

18. =? =?
D;@ N

19. =?

20. =?
x (km)
U
T
R
B
A
B
A
30°
Sekolah Tarigan
7
12
0
r2
r1
Dr
112,7°
–5
–5
3
8
y
x
Gambar 1.4
Ax dan Ay merupakan komponen-komponen
vektor A pada sumbu-x
dan sumbu-y.
Contoh 1.1
Contoh 1.2
Penulisan notasi vektor yang benar
adalah dengan tanda panah di atas atau
dengan huruf tebal.
P = P.
Dalam buku ini digunakan huruf tebal
sebagai penanda vektor.
x
y
Ay
A
Ax
0

21. 2. Perpindahan dan Jarak
(363 FC3;3@ D747F?@J3 E73: 6;73D=3@ 43:H3 B7CF43:3@ BAD;D;
?7?F@5F=3@ B7CB;@63:3@ 3E3F D753C3 ?3E7?3E;D 6;EF;D / (AD;D; 63@
B7CB;@63:3@ =76F3@J3?7CFB3=3@ 47D3C3@ G7=EAC *3@93E B7@E;@9 F@EF=
6;;@93E 43:H3 8;D;=3 ?7?4763=3@ B7@97CE;3@ B7CB;@63:3@ 63@ 3C3=
%;D3@J3 !4F B7C9; 63C; CF?3: =7 B3D3C F@EF= 473@3 *3EF 3?
47C;=FE@J3!4F=7?43;39;=7CF?3:%7@FCFEB7@97CE;3@B7CB;@63:3@
D73?3D3EF3?E7CD74FE!4F?7@933?;B7CB;@63:3@@A63BF@3C3=
J3@9 6;33?; !4F 3633: EAE3 B3@3@9 ;@E3D3@ D33E !4F 47C97C3= 4A3=
43;= 63C; CF?3: =7 B3D3C
3. Persamaan Kecepatan dan Kelajuan
7@63 3E3F D7D7AC3@9 6;=3E3=3@ 47C97C3= =3C7@3 BAD;D;@J3 47CF43:
3E3F ?7@933?; B7CB;@63:3@ *73;@ 47CB;@63: 97C3= ?7@93=;43E=3@
B7CF43:3@ H3=EF 3E3F D73@9 H3=EF 7@93@ 67?;=;3@ =7E;=3 47@63
47C97C3= E7C36; B7CF43:3@ BAD;D; D7E;3B D33E 3CE;@J3 BAD;D; ?7CFB3=3@
8F@9D; H3=EF (7C@J3E33@ E7CD74FE D753C3 ?3E7?3E;D 63B3E 6;EF;D D74393;
/E *33E 47@63 47CF43: BAD;D; 3E3F 47CB;@63: 633? D73@9 H3=EF E7C
E7@EF?F@5F3:47D3C3@=757B3E3@47@63E7CD74FE 3;@;?7@93=;43E
=3@ =757B3E3@ 63B3E 6;EFCF@=3@ 63C; 8F@9D; BAD;D;
a. Kecepatan dan Kelajuan Rata-Rata
#757B3E3@ C3E3C3E3 63C; D74F3: 47@63 J3@9 47C97C3= B363 D3EF
6;?7@D; D3?3 67@93@ B7CB;@63:3@ D74F3: 47@63 6;439; 67@93@ ;@E7CG3
H3=EF J3@9 6;9F@3=3@ D73?3 B7CB;@63:3@ E7CD74FE
3 E7CD74FE F93 47C3=F F@EF= 97C3= B363 6F3 6;?7@D; 63@ E;93
6;?7@D;*753C3?3E7?3E;D=757B3E3@C3E3C3E363B3E6;EF;DD74393;47C;=FE
Dr = r2 – r1
Ingatlah
4 Mudah dan Aktif Belajar Fisika untuk Kelas XI
P
#7E7C3@93@
3 =757B3E3@ C3E3C3E3 ?D
/ B7CB;@63:3@ 47@63 ?
D73@9 H3=EF D7=A@
;=3@63 B7C:3E;=3@(7CD3?33@
P

22. 63@
P67@93@
63@
G7=EAC =757B3E3@ C3E3C3E3 633? 6F3 6;?7@D; 3633:
3 ' (
P
Gambar 1.5
Kecepatan rata-rata v di antara A
dan B searah dengan arah r .
0
r1
r2
y
x
A, t1
B, t2
(AD;D;E;E;=63B3E6;EF;D=3@D74393;47C;=FE

23. P=? =?P=?

24. =?

25. =? =?
(AD;D;E;E;=63B3E6;EF;D633?@AE3D;G7=EACJ3;EF
P' (=?
36;3C3:B7CB;@63:3@=AE363C;D7=A3:+3C;93@3633:
E3@

27. =F36C3@!!
3C5E3@

28. N
Penulisan vektor satuan yang benar
adalah i, j, dan k atau i
, j
, dan k .

29. Ingatlah
Analisis Gerak 5
EAE33C3=
Contoh 1.3
;=7E3:F;G7=EACBAD;D;DF3EFB3CE;=7J3@947C97C3=3633:/
'P

30. (
67@93@633??7E7C63@633?D7=A@+7@EF=3@3:
3 BAD;D;47@63B363D33ED7=A@
4 =757B3E3@63@47D3C=757B3E3@C3E3C3E3D73?3D73@9H3=EFD7=A@:;@993
D7=A@
4 !
;=7E3:F;/
'P

31. (
3 -7=EACBAD;D;B3CE;=7B363D33ED7=A@3633:
/1
2'1P

32. 2('P(
4 -7=EACBAD;D;B3CE;=7B363D33E
D7=A@3633:
/1
2'1
P

33. 2(
'P
(
?3=3=757B3E3@C3E3C3E3B3CE;=73633:
3
/
/
/
i ( '
(
? ?
D
( '

34. (
) ?
D
'P
(?D
63BF@47D3C=757B3E3@C3E3C3E3@J33633:
| v |v
?D
3E3F
v
C

35. ?D
Tugas Anda 1.1
Berilah sebuah contoh persamaan
kecepatan sebagai fungsi dari
waktu yang berorde 2.
#7E7C3@93@
3 G7=EAC =757B3E3@ C3E3C3E3 ?D
x 47D3C =757B3E3@ C3E3C3E3 B363 DF?4F- ?D
y 47D3C =757B3E3@ C3E3C3E3 B363 DF?4F- ?D
7D3C =757B3E3@ C3E3C3E3 ?7?7@F:; B7CD3?33@
P
#73F3@ C3E3C3E3 47C=3;E3@ 67@93@ 3C3= 4F=3@ B7CB;@63:3@
*7C;@9 =73F3@ C3E3C3E3 D3?3 67@93@ =757B3E3@ C3E3C3E3 E3@B3 363
E3@63 3C3: ?;@FD 3E3F BAD;E;8 3?F@ :3 ;@; E7C36; B363 97C3= D3EF
3C3:393;?3@3=73F3@C3E3C3E363@=757B3E3@C3E3C3E3DF3EF47@63
;=3 97C3= 47C43;= 3C3: =7 E7?B3E 3H3
36; =73F3@ C3E3C3E3 63B3E 6;EF;D D74393; 47C;=FE
3FC3E3C3E3
t
b. Kecepatan dan Kelajuan Sesaat
#7E;=3 @63 47C3@9=3E =7 D7=A3: 43;= 67@93@ 47C33@ 47CD7B763
3E3F 67@93@ ?7@3;=; =7@63C33@ 4393;?3@3 @;3; =757B3E3@ 97C3= @63
+7@EF@J3 E;63= D7B3@3@9 B7C33@3@ ?7?;;=; @;3; =757B3E3@ J3@9 D3?3
;3; =757B3E3@ D73F 47CF43: D7E;3B D33E #757B3E3@ D7D33E 3E3F
=757B3E3@D333633:=757B3E3@97C3=47@636;DF3EFE;E;=B363;@E3D3@
@J3 7@93@ =3E3 3;@ =757B3E3@ D7D33E 63C; DF3EF 47@63 J3@9 47C97C3=
3633: =757B3E3@ J3@9 6;?;;=; 47@63 B363 ;@E7CG3 H3=EF ?7@67=3E; @A
Laju adalah besaran skalar, sehingga
ditulis dengan huruf v miring.

