{"id":1109,"date":"2015-08-02T14:35:06","date_gmt":"2015-08-02T14:35:06","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=1109"},"modified":"2024-08-22T17:47:17","modified_gmt":"2024-08-22T17:47:17","slug":"resolucao-comentada-dos-exercicios-de-composicao-de-movimentos","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/mecanica\/cinematica\/composicao-de-movimentos\/resolucao-comentada-dos-exercicios-de-composicao-de-movimentos\/","title":{"rendered":"Composi\u00e7\u00e3o de Movimentos – Resolu\u00e7\u00e3o"},"content":{"rendered":"

Resolu\u00e7\u00e3o comentada dos exerc\u00edcios de vestibulares sobre COMPOSI\u00c7\u00c3O DE MOVIMENTOS<\/b><\/span><\/span><\/span><\/p>\n

 <\/p>\n

01-<\/b><\/span><\/span>\u00a0R- A\u00a0 (veja teoria)<\/b><\/span><\/span><\/p>\n

02-<\/b><\/span><\/span>\u00a0Veja a figura abaixo:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Sen\u03b2=50\/250=0,2\u00a0 —\u00a0 \u03b2=aproximadamente 12<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R-D<\/b><\/span><\/span><\/p>\n

03-\u00a0<\/b><\/span><\/span>Aplicando Pit\u00e1goras no tri\u00e2ngulo abaixo:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

V<\/b><\/span><\/span>T<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>=(120)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0+ (90)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 V<\/b><\/span><\/span>T<\/b><\/span><\/span><\/sub>=150km\/h\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- A<\/b><\/span><\/span><\/p>\n

04-<\/b><\/span><\/span>\u00a0Ida do ninho para a \u00e1rvore (contra o vento)—-\u00a0 V<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=5m\/s\u00a0 —\u00a0 V<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=\u0394S\/\u0394t\u00a0 —\u00a0 5=75\/\u0394t\u00a0 —\u00a0 \u0394t=15s\u00a0 —\u00a0 volta da \u00e1rvore para o ninho (a favor do vento)\u00a0 —\u00a0 V<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=15m\/s\u00a0 —\u00a0 V<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>= \u0394S\/\u0394t\u00a0 —\u00a0 15=75\/\u0394t\u00a0 —\u00a0 \u0394t=5s\u00a0 —\u00a0 \u0394t<\/b><\/span><\/span>total<\/b><\/span><\/span><\/sub>=15 + 5\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>\u0394t<\/b><\/span><\/span>total<\/b><\/span><\/span><\/sub>=20s<\/b><\/span><\/span><\/p>\n

05-<\/b><\/span><\/span>\u00a0R:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

a)\u00a0\"\"\u00a0\u00a0\u00a0b)\u00a0\"\"<\/b><\/span><\/span><\/p>\n

06-<\/b><\/span><\/span>\u00a0O avi\u00e3o dever\u00e1 estar orientado na dire\u00e7\u00e3o sudeste, para que ele siga a rota norte-sul<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

 <\/p>\n

Sen\u03b2=V<\/b><\/span><\/span>v<\/b><\/span><\/span><\/sub>\/V<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=150\u221a3\/300\u00a0 —\u00a0 sen\u03b2=\u221a3\/2\u00a0 —\u00a0 \u03b2=60<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- D<\/b><\/span><\/span><\/p>\n

07-<\/b><\/span><\/span>\u00a0V<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>\u00a0\u2013 velocidade da caminhonete\u00a0 —\u00a0 velocidade do carro patrulheiro \u2013 V<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>=60km\/h\u00a0 —\u00a0 o radar do carro patrulha indica a velocidade relativa \u2013 V<\/b><\/span><\/span>r<\/b><\/span><\/span><\/sub>=30km\/h\u00a0 —\u00a0 V<\/b><\/span><\/span>r<\/b><\/span><\/span><\/sub>=V<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>\u00a0\u2013 V<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 30=V<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>\u00a0\u2013 60\u00a0 —\u00a0 V<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>=90km\/h\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- E<\/b><\/span><\/span><\/p>\n

