02 bab1

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

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

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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

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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;=

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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

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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

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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

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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

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197. ?D

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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

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202. 3C3=

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

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219. 7@6363@6;3EF:=3@63C;=7E;@99;3@J3@9D3?3 D753C347CD3?33@=3@E7E3B;47@636;3EF:=3@ D753C34743D63@47@636;3EF:=3@D753C3:AC;KA@E3 67@93@ =757B3E3@ 3H3 .3=EF J3@9 6;4FEF:=3@ =76F347@63F@EF=?7@53B3;E3@3:3633:97D7=3@ 67@93@F63C36;343;=3@ 3 6 4 7 5

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