36. (363 +! / ;=3E;E;=?7@67=3E;BAD;D;E;E;=G7=EACB7CB;@
63:3@/3=3@47C;?B;E67@93@93C;DD;@99F@9B363;@E3D3@6;
36; 3C3: =757B3E3@ D7D33E 3 D3?3 67@93@ 93C;D D;@99F@9 ;@E3D3@ 6;
E;E;= *753C3 ?3E7?3E;D =757B3E3@ D7D33E 6;EF;D
;? 3
;?
/ /
*77=AC4FCF@9E7C43@967@93@=757B3E3@ ?D633?3C3:

37. NE7C:363B3C3:
:AC;KA@E3+7@EF=3@3:=A?BA@7@=A?BA@7@=73F3@4FCF@963@E7@EF=3@F93
G7=EAC=757B3E3@@J3633?G7=EACD3EF3@
4 !
$F=;D3: E7C74;: 6FF G7=EAC =757B3E3@ 63@ =A?BA@7@=A?BA@7@ =757B3E3@
E7C:363BDF?4F-63@DF?4F- 393C?F63:6;B3:3?;
#73F3@B363DF?4F-
5AD

38. N
?D
?D
#73F3@B363DF?4F-
D;@

39. N
?D
?D
-7=EAC=757B3E3@4FCF@93633:
3 ' (
'
(?D
r'
r2
B', t2'
B, t2
r
Gambar 1.7
v
r''
r2''
B'', t2''
r2'
A, t1
v
Kecepatan sesaat suatu benda dapat
diperoleh dari garis singgung kurva
lintasan benda untuk satu dimensi.
v
Gambar 1.8
Grafik kecepatan sesaat pada bidang
xy atau bidang dua dimensi.
6 Mudah dan Aktif Belajar Fisika untuk Kelas XI
3
d
t dt
P
*753C3 ?3E7?3E;D D7DF3; 67@93@ E38D;C3@ 97A?7EC;D F@EF= EFCF@3@
EFCF@3@ B7CE3?3 63C; DF3EF 8F@9D; B363 DF3EF E;E;= 3633: 9C36;7@ 93C;D
D;@99F@9 =FCG3 6; E;E;= E7CD74FE 7@93@ 67?;=;3@ =757B3E3@ D7D33E
63C; DF3EF E;E;= ?3E7C; 63B3E 6;E7@EF=3@ D753C3 9C38;= 3B34;3 6;=7E3:F;
9C38;= B7CB;@63:3@ E;E;= ?3E7C; E7C:363B H3=EF
(7C:3E;=3@ +! / ,@EF= 9C38;= B7CB;@63:3@ E7C:363B H3=EF
633?97C3=D3EF6;?7@D;63B3E6;=7E3:F;47D3C=757B3E3@D7D33E47@63;=3
93C;D D;@99F@9 =FCG3 6; DF3EF E;E;= ?7?47@EF= DF6FE E7C:363B DF?4F-
47D3C =757B3E3@ D7D33E 47@63 E7CD74FE 63B3E 6;EF;D D74393; 47C;=FE
E3@
P
+! / ?7?B7C;:3E=3@ D74F3: E;E;= ?3E7C; B363 4;63@9
-7=EAC =757B3E3@ D7D33E@J3 3633: 3 ' ( D763@9=3@ 47D3C@J3
?7@36; =73F3@ D7D33E #73F3@ D7D33E 6;B7CA7: 67@93@ B7CD3?33@
P
63BF@ 3C3: =757B3E3@ D7D33E B363 4;63@9 67@93@ ?7;:3E =757B3E
3@B363DF?4F-3633:63@=757B3E3@B363DF?4F- 3633: D7:;@993
3=3@ 6;B7CA7:
E3@
P
36; =73F3@ B363 DF?4F- 63@ DF?4F- 3633:
5AD 63@ D;@
P

40. #7E7C3@93@
=73F3@ B363 DF?4F- ?D
=73F3@ B363 DF?4F- ?D
x
y
vy
vx
v sin
v cos
x
garis
singgung
t
y = x(t)
v sesaat
y
x
vy j
v = 20 m/s
vx i
37°
Gambar 1.6
Ketika B makin dekat dengan A atau
lim
t
0
, kecepatan sesaat v di A
menyinggung lintasan di A.
r1
y
0 x
Contoh 1.4
0
0
0

41. Analisis Gerak 7
3?43C6;D3?B;@93633:9C38;=B7CB;@63:3@
D74F3:D7B763J3@947C97C3=E7C:363BH3=EF
+7@EF=3@=757B3E3@D7B763B363D33E
3

42. D7=A@
4 D7=A@63@
5
D7=A@
Contoh 1.5
x (m)
4 !
*7B3@3@99C38;=97C3=P97C3=D7B7636;439;633?E;9393C;DFCFDJ3;EF93C;D
93C;D63@93C;D
3 (363D33E

43. D7=A@9C38;=97C3=PD7B76347C363B36393C;DFCFD
7D3C=757B3E3@D7B3@3@993C;DE7CD74FE3633:
E3@
?
D
?D
4 (363D33ED7=A@9C38;=97C3=PD7B76347C363B36393C;DFCFD
7D3C=757B3E3@@J33633:
E3@
D
D7B763E;63=47C97C3=
5 (363D33E
D7=A@D7B76347C363B36393C;DFCFD#757B3E3@D7B763
?7CFB3=3@=7?;C;@93@93C;DJ3;EF
E3@
?
D
P
?D
+3@63@793E;8?7@F@F==3@D7B76347C43;=3C3:
6 0
3
30
A
B C
D
6 8 10 14 15 t (s)
Contoh 1.6
*74F3:B3CE;=747C97C3=67@93@B7CD3?33@;@E3D3@/

44. P
?67@93@
633??7E7C63@633?D7=A@+7@EF=3@=757B3E3@B3CE;=7=7E;=3D7=A@
4 !
#757B3E3@6;B7CA7:63C;6;87C7@D;3B7CD3?33@BAD;D;7@93@?7?3DF==3@H3=EF
6;B7CA7:
c. Menghitung Posisi dari Kecepatan
+73: @63 =7E3:F; 43:H3 =757B3E3@ ?7CFB3=3@ EFCF@3@ B7CE3?3
63C; 8F@9D; BAD;D; J3;EF 3
/
' ( *753C3?3E7?3E;D BAD;D;
D74F3: B3CE;=7 63B3E 6;B7CA7: 63C; 8F@9D; =757B3E3@@J3 ?73F; BCAD7D
;@E79C3D;
7D3C =757B3E3@ 633? 3C3: DF?4F-
P
P
/

45. P
'

46. '?D

47. '

48. ''?D
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.
0

49. 7D3C =757B3E3@ 633? 3C3: DF?4F-
P
*77=AC =7;@5; 47C33@ 6; 3E3D CF?BFE B363 4;63@9 $7E3= 3H3 =7;@5; B363
=AAC6;@3E

50. ?#A?BA@7@=757B3E3@@J33633:
63@

51. ;=3
63@633??D63@633?D7=A@E7@EF=3@3:
3 G7=EACBAD;D;=7;@5;63@
4 BAD;D;=7;@5;B363D33ED7=A@
4 !
3 #AAC6;@3E3H3

52. ?63@47D3C=A?BA@7@=757B3E3@@J33633:
63@

53. D7:;@993

54. -7=EACBAD;D;=7;@5;3633:
/' (

55. '

56. (
4 #AAC6;@3E=7;@5;B363D33ED7=A@3633:

57. ?
?
?

58. ?
??
?
36;G7=EACBAD;D;B363D33ED7=A@3633:
/
?'
?(
d. Menghitung Perpindahan dan Jarak dari Grafik Kecepatan
terhadap Waktu
C38;= =757B3E3@ E7C:363B H3=EF 63C; 97C3= DF3EF 47@63 63B3E
6;;:3EB363 +! / *753C39C38;=F3D637C3:J3@96;3CD;CJ3;EF
637C3: J3@9 6;43E3D; 9C38;= 47D3C =757B3E3@ D74393; 8F@9D; H3=EF
67@93@ DF?4F :AC;KA@E3 3633: BAD;D; 63C; 47@63
;=3D74F3:47@6347C97C3=?7@7?BF:93C;DFCFDE3@B347C43;=3C3:
47D3C B7CB;@63:3@ D73F D3?3 67@93@ 3C3= J3@9 6;E7?BF: 47@63 =3@
E7E3B; F@EF= 47@63 J3@9 47C97C3= FCFD 63@ D7D33E =7?F6;3@ 47C43;=
3C3: 3=3@ ?7?;;=; 47D3C 3C3= J3@9 47C4763 67@93@ 47D3C B7CB;@63:3@
(7C:3E;=3@ 93?43C +! / ! $F3D 637C3: J3@9 6;3CD;C
?7@F@F==3@ 47D3C@J3 3C3= 7@93@ ?7@9;@E79C3=3@ B7CD3?33@
93C;D@J3 6;B7CA7: 47D3C 3C3= D74393; 47C;=FE
t
t
8 Mudah dan Aktif Belajar Fisika untuk Kelas XI

59. +3@63?FE3=6;9F@3=3@F@EF=?7?3DE;=3@43:H347D3C3C3=D73F
47CE3@63 BAD;E;8 63BF@ F@EF= ?7@9:;EF@9 47D3C B7CB;@63:3@@J3
6;9F@3=3@ B7CD3?33@ 47C;=FE

60. Gambar 1.9
(a) Grafik fungsi kecepatan (v)
terhadap waktu (t).
(b) Grafik v–t untuk gerak benda
yang berbalik arah.
Contoh 1.7
P
t1
t2 t3
v
t
4
0
v (t)
t
2
1
v(t)dt
t1 t2
a (t)
3
0

61. –4 –3 –2 1 2 3 4 5
v
t
Contoh 1.8 v (m/s)
8
7
6
5
4
3
Analisis Gerak 9
#757B3E3@B3CE;=7J3@947C97C3=FCFD?7?7@F:;B7CD3?33@3

62. PP'
67@93@47D3C633??D63@633?D7=A@+7@EF=3@B7CB;@63:3@63@3C3=J3@9
6;E7?BF:B3CE;=7F@EF=D73@9H3=EF3@E3C3D7=A@63@
D7=A@
4 !
3?43C=3@9C38;=