08-<\/b><\/span><\/span>\u00a0a) V=8 + 5\u00a0 —\u00a0 V=13km\/h<\/b><\/span><\/span><\/p>\n

b) V=8 \u2013 5\u00a0 —\u00a0 V=3km\/h<\/b><\/span><\/span><\/p>\n

09-<\/b><\/span><\/span>\u00a0V<\/b><\/span><\/span>b<\/b><\/span><\/span><\/sub>= \u0394S\/\u0394t\u00a0 —\u00a0 8= 3,2\u0394t\u00a0 —\u00a0 \u0394t=0,4h\u00a0 —\u00a0 nesse tempo, devido \u00e0 correnteza ele se deslocou\u00a0 —\u00a0 V<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>= \u0394S\/\u0394t\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

5= \u0394S\/0,4\u00a0 —\u00a0 \u0394S=2km\u00a0 —\u00a0 chegar\u00e1 no<\/b><\/span><\/span>\u00a0ponto C<\/b><\/span><\/span>.<\/b><\/span><\/span><\/p>\n

10-\u00a0<\/b><\/span><\/span>V<\/b><\/span><\/span>subida<\/b><\/span><\/span><\/sub>= 8 \u2013 V\u00a0 —\u00a0 t<\/b><\/span><\/span>subida<\/b><\/span><\/span><\/sub>=t<\/b><\/span><\/span>s\u00a0\u00a0<\/b><\/span><\/span><\/sub>—\u00a0 V<\/b><\/span><\/span>subida<\/b><\/span><\/span><\/sub>=d\/t<\/b><\/span><\/span>s<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 8 \u2013 V=d\/t<\/b><\/span><\/span>s<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>t<\/b><\/span><\/span>s<\/b><\/span><\/span><\/sub>=d\/(8 \u2013 V)<\/b><\/span><\/span>\u00a0 —\u00a0 V<\/b><\/span><\/span>descida<\/b><\/span><\/span><\/sub>=2 + V\u00a0 —\u00a0 V<\/b><\/span><\/span>descida<\/b><\/span><\/span><\/sub>=d\/t<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 2 + V=d\/t<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0t<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>=d\/(2 + V)<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>\u00a0—\u00a0\u00a0<\/b><\/span><\/span>t<\/b><\/span><\/span>s<\/b><\/span><\/span><\/sub>\u00a0+ t<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>=10min\u00a0 —\u00a0 t<\/b><\/span><\/span>s<\/b><\/span><\/span><\/sub>\u00a0+ t<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>=600\u00a0 —\u00a0 d\/(8 \u2013 V) + d\/(2 + V)=600\u00a0 —\u00a0 d(2 + V) + d(8 \u2013 V)=600.(8 \u2013 V).(2 + V)\u00a0 —\u00a0 2d + Vd + 8d \u2013 Vd = 600.(16 + 8V \u2013 2V\u00a0 – V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>)\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0d=960 + 360V \u2013 60V<\/b><\/span><\/span>2\u00a0\u00a0<\/b><\/span><\/span><\/sup>I\u00a0\u00a0<\/b><\/span><\/span>—\u00a0 esta \u00e9 uma equa\u00e7\u00e3o do segundo grau cujo gr\u00e1fico \u00e9 uma par\u00e1bola e da qual se quer determinar o valor m\u00e1ximo para d, que ocorre no v\u00e9rtice da par\u00e1bola, de\u00a0 valor V<\/b><\/span><\/span>m\u00e1ximo<\/b><\/span><\/span><\/sub>=-B\/A, onde A=-60 e B=360 (veja I)\u00a0 —\u00a0 V<\/b><\/span><\/span>m\u00e1ximo<\/b><\/span><\/span><\/sub>=-B\/A=-360\/-60\u00a0 —\u00a0 V<\/b><\/span><\/span>m\u00e1ximo<\/b><\/span><\/span><\/sub>=3, que substitu\u00eddo em I, nos fornece a dist\u00e2ncia m\u00e1xima percorrida\u00a0 —\u00a0 d<\/b><\/span><\/span>m\u00e1ximo<\/b><\/span><\/span><\/sub>=960 +360.3 \u2013 60.3<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 d<\/b><\/span><\/span>m\u00e1ximo<\/b><\/span><\/span><\/sub>=1500m<\/b><\/span><\/span>\u00a0 —\u00a0 R- B<\/b><\/span><\/span><\/p>\n