63. PPE7C74;:63:FF67@93@B7C:;EF@93@47C;=FE
O +;E;=BAEA@9E7C:363BDF?4F-6;B7CA7:63C;

64. PP
PP

65. P

66. 63@P
O +;E;=BF@53=9C38;=

67. D7:;@993=757B3E3@6;
D3633:

68. P
PP
?D
,@EF=D73@9H3=EFD7=A@:;@993
D7=A@6;B7CA7:
O (7CB;@63:3@

69. PP

70. P

71. P
P

72. P

73. P?

74. O 3C3= P

76. P

78. P

82. P

89. P

93. ?
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 47@63 J3@9 ?7@63B3E 93J3
3=3@ ?7@933?; B7CF43:3@
=757B3E3@ D7:;@993 47@63 E7CD74FE ?7?;;=; B7C57B3E3@ *3?3 :3@J3
67@93@ =757B3E3@ B363 B7C57B3E3@ 6;=7@3 F93 ;DE;3: B7C57B3E3@ C3E3
C3E363@B7C57B3E3@D7D33E'7:=3C7@3=757B3E3@E7C?3DF=47D3C3@G7=EAC
?3=3B7C57B3E3@F93?7CFB3=3@47D3C3@G7=EACJ3@9@;3;@J3?7CFB3=3@
EFCF@3@ B7CE3?3 63C; =757B3E3@ ,@EF= 47@63 J3@9 47C97C3= G7CE;=3 =7
3E3D B7C57B3E3@ J3@9 6;?;;=; 3633: B7C57B3E3@ 9C3G;E3D; 63@ 47C@;3;
@793E;8 D763@9=3@ F@EF= 97C3= 3EF: B7C57B3E3@@J3 47C@;3; BAD;E;8
a. Percepatan Rata-Rata
(7C57B3E3@C3E3C3E3B36397C3=6F36;?7@D;?7?;;=;B7@97CE;3@D3?3
67@93@ B7C57B3E3@ C3E3C3E3 B363 97C3= D3EF 6;?7@D; J3;EF :3D; 439;
B7CF43:3@ =757B3E3@ E7C:363B ;@E7CG3 H3=EF
+! / ?7?B7C;:3E=3@ 9C38;= :F4F@93@ =757B3E3@ E7C:363B
H3=EF (363 D33E
47@63 47C363 6; E;E;= 67@93@ =757B3E3@ J3@9
6;?;;=;
(363 D33E 47@63 47C363 6; E;E;= 67@93@ =757B3E3@ J3@9
6;?;;=; (7C57B3E3@ C3E3C3E3 47@63 63C; D3?B3; 3633:
P
#7E7C3@93@
B7C57B3E3@ C3E3C3E3 ?D
B7CF43:3@ =757B3E3@ ?D
D73@9 H3=EF D
v2
v1 A
Gambar 1.10
Percepatan rata-rata.
B
t1
t2
t
v
a =
v
t
t (s)
–1
2
–1
0
–2
–3
–4
–5
–6
–7
–8
–9
–10
–11
–12
1
v = 3t2 – 6t – 9

94. -7=EAC B7C57B3E3@ D74F3: 47@63 J3@9 47C97C3= B363 4;63@9 J3;EF
B363DF?4F:AC;KA@E3DF?4F-63@DF?4FG7CE;=3DF?4F- @;3;@J3
3633:
0
lim
t t
v a
t2
v a
v1
10 Mudah dan Aktif Belajar Fisika untuk Kelas XI
a'a (
P
63BF@ 47D3C B7C57B3E3@ C3E3C3E3 ?7?7@F:; B7CD3?33@
2 2
ax ay
P
#7E7C3@93@
B7C57B3E3@ C3E3C3E3 ?D
a 47D3C B7C57B3E3@ B363 DF?4F- ?D
a 47D3C B7C57B3E3@ B363 DF?4F- ?D
b. Percepatan Sesaat
*7B7CE;B363=757B3E3@D7D33EB7C57B3E3@D7D33E97C3=D74F3:B3CE;
=7 ?7?4FEF:=3@ D73@9 H3=EF J3@9 D3@93E D;@9=3E J3;EF
?7@67=3E; @A 36; B7C57B3E3@ D7D33E ?7CFB3=3@ EFCF@3@ B7CE3?3 63C;
B7CD3?33@ =757B3E3@ F@EF= D73@9 H3=EF ?7@67=3E; @A
;?
;?
d3
3t
dt
P
(7C57B3E3@D7D33E?7CFB3=3@EFCF@3@=76F363C;8F@9D;BAD;D;=3C7@3
3
/
'7: =3C7@3 ;EF
3
/
P
+3:F=3: @63 53C3 ?7@F@F==3@ BCAD7D ;?;E 633? ?7@7@EF=3@
B7C57B3E3@D7D33E47C63D3C=3@9C38;=,@EF=?7@97E3:F;@J3B7C:3E;=3@
+! /
(363 +! / 63@ +! / ! G7=EAC 3
63@ 3
?7CFB3=3@ G7=EAC =757B3E3@ B363 D33E
63@ D763@9=3@ 3 ?7CFB3=3@
B7CF43:3@ =757B3E3@
6;4F3E E7E3B D763@9=3@ 6;4F3E ?7@67=3E;
D7:;@993 7C63D3C=3@678;@;D;
3
?7?;;=; 3C3: J3@9 D3?3
67@93@ 3'7:=3C7@3 ;EF ?7?;;=;3C3:J3@9 D3?367@93@3=7E;=3
?7@67=3E;@A(363 D33E 6;53B3; D7B7CE;6;EF@F==3@ +! /
63@3
6;EF@F==3@67@93@63@3D7:;@993 ?7@36; -7=EAC
B7C57B3E3@ D7D33E D73F ?7@36; D;D; 7@9=F@9 ;@E3D3@ E;E;= ?3E7C;
D763@9=3@ G7=EAC =757B3E3@ 3 E7E3B ?7@J;@99F@9 ;@E3D3@ 6;
;=3B3CE;=747C97C3=B3634;63@9 6;63B3E=A?BA@7@=A?BA@7@
B7C57B3E3@ 47C;=FE ;@;
3 ' ( d d
dt dt
' ( ' ( x y dv dv
dt dt
@63 E7@EF E73:?7@97E3:F; 43:H3 63@
7@93@ 67?;=;3@
63@
2 2
2 2
d x d y
dt dt
D7:;@993 ' (
P
Gambar 1.11
Percepataan sesaat a =
v
merupakan percepatan pada saat
t2 – t1 menuju nol atau t menuju
nol.
3
4
5
y
x
v2
v1
v1 t1
v2
t
v2
v1
t
y
x
a
t1
v2
t2
0
0
y
x
t
0
v

95. t
Analisis Gerak 11
*74F3:B3CE;=747C97C3=67@93@B7CD3?33@=73F3@
P67@93@
633??D63@633?D7=A@+7@EF=3@3:
3 47D3CB7C57B3E3@C3E3C3E397C3=B3CE;=7F@EF=DD3?B3;67@93@
D
4 47D3CB7C57B3E3@3H3B3CE;=763@
5 47D3CB7C57B3E3@B3CE;=7B363D33E

96. D7=A@
4 !
3 (7CD3?33@=757B3E3@3633:
P
,@EF=
D
P ?D
,@EF=
D
P
?D
a
?D
4 (7CD3?33@B7C57B3E3@6;B7CA7:63C;EFCF@3@B7CE3?3B7CD3?33@=757B3E3@J3;EF
d
dt
P P
7D3CB7C57B3E3@3H3B3CE;=7B363D33E 3633:
P PP?D
5 7D3CB7C57B3E3@B3CE;=7B363D33E

97. D7=A@3633:
P

98. P?D
Tugas Anda 1.3
Jika Anda naik mobil dan duduk
di sebelah sopir, coba perhatikan
bagaimana gerakan speedometer
mobil. Diskusikan bersama teman
Anda bagaimana pengaruhnya
terhadap percepatan mobil.
c. Persamaan Kecepatan dari Percepatan Fungsi Waktu
,@EF= 97C3= 47@63 B363 D3EF 4;63@9 =757B3E3@ 6;678;@;D;=3@ D74393;
B7CF43:3@ BAD;D; 633? D73@9 H3=EF E7CE7@EF '7: =3C7@3 B7C57B3E3@
63@ =757B3E3@ ?7CFB3=3@ 47D3C3@ G7=EAC ?3=3 =757B3E3@ 63B3E F93
6;B7CA7: 63C; 8F@9D; B7C57B3E3@ 67@93@ ?7EA67 ;@E79C3D; 47C;=FE
d3
dt 3E3F 0
d3
v
v
0
dt
t
P
P 0
dt
t
P
#7E7C3@93@
=757B3E3@ 47@63 B363 D33E D7=A@ ?D
=757B3E3@ 3H3 47@63 B363 D33E ?D
B7C57B3E3@ 47@63 ?D
C38;= 47D3C B7C57B3E3@ E7C:363B H3=EF 63C; 97C3= D74F3: 47@63
63B3E 6;;:3E B363 +! /
$F3D 637C3: J3@9 6;3CD;C J3;EF 637C3:
J3@9 6;43E3D; A7: 9C38;= 47D3C B7C57B3E3@ D74393; 8F@9D; H3=EF 3
67@93@ DF?4F :AC;KA@E3 3633: B7CF43:3@ =757B3E3@ 97C3= 47@63
;=3 D74F3: 47@63 47C97C3= B363 6F3 6;?7@D; J3;EF B363 4;63@9
:AC;KA@E3 63@ 4;63@9 G7CE;=3 47D3C =A?BA@7@=A?BA@7@ =757B3E3@
E7C:363B DF?4F 63@ DF?4F ?7?7@F:; B7CD3?33@ 47C;=FE
x dt 63@
y dt
P
Gambar 1.12
Grafik fungsi percepatan (a)
terhadap waktu (t).
t
a
t
0
adt
Contoh 1.9
0
63BF@ 47D3C B7C57B3E3@ D7D33E 3633:
P
63BF@ 3C3: B7C57B3E3@ D7D33E E7C:363B DF?4F- 63B3E 6;E7@EF=3@
67@93@ B7CD3?33@ E3@
P

99. Contoh 1.10
*74F3:B3CE;=747C97C3=D7B3@3@9DF?4F67@93@B7CD3?33@B7C57B3E3@

100. ;=3
B363 ?7?;;=;47D3C=757B3E3@?DE7@EF=3@=73F3@B3CE;=7B363D33E
3 D63@
4
D
4 !
#73F3@3H3B3CE;=7?D?3=3B7CD3?33@47D3C=73F3@

101. *74F3:4A36;7?B3C=3@B3634;63@9 #A?BA@7@B7C57B3E3@4A3B3633C3:
:AC;KA@E33633: '?D63@=A?BA@7@B7C57B3E3@633?3C3:G7CE;=3
P(?D(363D33E 4A347C3636;BFD3E=AAC6;@3E 67@93@
=A?BA@7@=A?BA@7@=757B3E3@3H3@J33633:3 I'?D63@3 J(?D
3 +F;D=3@G7=EAC=757B3E3@63@G7=EACBAD;D;D74393;8F@9D;H3=EF
4 7C3B3E;@99;?3=D;?F?J3@96;53B3;4A3
5 +7@EF=3@3C3=E7C3F:J3@96;53B3;4A3
4 !
3 -7=EAC=757B3E3@63B3E6;B7CA7:67@93@B7CD3?33@
Informasi
untuk Anda
Information for You
12 Mudah dan Aktif Belajar Fisika untuk Kelas XI
'
'

102. '?D
(
(
P(?D
-7=EAC=757B3E3@@J33633:
'(L

103. '
P(M?D
-7=EACBAD;D;6;B7CA7:63C;B7CD3?33@
'

104. '
'
(
(P

105. (
/' (
'

106. (

107. 4 +;@99;?3=D;?F?J3@96;53B3;=7E;=3 3633:
P
P
7@93@B7C:;EF@93@?3E7?3E;=36;B7CA7:D+;@99;?3=D;?F?6;B7CA7:
?73F;B7CD3?33@
DP

108. P

109. ?
5 A3=7?43;=7E3@3:47C3CE;
P

110. P

111. P

112. PP
7@93@B7C:;EF@93@?3E7?3E;=36;B7CA7:
D7=A@3C3=E7C3F:63B3E
6;:;EF@9?73F;B7CD3?33@
D
?
3 (363D33ED?3=3
?D
36;=73F3@3H3B3CE;=7B363D33EJ3;EF?D
4 (363D33E
D?3=3
?D
36;=73F3@3H3B3CE;=7B363D33E
D3633:?D
Contoh 1.11
Pada 26 September 1993, seorang
mekanik mesin diesel bernama Dave
Munday yang untuk kedua kalinya
melakukan aksi jatuh bebas setinggi
48 m di air terjun Niagara yang
berada di wilayah Kanada. Pada
aksinya itu, ia menggunakan sebuah
bola baja yang diberi lubang udara
supaya ia bisa bernapas ketika berada
di dalamnya. Munday sangat
memperhatikan faktor keselamatan
pada aksinya itu karena sudah 4
orang yang tewas ketika melakukan
aksi serupa. Oleh karena itu, ia
memperhitungkan aspek fisika
(terutama gerak lurus) dan aspek
teknis dari bola baja yang
digunakannya.
On September 26th, 1993, Dave Munday
a diesel mechanic went over the
Canadian edge of Niagara Falls for
the second time. Freely falling 48 m to
the water (and rocks) below. On this
attempt, he rode in a steel ball with a
hole of air. Munday keep on surviving
this plunge that had killed four other
stuntman, had done considerable
research on the physics (motion along
a straight line) and engineering
aspects of the plunge.
Sumber: Fundamental of Physics, 2001

113. Analisis Gerak 13
5. Perpaduan Dua Vektor
(7CB36F3@ 3@E3C3 6F3 G7=EAC 3=3@ ?7@9:3D;=3@ G7=EAC B7CB36F3@
3E3F G7=EAC C7DFE3@ A@EA: G7=EAC B7CB36F3@ 3E3F G7=EAC C7DFE3@ ;@;
63B3E@63F?B3;B363D74F3:B7C3:FJ3@9D763@9?7@J747C3@9;DF@93;
67@93@ =757B3E3@ E7E3B
%;D3=3@=757B3E3@3;C3@DF@93;6;@J3E3=3@67@93@
63@=757B3E3@
B7C3:F 6;@J3E3=3@ 67@93@ D7B7CE; 6;EF@F==3@ B363 +! /
(7CB36F3@ 3@E3C3 6F3 4F3: 97C3= FCFD 47C3EFC3@ E7CD74FE ?7?47@EF=
C7DFE3@6F3G7=EAC J3;EF 1 27D3C C7DFE3@=757B3E3@@J33633:
5AD
P
63BF@3C3=J3@96;E7?BF:B7C3:F=3C7@3C7DFE3@=757B3E3@@J33633:
#7E7C3@93@
47D3C C7DFE3@ =757B3E3@ =76F3 97C3= ?D
3C3= ?
H3=EF E7?BF: D
Tugas Anda 1.4
Buatlah kelompok diskusi kecil.
Apakah Anda dan teman-teman
dapat memprediksi gerak seperti
apa yang dihasilkan oleh
perpaduan dua buah gerak lurus
berubah beraturan? Setelah
selesai diskusi, kemukakan
pendapat kelompok Anda di
depan kelas.
Gambar 1.13
v adalah vektor resultan dari v1
(kecepatan aliran sungai) dan v2
(kecepatan perahu).
Contoh 1.12
*74F3:B7C3:F:7@63=?7@J747C3@9;DF@93;67@93@=757B3E3@

114. ?D63@?7?47@EF=
DF6FE NE7C:363B3C3:3CFDDF@93;J3@9?7@93;C67@93@=757B3E3@
?D
3 3?43C=3@;@E3D3@B7C3:F
4 +7@EF=3@C7DFE3@=757B3E3@B7C3:F
5 7C3B3=3:B7CB;@63:3@B7C3:FD3?B3;=7D747C3@9D7E73:D7=A@
4 !
;=7E3:F;
.

115. ?D N
?D
3 $;@E3D3@B7C3:F
743CDF@93; 3 =757B3E3@C7DFE3@
3. =757B3E3@B7C3:F 30 =757B3E3@3;C3@DF@93;
4 3 3.30
5AD

116. 1

117. 5AD 2
1
2
?D
36;47D3CC7DFE3@=757B3E3@B7C3:F3633: ?D
5 ?DQD

118. ?
36;B3@3@9;@E3D3@J3@96;E7?BF:B7C3:FD7E73:D7=A@3633:

119. ?
Kata Kunci
• arah vektor
• besar vektor
• jarak
• kecepatan
• kelajuan
• percepatan
• perpindahan
• posisi
• vektor satuan
v
v2 (kecepatan perahu)
v1 (aliran sungai)
vp
vs
v
s

120. Tes Kompetensi Subbab A
$/( ) ,* # * +!2)2* 1' ,
bola (A) bola (B)
Sumber: Physics for Scientist
Engineers, 2000
14 Mudah dan Aktif Belajar Fisika untuk Kelas XI
3 +7@EF=3@G7=EACBAD;D;63@3C3=47@6363C;E;E;=
3D3B363D33ED7=A@
4 +7@EF=3@B7CB;@63:3@63@=757B3E3@C3E3C3E3
47@63633?D73@9H3=EF
D7=A@D3?B3;67@93@

121. D7=A@
5 +FCF@=3@B7CD3?33@F?F?=757B3E3@47@63
6 +7@EF=3@=73F3@47@63B363D33ED7=A@
*74F3:43EF6;7?B3C=3@G7CE;=367@93@B7CD3?33@
P67@93@ 633??7E7C63@633?
D7=A@+7@EF=3@
3 47D3CB7C57B3E3@43EF63@
4 E;@99;?3=D;?F?J3@96;53B3;43EF
*74F3:CA=7E47C97C3=B3634;63@9 (7C57B3E3@
CA=7E?7?;;=;=A?BA@7@
?D63@
P
E?D67@93@633?D7=A@(363D33E CA=7E
47C3636;E;E;=BFD3E=AAC6;@3E 67@93@=A?BA@7@
=757B3E3@3H3
?D63@ ?DD7:;@993
D753C3G7=EAC6;EF;D=3@
'(?D
3 J3E3=3@ G7=EAC =757B3E3@ 63@ BAD;D; D74393;
8F@9D;63C;H3=EF
4 7C3B3E;@99;?3=D;?F?J3@96;53B3;CA=7E
5 7C3B33C3=E7C3F:J3@96;53B3;CA=7E
*74F3:B3CE;=747CB;@63:63C;BAD;D;/
=7BAD;D;/
;=3 BAD;D; /
' P ( 63@ BAD;D; /
'

122. (
E7@EF=3@3:
3 G7=EACB7CB;@63:3@@J363@
4 47D3CB7CB;@63:3@@J3
*74F3:B3CE;=747C97C3=47C63D3C=3@8F@9D;=757B3E3@
67@93@633??D63@633?
D7=A@;=3=A@DE3@E3

123. ?D

124. P?D63@
?DE7@EF=3@3:
3 B7C57B3E3@C3E3C3E3B3CE;=7F@EF=D73@9H3=EF
D7=A@D3?B3;D7=A@
4 B7C57B3E3@3H3B3CE;=763@
5 B7C57B3E3@B3CE;=7B363D33ED7=A@

125. (7CD3?33@97C3=D74F3:B3CE;=76;@J3E3=3@A7:8F@9D;