11-<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Sen30<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=V<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>\/V<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 1\/2 =60\/V<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 V<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=120km\/h\u00a0 —<\/b><\/span><\/span>\u00a0 R- C<\/b><\/span><\/span><\/p>\n

12-\u00a0<\/b><\/span><\/span>d \u2013 comprimento da escada rolante\u00a0 —\u00a0 parado na escada\u00a0 —\u00a0 V<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>=d\/t=d\/10\u00a0 —\u00a0 V<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>=d\/10\u00a0 —\u00a0 subindo a escada\u00a0 —\u00a0 V<\/b><\/span><\/span>h<\/b><\/span><\/span><\/sub>\u00a0\u2013 V<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>=d\/15\u00a0 —\u00a0 V<\/b><\/span><\/span>h<\/b><\/span><\/span><\/sub>\u00a0\u2013 d\/10=d\/15\u00a0 —\u00a0 V<\/b><\/span><\/span>h<\/b><\/span><\/span><\/sub>=d\/15 + d\/10\u00a0 —\u00a0 V<\/b><\/span><\/span>h<\/b><\/span><\/span><\/sub>=d\/6\u00a0 —\u00a0 descendo a escada\u00a0 —\u00a0 V<\/b><\/span><\/span>h<\/b><\/span><\/span><\/sub>\u00a0+ V<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>=d\/t\u00a0 —\u00a0 d\/6 + d\/10=d\/t\u00a0 —\u00a0 10dt + 6dt=60d\u00a0 —\u00a0 t=60\/16\u00a0 —\u00a0 t=3,75s\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- B<\/b><\/span><\/span><\/p>\n

13-<\/b><\/span><\/span>\u00a0Observe as figuras abaixo:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

a) No tri\u00e2ngulo ABC\u00a0 —\u00a0 sen\u03b8=300\/500\u00a0 —\u00a0 sen\u03b8=0,6\u00a0 —\u00a0 cos\u03b8=400\/500\u00a0 —\u00a0 cos\u03b8=0,8\u00a0 —\u00a0 na figura da direita\u00a0 —\u00a0 sen\u03b8=V<\/b><\/span><\/span>arr<\/b><\/span><\/span><\/sub>\/4,5\u00a0 —\u00a0 0,6 =V<\/b><\/span><\/span>arr<\/b><\/span><\/span><\/sub>\/4,5\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>V<\/b><\/span><\/span>arr<\/b><\/span><\/span><\/sub>=2,7m\/s<\/b><\/span><\/span><\/p>\n

b) cos\u03b8=V<\/b><\/span><\/span>res<\/b><\/span><\/span><\/sub>\/4,5\u00a0 — 0,8=V<\/b><\/span><\/span>res<\/b><\/span><\/span><\/sub>\/4,5\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>V<\/b><\/span><\/span>res<\/b><\/span><\/span><\/sub>=3,6m\/s<\/b><\/span><\/span><\/p>\n

14-<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>R- B<\/b><\/span><\/span>\u00a0 (veja teoria)<\/b><\/span><\/span><\/p>\n

15-<\/b><\/span><\/span>\u00a0a) veja figura abaixo<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