126. 67@93@633??7E7C63@633?D7=A@
3 ;EF@93: B7C57B3E3@ C3E3C3E3 633? D73@9
H3=EF

127. D7=A@D3?B3;67@93@
D7=A@
4 ;EF@93:=757B3E3@D7D33EB363D7=A@
5 ;EF@93:B7C57B3E3@D7D33EB363D7=A@
(AD;D;D74F3:47@636;@J3E3=3@A7:/

128. P
'
(67@93@/633??7E7C63@633?D7=A@
B. Gerak Parabola
+3:F=3:@63J3@96;?3=DF667@93@97C3=B3C34A3%7@FCFE3;7A
97C3=B3C34A363B3E6;B3@63@9D74393;:3D;B7CB36F3@97C3=FCFD47C3EFC3@
B363 DF?4F :AC;KA@E3 DF?4F- 63@ 97C3= FCFD 47CF43: 47C3EFC3@ B363
DF?4F G7CE;=3 DF?4F- D753C3 E7CB;D3: *7E;3B 97C3= E;63= D3;@9
?7?7@93CF:; 34F@93@ 63C; =76F3 97C3= E7CD74FE 47CFB3 97C3= B3C34A3
7C3= B3C34A3 6;D74FE F93 D74393; 97C3= B7FCF 33? =7:;6FB3@ D7:3C;
:3C; :3 E7CD74FE 63B3E @63 F?B3; ?;D3@J3 =7E;=3 @63 ?7@7@63@9
4A367@93@DF6FEE7@63@93@E7CE7@EF3E3F NE7C:363B4;63@9:AC;KA@E3
?3=3 ;@E3D3@ 97C3= 4A3 3=3@ 47C47@EF= B3C34A3
(7C:3E;=3@ +! /

129. 6F3 4F3: 4A3 6;7B3D=3@ 63C; =7E;@99;3@
E7CE7@EF A3 6;7B3D=3@ 63C; =73633@ 6;3? 33? :3 ;@; 97C3=
4A3 3633: 97C3= 3EF: 4743D J3@9 6;B7@93CF:; A7: B7C57B3E3@
9C3G;E3D; F?; (363 D33E J3@9 47CD3?33@ 4A3 6;47C; 93J3 :AC;KA@E3
67@93@ B793D $;@E3D3@ 4A3 D7B7CE; B363 +! /

130. 47CFB3
B3C34A3 (363 3H3@J3 =76F3 4A3 E7C7E3= B363 =7E;@99;3@ J3@9 D3?3
=7?F6;3@ 6;7B3D=3@ B363 D33E J3@9 D3?3 3B3E=3: @63 ?7@743=
4A3 ?3@3 J3@9 ?7@J7@EF: E3@3: 74;: 57B3E
+7C@J3E3 H3=EF J3@9 6;4FEF:=3@ A7: =76F3 4A3 :;@993 ?7@J7@EF:
E3@3: 3633: D3?3 7@93@ 67?;=;3@ =757B3E3@ ?7@63E3C 3=;43E 363@J3
93J3E7C:363B4A3E;63=?7@36;=3@4A3?7@J7@EF:E3@3:74;:57B3E
7C3= 47@63 J3@9 D7C;@9 6;36;=3@ D74393; 5A@EA: 97C3= B3C34A3 3633:
97C3= B7FCF J3@9 3C3:@J3 ?7?47@EF= DF6FE 77G3D; E7CE7@EF E7C:363B
B7C?F=33@ F?;
Gambar 1.14
Dua buah bola dilepaskan dari
ketinggian yang sama dengan
perlakuan yang berbeda.

131. Gambar 1.15
Lintasan gerak parabola. Nilai a
selalu negatif karena ditetapkan
arah positif adalah arah ke atas, dan
arah gravitasi selalu ke bawah.
Sumber: Conceptual Physics, 1998
Analisis Gerak 15
(7C:3E;=3@ 93?43C 47C;=FE
v
vx
(363 +! /
E7C;:3E=757B3E3@B363DF?4F-J3;EF367@93@
3C3:63@47D3C@J3D73F=A@DE3@63BF@=757B3E3@B363DF?4F- J3;EF
3 3C3: 63@ 47D3C@J3 6;B7@93CF:; A7: B7C57B3E3@ 9C3G;E3D; -7=EAC
C7DFE3@ =757B3E3@ 3 D7E;3B D33E 6; =AAC6;@3E ?7CFB3=3@ C7DFE3@
63C;=A?BA@7@G7=EAC63@
;=3 47D3C =757B3E3@ 3H3 B7FCF 3633: 47D3C =757B3E3@ 633?
3C3: DF?4F- 3633: 63@ 47D3C =757B3E3@ B363 DF?4F- 3633:
6;B7CA7: B7CD3?33@B7CD3?33@ 47C;=FE ;@;
(7CD3?33@ B363 DF?4F-
7D3C =757B3E3@ B363 DF?4F- 3633:
5AD
P
3C3= B363 DF?4F- 3633:
5AD
P
(7CD3?33@ B363 DF?4F-
7D3C =757B3E3@ B363 DF?4F- 67@93@
P 3633:
D;@
P

132. 3C3= B363 DF?4F- 3633:
D;@ 1
2
P

133. #7E7C3@93@
47D3C =757B3E3@ B363 DF?4F- ?D
47D3C =757B3E3@ B363 DF?4F- ?D
A 47D3C =757B3E3@ 3H3 ?D
47D3C B7C57B3E3@ 9C3G;E3D; ?D
DF6FE 77G3D;
3C3= E7?BF: ?7@63E3C ?
3C3= E7?BF: G7CE;=3 ?
7C63D3C=3@ B7CD3?33@B7CD3?33@ E7CD74FE 63B3E 6;D;?BF=3@
43:H3=A?BA@7@97C3=B363DF?4F- G7CE;=3D3@93E6;B7@93CF:;A7:
B7C57B3E3@ 9C3G;E3D; F?; D763@9=3@ 47D3C =757B3E3@ 633? 3C3:
?7@63E3C D73F E7E3B 7C3= B363 DF?4F- 3633: 97C3= FCFD 47CF43:
47C3EFC3@ 63@ DF?4F- 3633: 97C3= FCFD 47C3EFC3@
Tokoh
Galileo Galilei
(1564–1642)
Galileo Galilei lahir di kota Pisa, Italia,
pada 15 Februari 1564. Ia belajar
kedokteran di Universitas Pisa. Oleh
karena ia lebih tertarik dengan ilmu
alam, ia tidak melanjutkan belajar
kedokterannya, tetapi belajar
matematika. Ia meninggalkan Pisa
untuk belajar di Universitas Padua. Ia
adalah pendukung teori Heliosentris
yang menyatakan bahwa Matahari
adalah pusat tata surya. Penemuan
Galileo yang terkenal adalah
teleskop dan menemukan
pegunungan di Bulan serta satelit di
Yupiter. Oleh karena hobinya
mengamati benda-benda langit
termasuk Matahari, ia menderita
kebutaan pada usia 74 tahun. Ia
meninggal dunia 4 tahun kemudian
pada usia 78 tahun.
y
v0 sin
O(0,0)
v0
vy
vx
v0 cos
vy = 0
ay = –g
vy
vx
v
v
vy
x
vx

134. Contoh 1.13
*7AC3@9D;DH3?77?B3C4A367@93@=757B3E3@
?DB3633C3:J3@9?7?47@EF=
DF6FE

135. NE7C:363BE3@3:D;@

136. +7@EF=3@
3 47D3C=757B3E3@4A3D7E73: D63@
4 47D3C=76F6F=3@4A3D7E73: D;=3B7C57B3E3@9C3G;E3D;
?D
4 !
;=7E3:F;
?D

137. N
D;@ D;@

138. N
5AD
?D
3 7D3C=A?BA@7@=757B3E3@B363D33E D7=A@3633:
5AD D;@
?D ?D
?D ?D
7D3C=757B3E3@B363D33E D7=A@3633:
?D
P E
?DP
?D D

139. ?D
2 2
tx ty m/s m/s

140. ?D
36;47D3C=757B3E3@4A3D33E D3633:
?D
4 #76F6F=3@4A3D33E D7=A@3633:
?D D

141. ?
P
16 Mudah dan Aktif Belajar Fisika untuk Kelas XI
?D DP
?D D
P
??
36;=76F6F=3@4A3B363D33E D3633:B363=AAC6;@3E

142. ??
1. Tinggi Maksimum dan Jarak Terjauh
,@EF= ?7@7@EF=3@ E;@99; ?3=D;?F? 63@ 3C3= E7C3F: J3@9 63B3E
6;53B3; A7: 47@63 J3@9 47C97C3= 67@93@ ;@E3D3@ B3C34A3 B7CF
6;E7@EF=3@63:FFH3=EFJ3@96;B7CF=3@F@EF=?7@53B3;E;E;=E7CE;@99;
63@ 3C3= E7C3F:@J3
a. Waktu untuk Mencapai Tinggi Maksimum
#7E;=3 ?7@53B3; E;@99; ?3=D;?F? 47D3C =757B3E3@ 47@63 J3@9
47C97C3= B3C34A3 B363 3C3: G7CE;=3 D3?3 67@93@ @A CE;@J3
47@63D7?B3E47C:7@E;4747C3B3D33ED747F?3=:;C@J3EFCF@36;H3=EF
E7?BF: 47@63 D33E ?7@53B3; E;@99; ?3=D;?F? 3633:
D;@
D;@P
?3=D D;@
P

143. b. Waktu untuk Mencapai Jarak Terjauh
(363 97C3= B3C34A3 3C3=E7C3F:3633: 3C3= J3@9 6;E7?BF: 47@63
D33E =7?43; =7 B7C?F=33@ E3@3: (363 =A@6;D; E7CD74FE 3C3= ?7@FCFE
DF?4F- 63@ 3C3= ?7@FCFE DF?4F- 3633: ?3=D;?F? '7: =3C7@3
;EF 6;9F@3=3@ =A@6;D; =76F3 J3;EF 36; H3=EF F@EF= ?7@53B3;
3C3= E7C3F: 3633:

144. Analisis Gerak 17
I?3=D
D;@
D;@
D;@
P

146. c. Tinggi Maksimum
+;@99; ?3=D;?F? J3@9 63B3E 6;53B3; A7: 47@63 J3@9 ?73=F=3@
97C3=B3C34A363B3E6;53C;67@93@?7@99F@3=3@H3=EFF@EF=?7@53B3;
E;@99;?3=D;?F?=7B7CD3?33@97C3=633?DF?4F- 7@93@?73=F=3@
DF4DE;EFD;$/0 + , 6 =7633?$/0 + , 66;B7CA7:E;@99;
?3=D;?F? D74393; 47C;=FE
?3=D D;@
D;@
D;@
D;@
P
D;@
?3=D
D;@
P

147. d. Jarak Terjauh pada Sumbu-x
7@93@?73=F=3@DF4DE;EFD;$/0 + , 6=7633?$/0 + ,
6
?3=3 6;B7CA7: 3C3= E7C3F: B363 DF?4F- D74393; 47C;=FE
?3=D 5AD ?3=D
5AD
D;@
(
D;@ 5AD
?3=D
D;@
P

148. 2. Jarak Terjauh dan Pasangan Sudut Elevasi
(7C:3E;=3@ +! / J3@9 ?7@F@F==3@ 9C38;= ;@E3D3@ B3C34A3
J3@9 6;E7?BF: D74F3: 47@63 67@93@ =73F3@ 3H3 J3@9 D3?3 E7E3B;
67@93@ DF6FEDF6FE 77G3D; J3@9 47C4763 @63 63B3E ?7@J;?BF=3@
43:H3B3D3@93@DF6FE77G3D;J3@947CF?3: NJ3;EFN63@
ND7CE3
N 63@

149. N 3=3@ ?7@9:3D;=3@ 3C3= E7C3F: J3@9 D3?3 (363 93?43C
E7CD74FE E7C;:3E 43:H3 3C3= E7C3F: ?7@53B3; :3C93 ?3=D;?F? F@EF=
DF6FE77G3D;
N7C63D3C=3@B7CD3?33@B7CD3?33@97C3=B3C34A33C3=
E7C3F: 6;B7CA7: ;=3 D;@
3E3F
N
Gambar 1.16
Grafik lintasan sebuah benda
dengan sudut elevasi berbeda dan
kecepatan awal yang sama.
*74F3:47@636;7B3D=3@63C;B7D3H3EE7C43@9J3@9D763@9E7C43@9?7@63E3C67@93@
=757B3E3@
?D63@47C363B363=7E;@99;3@ ?6;3E3DE3@3:;=3
?D
47C3B3=3:
3 H3=EFJ3@96;B7CF=3@47@63:;@993E;436;E3@3:
4 3C3=?7@63E3C3EF:@J347@6363@
5 =757B3E3@47@63D747F??7@J7@EF:E3@3:
y jarak vertikal (m)
x jangkauan
(meter)
75°
60°
45°
15°
Contoh 1.14
Tantangan
untuk Anda
Apakah tendangan bebas yang
biasa dilakukan oleh pemain
sepakbola seperti David Beckham
termasuk gerak parabola?
30°

150. 4 !
3 7C3=G7CE;=347@63D3?367@93@97C3=3EF:4743D
?
?D
??D
D
36;47@63E;436;E3@3:D7E73:
D7=A@
4 3C3=?7@63E3C6;B7CA7:?73F;B7CD3?33@
F34F3:B7FCF67@93@3@9=3F3@

151. ?7?4FEF:=3@H3=EF
63@F@EF=?7@53B3;
=7E;@99;3@D7?F3B7FCF=7?43;=7E3@3:F=E;=3@43:H3

152. Ingatlah
b b ac
18 Mudah dan Aktif Belajar Fisika untuk Kelas XI
4 !

153. 5AD 3E3F5AD

154. D;@
3E3FD;@
7@93@?7?3DF==3@:3C93D;@ 63@5AD B363CF?FD5AD D;@
@633=3@?7?B7CA7:
P

155. ,@EF=?7@63B3E=3@@;3;
63@@63:3CFD?7@53C;@;3;3=3C3=3CB7CD3?33@
E7CD74FE67@93@?7@99F@3=3@CF?FD47C;=FE
t
34P
5
)
t
t
t
t

156. E7C4F=E;
?D
D
?
36;3C3=?7@63E3C47@63
?
5 #757B3E3@47@63D747F??7@J7@EF:E3@3:63B3E6;B7CA7:67@93@CF?FD
G7=EACC7DFE3@=757B3E3@J3;EF
?D
P
?D
DP
?D
?D
?D
?D
36;47D3C=757B3E3@47@63D747F??7@J7@EF:E3@3:
?D
Contoh 1.15
Tantangan
untuk Anda
Buktikan bahwa jika kecepatan
awalnya sama, pasangan sudut
elevasi yang hasil jumlahnya 90°
akan menghasilkan jarak terjauh
yang sama. Kemudian, apakah
waktu tempuh keduanya sama?
Untuk mencari nilai akar-akar dari y
= ax2 + bx + c digunakan rumus abc
berikut ini.
x1,2 =
2 4
2
a

157. Contoh 1.16 Kata Kunci
10 m
Analisis Gerak 19
*74F3:B7FCF6;E7?43==3@67@93@DF6FE77G3D;

158. N+7@EF=3@B7C43@6;@93@3@E3C3
E;@99;?3=D;?F?63@3C3=E7C3F:J3@96;53B3;B7FCF
4 !
3C;B7CD3?33@E;@99;?3=D;?F?63@3C3=:AC;KA@E3E7C3F:6;B7CA7:
D;@
D;@
D;@
Tes Kompetensi Subbab B
$/( ) ,* # * +!2)2* 1' ,
*74F3:47@636;7?B3C=3@67@93@DF6FE77G3D;

159. N
63@ =757B3E3@ 3H3@J3
?D ;=3
?D
E7@EF=3@3:
3 H3=EFE7?BF:47@636;F63C3
4 E;@99;?3=D;?F?J3@96;53B3;63@
5 3C3=?7@63E3CE7C3F:
F=E;=3@ 43:H3 47@63 J3@9 6;7?B3C=3@ 67@93@
DF6FE 77G3D; 3=3@ ?7@53B3; E;E;= E7C3F: ;=3
N

160. *74F3:47@636;7?B3C=3@67@93@DF6FE77G3D; N
63@=757B3E3@
?D ;EF@947D3C63@3C3:=757B3E3@
D7E73:
3

161. D7=A@63@
4

162. D7=A@
(AD;D;4FCF@9E7C7E3=B363=AAC6;@3E ?*7AC3@9
3@3=?7A@E3C=3@43EF67@93@?7@99F@3=3@=3E3
B7@J3 B363 DF6FE 77G3D;

163. N =7 3C3: 4FCF@9
7C3B3=3:=757B3E3@J3@9:3CFD6;47C;=3@393C43EF
?7@97@3;4FCF@9E7CD74FE
33?B7C?3;@3@D7B3=4A3(FEF?7@7@63@94A3
67@93@DF6FE77G3D; D7:;@993?7@53B3;=7E;@9
9;3@?3=D;?F?
?7E7C7C3B33?34A3:3CFD6;
EF@99F:;@993E;43=7?43;6;E3@3:
?D
*74F3:D3D3C3@E7C7E3=B3633C3=
?63@B363
=7E;@99;3@
? 7C3B3 DF6FE 77G3D; B7FCF 393C
E7?43=3@@J3E7B3E?7@97@3;D3D3C3@;=3=757B3E3@
3H3@J3 ?D
100 m
*74F3:D3D3C3@E7C7E3=B363=AAC6;@3E ?
*7D7AC3@9?77?B3C=3@43EF67@93@DF6FE

164. N=7
3C3:D3D3C3@E7CD74FE63C;BFD3E=AAC6;@3E7C3B3=3:
=757B3E3@J3@9:3CFD6;47C;=3@393C43EF;EFE7B3E
?7@97@3;D3D3C3@
*74F3:B7D3H3EE7?BFC47C97C3=67@93@=757B3E3@
?D63@3C3:@J3?7?47@EF=DF6FE N67@93@
4;63@963E3C#7E;=347C363B363=7E;@99;3@ ?
63C;3C3:G7CE;=3D74F3:4A?6;7B3D=3@63C;
B7D3H3EE7CD74FE;=34A?3EF:6;E;E;=E7@EF=3@
3C3=
2
:
D;@
D;@
: D;@
60o
2
:

165. 1 :

166. 1:

167. 36;B7C43@6;@93@@J33633:1:

168. • jarak terjauh
• sudut elevasi
• tinggi maksimum
60°
800 m
A B

169. 3
4
5
C. Gerak Melingkar
;#73D/@63E73:?7?B733C;97C3=?7;@9=3C47C3EFC3@(363
DF4434 ;@; 3=3@ 6;43:3D 97C3= ?7;@9=3C D753C3 F?F? 67@93@
?7@99F@3=3@ =3;63: G7=EAC (7C:3E;=3@ +! / *74F3:
B3CE;=7 47C97C3= ?7;@9=3C B363 4;63@9 67@93@ 3C;3C; E7C:363B
BFD3E =AAC6;@3E @63 E7@EF E73: ?7@97E3:F; 43:H3 633? 97C3=
?7;@9=3CG7=EAC=757B3E3@97C3=B3CE;=73D73F?7@J;@99F@9;@E3D3@
97C3= B3CE;=7 63@ E793= FCFD 3C;3C; ;@E3D3@ B3CE;=7
; E;E;= ( G7=EAC =757B3E3@ 3 ?7?47@EF= DF6FE E7C:363B 93C;D
G7CE;=3 63@ 47D3C@J3 D3?3 67@93@ DF6FE J3@9 6;47@EF= A7: 3C;3C;
/ 63@ DF?4F '7: =3C7@3 ;EF =A?BA@7@ G7=EAC 3 6;@J3E3=3@ 67@93@
B7CD3?33@
20 Mudah dan Aktif Belajar Fisika untuk Kelas XI
3' (
3PD;@ '5AD (
P

170. 63BF@BAD;D;B3CE;=7B3634;63@9 6;@J3E3=3@67@93@B7CD3?33@
/' (
P

171. 7@93@ 67?;=;3@ G7=EAC =757B3E3@ 3 63B3E 6;@J3E3=3@ 67@93@
B7CD3?33@
3 ' (
P

172. @63 E7@EF ?7@97E3:F; 43:H3 B7CD3?33@ 97C3= B3CE;=7 6;63B3E=3@
63C;EFCF@3@B7CE3?3B7CD3?33@=757B3E3@3E3FEFCF@3@=76F3B7CD3?33@
BAD;D; B3CE;=7 '7: =3C7@3 ;EF B7CD3?33@ B7C57B3E3@ ;@73C 633? 97C3=
?7;@9=3C 6;@J3E3=3@ 67@93@ B7CD3?33@
3
' (
P

173. 3C;$/0 + , 6@;3;
?7CFB3=3@=A?BA@7@=757B3E3@
633? 3C3: DF?4F- 63@
?7CFB3=3@ =A?BA@7@ =757B3E3@ 633?
3C3: DF?4F- (7C:3E;=3@ +! / ! 3C; 93?43C E7CD74FE @63
3=3@ ?7@97E3:F; 43:H3 P D;@ 63@ D;@ 7@93@ 67?;=;3@
$/0 + , 6 63B3E 6;@J3E3=3@ 67@93@ B7CD3?33@
5AD D;@
' (
P
-7=EAC B7C57B3E3@ 63@ =A?BA@7@ G7=EAC B7C57B3E3@ 633? 97C3=
?7;@9=3C 6;EF@F==3@ A7: +! / D763@9=3@ 47D3C G7=EAC
B7C57B3E3@ 6;@J3E3=3@ 67@93@ B7CD3?33@
5AD D;@
P
,@EF= ?7@7@EF=3@ 3C3: 63B3E @63 E7@EF=3@ ?73F; DF6FE
D7B7CE; 6;EF@F==3@ +! /
y v
P
yP
xP x
y v vy
vx
x
y
x
ay
ax
a
Gambar 1.17
Sebuah partikel bergerak melingkar.
(a) Posisi dan kecepatan partikel
pada saat tertentu.
(b) Komponen-komponen vektor
kecepatan.
(c) Percepatan gerak partikel dan
komponen-komponennya.

174. Analisis Gerak 21
D;@
E3@ E3@
5AD
36; 3CE;@J3B7C57B3E3@ 3C3:@J3 D7B3@3@9 3C;3C; ?7@FF
BFD3E ;@9=3C3@ +! / (7C57B3E3@ E7CD74FE 6;@3?3=3@
Rangkuman
-7=EACBAD;D;?7@F@F==3@BAD;D;B3CE;=763C;E;E;=
3D3=7BAD;D;B3CE;=7E7CD74FEJ3;EF
/'(
(7CB;@63:3@?7CFB3=3@B7CF43:3@G7=EACBAD;D;/
D73?3D73@9H3=EF
//P/
' (
67@93@P
63@ P
#757B3E3@C3E3C3E33633:47D3C@J3B7CB;@63:3@
633?D73@9H3=EFE7CE7@EF
/ / /
3
#757B3E3@ D7D33E 3633: =757B3E3@ C3E3C3E3 F@EF=
D73@9H3=EF?7@67=3E;@A
;? ;?
/ /
3 3
-7=EACBAD;D;63B3E6;E7@EF=3@63C;=757B3E3@D7D33E
67@93@53C3;@E79C3D;
v
67@93@3633:G7=EACBAD;D;B363

175. (7C57B3E3@C3E3C3E33633:B7CF43:3@=757B3E3@
633?D7E;3BD3EF3@H3=EF
3 3 3
(7C57B3E3@D7D33E3633:B7C57B3E3@C3E3C3E3F@EF=
D73@9H3=EF?7@67=3E;@A
;? ;?
3 3
#757B3E3@D7D33E63B3E6;E7@EF=3@63C;B7C57B3E3@
D7D33E67@93@53C3;@E79C3D;
67@93@3633:=757B3E3@B363
D33E
7C3=B3C34A3?7CFB3=3@B7CB36F3@63C;$B363
DF?4F-63@$B363DF?4F- D7:;@993
E7E3B63@
63@
67@93@
P
(36397C3=B3C34A3E;E;=E7CE;@99;6;53B3;B363D33E
7@93@ ?7?3DF==3@ DJ3C3E ;@; 63B3E
6;E7@EF=3@
?3=D D;@
?3=D
D;@
+;E;=B3;@93F:6;53B3;B363D33E 7@93@
?7@99F@3=3@D;83ED;?7EC;B3C34A36;B7CA7:
?3=D ?3=D D;@
?3=D
D;@
(36397C3=?7;@9=3C47D3CB7C57B3E3@D7@EC;B7E3
6;@J3E3=3@67@93@B7CD3?33@
63BF@DF6FEJ3@96;47@EF=3C3:G7=EACB7C57B3E3@
D7@EC;B7E36;@J3E3=3@67@93@B7CD3?33@
E3@
(7C57B3E3@ D7@EC;B7E3 3633: B7C57B3E3@ 97C3=
B3CE;=7 J3@9 47C97C3= ?7;@9=3C 63@ 3C3:@J3
?7@FF=7BFD3E;@9=3C3@
Tes Kompetensi Subbab C
$/( ) ,* # * +!2)2* 1' ,
*74F3:B3CE;=747C97C3=?7;@9=3C67@93@=757B3E3@
=A@DE3@633?4;63@9 #7E;=3B3CE;=747C3636;BAD;D;
P ? =757B3E3@@J3 P
?D( +7@EF=3@3:
=757B3E3@63@B7C57B3E3@B3CE;=7B363BAD;D; ?
*74F3: D3E7;E F?; 47C97C3= B363 AC4;E@J3 J3@9
47C3C3=
=? 6; 3E3D B7C?F=33@ F?; 67@93@
B7C;A67?7@;E+7@EF=3@3:=757B3E3@63@47D3C
B7C57B3E3@D7@EC;B7E363C;D3E7;EE7CD74FE

176. *7AC3@93DECA@3FE47C97C3=633?;@E3D3@?7;@9=3C
47C3C;3C;?
3 +7@EF=3@=757B3E3@97C3=3DECA@3FE;=347D3C
B7C57B3E3@D7@EC;B7E3@J3
4 7C3B3=3: F?3: BFE3C3@ J3@9 6;:3D;=3@
47C63D3C=3@B7C57B3E3@E7CD74FE
5 +7@EF=3@B7C;A6763C;97C3=3DECA@3FEE7CD74FE

177. ?7?43:3D D753C3 G7=EAC ?7?43:3D D753C3 G7=EAC
Refleksi
22 Mudah dan Aktif Belajar Fisika untuk Kelas XI
$/ )$,#
63B3E
47C47@EF=
7C3=$FCFD 7C3=(3C34A3 7C3=%7;@9=3C
(AD;D; #757B3E3@
)3E3C3E3
(7C57B3E3@
)3E3C3E3
#757B3E3@
*7D33E
(7C57B3E3@
*7D33E
Peta Konsep
7C3=$FCFD
7C3EFC3@$
(AD;D; #757B3E3@ (7C57B3E3@
7C3=$FCFD7CF43:
7C3EFC3@$
+;E;=+7CE;@99; ?3=D .3=EF*3?B3;6;
+;E;=+7CE;@99; ?3=D
.3=EF*3?B3;6;
+;E;=+7C3F:?3=D
+;E;=+7C3F:?3=D
Setelah mempelajari bab ini, Anda tentu dapat
membedakan antara besaran vektor dan besaran skalar
yang ada pada konsep gerak. Anda juga dapat
menentukan persamaan besaran fisika dari persamaan
yang diketahui dengan menggunakan operasi integral
dan diferensial. Dari materi bab ini, bagian manakah yang
Anda anggap sulit?
Dengan mempelajari bab ini, Anda dapat
menentukan bentuk lintasan gerak suatu benda. Pada
gerak parabola, titik terjauh dan titik tertinggi dapat
ditentukan dari persamaan gerak dan waktunya. Nah,
sekarang coba Anda sebutkan manfaat lain mempelajari
bab ini.
?7?43:3D
DF?4F- DF?4F-
B363 B363

178. y = –t + 6
Analisis Gerak 23
Tes Kompetensi Bab 1
'*'* 0 * 0 12( 4 ! ,5 ,%. *',%1$. 1# ,)$/( ) ,* . # !2)2* 1' ,
(363 D33E
D77=AC 4FCF@9 ?7?;;=; BAD;D; 6;
=AAC6;@3E
??63@B363D33E
D?7?;;=;
=AAC6;@3E??33?D73@9H3=EFE7CD74FE
47D3C=757B3E3@C3E3C3E34FCF@93633:
3
?D 6 ?D
4
?D 7 ?D
5 ?D
(7CD3?33@ G7=EAC BAD;D; D74F3: ?3E7C; 6;@J3E3=3@
67@93@/

179. P'

180. (;=3D3EF3@/633?
?7E7C63@633?D7=A@47D3CB7C57B3E3@?3E7C;E7B3E
D7E73:D7=A@63C;3H3B7@93?3E3@3633:
3 ?D 6 ?D
4
?D 7
?D
5 ?D
!1 , 0

181. *74F3:B7FCFJ3@96;E7?43==3@67@93@DF6FE77G3D;
E7CE7@EF?7?;;=;B7CD3?33@G7=EACBAD;D;
/

182. '

184. (67@93@633?D7=A@63@
/633??7E7C3C3=E7C3F:J3@963B3E6;53B3;A7:
B7FCF633?3C3:DF?4F3633:
3

185. ? 6 ?
4

186. ? 7

187. ?
5
?
(AD;D;D74F3:47@63J3@96;7?B3C=3@G7CE;=3=73E3D
6;@J3E3=3@67@93@B7CD3?33@
P67@93@
633??7E7C63@633?D7=A@#757B3E3@3H363@
E;@99;?3=D;?F?J3@963B3E6;53B3;47@63?3D;@9
?3D;@93633:
3 ?D63@
? 6
?D63@ ?
4
?D63@ ? 7 ?D63@
?
5 ?D63@
?
(7C:3E;=3@93?43C47C;=FE
*74F3: B3CE;=7 J3@9 47C97C3= D7B3@3@9 DF?4F-
?7?;;=;9C38;==757B3E3@E7C:363BH3=EFD7B7CE;
E3?B3=B36393?43CE7CD74FE;=3B363D33E
D7=A@B3CE;=747C363B363

188. ?BAD;D;B3CE;=7B363
D33ED7=A@47C363B363
3 P? 6 P
?
4 ? 7 ?
5
?
*74F3: B3CE;=7 47C97C3= FCFD 63C; =73633@ 6;3?
67@93@ B7CD3?33@ B7C57B3E3@ 7D3C
=757B3E3@B3CE;=7D7E73:D7=A@3633:
3
?D 6
?D
4
?D 7
?D
5
?D
*74F3:B3CE;=747C97C3=D7B3@3@9DF?4F-67@93@
47D3CB7C57B3E3@
67@93@
633??D;=3
B363D33E =757B3E3@
?D=757B3E3@B3CE;=7
B363D33ED3633:
3 ?D 6 ?D
4
?D 7 ?D
5 ?D
*74F3:B3CE;=747C?F3E3@;DEC;=?F3?F347C97C3=
FCFD 63@ ?7?;;=; =757B3E3@
?D (3CE;=7
?7@933?; B7C57B3E3@ J3@9 6;@J3E3=3@ 67@93@
B7CD3?33@
P
?D67@93@3633:H3=EF
3?3@J393J3;DEC;=47=7C3#757B3E3@B3CE;=7D7E73:
93J347=7C3D73?3
D7=A@3633:
3
?D 6

189. ?D
4 ?D 7
?D
5

190. ?D
(7C:3E;=3@9C38;=47C;=FE
6
(363D33E D74F3:B3CE;=747C363B363

191. ?
63@47C97C3=67@93@=757B3E3@3H3
?D67@93@
3C3: DF?4F- BAD;E;8 (7C57B3E3@ B3CE;=7 47CF43:
E7C:363BH3=EFD7B7CE;E3?B3=B36393?43C(AD;D;
47@63B363D7=A@3633:
3
? 6

192. ?
4
? 7
?
5
?
(7C3:F?AEAC;=347C97C3=D73C3:67@93@3C3:3CFD
DF@93;?7?;;=;=73F3@?D67@93@=757B3E3@3CFD
DF@93;

193. ?D;=347C97C3=E793=FCFDE7C:363B3C3:
3CFDB7C3:F?F@9=;@:3@J3?3?BF47C97C3=67@93@
=73F3@
3

194. ?D 6 ?D
4
?D 7 ?D
5 ?D
(7C3:F?AEAC?7?7CF=3@H3=EF
?7@;EF@EF=?7
@J747C3@9; 63@3F J3@9 743C@J3
?7E7C 63@3F
E;63=47C3CFD.3=EFJ3@96;4FEF:=3@B7C3:FF@EF=
?7@J747C3@9; 63@3F ;=3 6; 63@3F E;?4F 3CFD 3;C
47C=757B3E3@

195. ?7E7C?7@;EE793=FCFDE7C:363B3C3:
B7C3:F3633:
vx
t
4
0
–2
2 6
ax (m/s2)
t (s)
3
0 3 6

196. 4 !* .$/1 ,5 ,!$/')21','#$,% ,1$. 1
*74F3:B3CE;=747C97C3=67@93@B7CD3?33@8F@9D;
=757B3E3@
67@93@633??D63@
633?D7=A@;=3

197. ?D

198. P?D63@?D
E7@EF=3@3:
3 B7C57B3E3@C3E3C3E3B3CE;=7F@EF=D73@9H3=EF
D7=A@D3?B3;D7=A@
4 B7C57B3E3@3H3B3CE;=763@
5 B7C57B3E3@B3CE;=7B363D33ED7=A@
(7CD3?33@97C3=DF3EFB3CE;=76;@J3E3=3@A7:8F@9D;
67@93@633??7E7C63@633?D7=A@
3 +7@EF=3@ =757B3E3@ C3E3C3E3 633? D73@9
H3=EFD7=A@D3?B3;67@93@

199. D7=A@
4 ;EF@9=757B3E3@D7D33EB363D33E
D7=A@

200. (7FCF?7C;3?6;E7?43==3@67@93@=757B3E3@3H3
?D63@DF6FE77G3D; 67@93@
5AD
;=3
B7C57B3E3@9C3G;E3D;
?DE7@EF=3@3:
3 =757B3E3@B7FCF?7C;3?D7E73: D7=A@
4 BAD;D;B7FCF?7C;3?D7E73: D7=A@
24 Mudah dan Aktif Belajar Fisika untuk Kelas XI
5 E;@99;?3=D;?F?63@
6 3C3=E7?43=3@?3=D;?F?
*74F3:B7FCF6;E7?43==3@63C;3E3B976F@9J3@9
E;@99;@J3
?7E7C67@93@=757B3E3@3H3 ?D
5A@6A@9 =7 3E3D 67@93@ DF6FE 77G3D; 6; ?3@3

201. 5AD
;=3
?DE7@EF=3@
3 H3=EFJ3@96;4FEF:=3@B7FCFF@EF=D3?B3;6;
E3@3:63@
4 3C3=E7C3F:J3@96;53B3;B7FCF63C;63D3C976F@9
D33E?7@53B3;E3@3:
(363B7C?3;@3@D7B3=4A3D7AC3@93@3=?7@7C;?3
63@?7@7CFD=3@4A367@93@?7@99F@3=3@=7B33
A3?7@;@993=3@=7B333@3=E7CD74FE67@93@3F
?D*F6FE77G3D;E7C:363B:AC;KA@E33633:
6;?3@3
E3@
;=34A3D3?B3;6;E3@3:B363

202. 3C3=

203. ?7E7C47C3B3E;@99;3@3=;EF
?D
3 ?7@;E 6
?7@;E
4 ?7@;E 7
?7@;E
5
?7@;E
(7C43@6;@93@3@E3C33C3=E7?43=3@6F34F3:B7FCF
J3@96;E7?43==3@63C;D74F3:D7@3B3@67@93@DF6FE
77G3D;

204. N63@ N3633:
3 3 :1 6
4 1 : 3 7

205. 5

206. *74F3:4A36;E7@63@9?7@9;=FE;97C3=B3C34A3;=3
E;@99;?3=D;?F?J3@963B3E6;53B3;3633:
?63@
?DH3=EFJ3@96;B7CF=3@A7:4A3D73?36;
F63C33633:
3

207. D7=A@ 6
D7=A@
4 D7=A@ 7
D7=A@
5 D7=A@
(7FCF6;E7?43==3@63C;E3@3:63E3C67@93@=757
B3E3@3H3
?D*F6FE77G3D;
E3@

208. 63@
B7C57B3E3@9C3G;E3D;
?D7D3C=757B3E3@B7FCF
D7E73:D7=A@63C;E7?43=3@3633:
3 ?D 6

209. ?D
4
?D 7
?D
5
?D
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?D
B363DF6FE77G3D; 5AD

210. 63@
?D
(363DF3EFD33E=757B3E3@43EF=73C3:DF?4F- 3633:

211. ?D#7E;@99;3@43EFB363D33EE7CD74FE3633:
3 ? 6
?D
4

212. ?D 7
?D
5

213. ?D
F3B7FCF6;E7?43==3@63C;D74F3:D7@3B3@3C3=
E7?43=3@3=3@D3?3;=3DF6FE77G3D;@J3
3
N63@ N 6

214. N63@

215. N
4

216. N63@
N 7

217. N63@N
5
N63@ N
*74F3:B7FCF6;E7?43==3@67@93@DF6FE77G3D; N
63@=757B3E3@3H3@J3
?D*7E73:?7@53B3;E;E;=
BF@53==757B3E3@B7FCF?7@36;
3 6 ?D
4 ?D 7
?D
5
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