 <\/p>\n

(V<\/b><\/span><\/span><\/span>p-s<\/b><\/span><\/span><\/span><\/sub>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup>\u00a0= (V<\/b><\/span><\/span><\/span>p-est<\/b><\/span><\/span><\/span><\/sub>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup>\u00a0<\/b><\/span><\/span><\/span><\/sub>+ (V<\/b><\/span><\/span><\/span>est-s<\/b><\/span><\/span><\/span><\/sub>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup>\u00a0 —\u00a0 \u00a0(V<\/b><\/span><\/span><\/span>p-s<\/b><\/span><\/span><\/span><\/sub>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup>=(1,5)<\/b><\/span><\/span><\/span>2\u00a0<\/b><\/span><\/span><\/span><\/sup>\u00a0+ (2,0)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>p-s<\/b><\/span><\/span><\/span><\/sub>=2,5m\/s<\/b><\/span><\/span><\/span><\/p>\n

b) o tempo de travessia depende apenas da velocidade perpendicular \u00e0 esteira (1,5m\/s) e da largura da mesma (3m) \u00a0—\u00a0<\/b><\/span><\/span><\/p>\n

V=\u0394S\/\u0394t\u00a0 —\u00a0 1,5=3\/\u0394t\u00a0 —\u00a0 \u0394t=2s\u00a0 —\u00a0 substituindo esse tempo em V<\/b><\/span><\/span>p-s<\/b><\/span><\/span><\/sub>= \u0394S\/\u0394t\u00a0 —\u00a0 2,5= \u0394S\/2\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0\u0394S=5,0m<\/b><\/span><\/span><\/p>\n

16-<\/b><\/span><\/span>\u00a0Se voc\u00ea estiver no barco I voc\u00ea ver\u00e1 o barco II se aproximar de voc\u00ea e, se voc\u00ea estiver no barco II voc\u00ea ver\u00e1 o barco I se aproximar de voc\u00ea\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/p>\n

17-\u00a0<\/b><\/span><\/span>Como a dist\u00e2ncia entre as duas b\u00f3ias n\u00e3o varia, elas est\u00e3o paradas uma em rela\u00e7\u00e3o a outra. Ent\u00e3o, o menino deve nadar diretamente de uma para outra\u00a0 —<\/b><\/span><\/span>\u00a0 R- A<\/b><\/span><\/span><\/p>\n

18-<\/b><\/span><\/span>\u00a0Como o observador est\u00e1 em repouso no navio, a dist\u00e2ncia entre ele e o navio n\u00e3o varia e como o p\u00e1ssaro voa na dire\u00e7\u00e3o leste-oeste em rela\u00e7\u00e3o ao navio e consequentemente \u00e0 pessoa, esta o ver\u00e1 voando na dire\u00e7\u00e3o leste-oeste com velocidade de 20m\/s\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- C<\/b><\/span><\/span><\/p>\n

19-<\/b><\/span><\/span>\u00a0Velocidade do barco em rela\u00e7\u00e3o \u00e0s margens\u00a0 —\u00a0 V<\/b><\/span><\/span>b-m<\/b><\/span><\/span><\/sub>=14 + 4=18km\/h=18\/3,6=5m\/s\u00a0 —\u00a0 para atravessar totalmente a ponte o barco percorre \u0394S=D + L=25 + 80=105m\u00a0 —\u00a0 V<\/b><\/span><\/span>b-m<\/b><\/span><\/span><\/sub>=\u0394S\/\u0394t\u00a0 —\u00a0 5=105\/\u0394t\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0\u0394t=21s<\/b><\/span><\/span><\/p>\n

20-<\/b><\/span><\/span>\u00a0Para calcular o tempo que o barco demora para atravessar o rio usa-se apenas a velocidade do barco em rela\u00e7\u00e3o \u00e0 \u00e1gua V<\/b><\/span><\/span>b<\/b><\/span><\/span><\/sub>=3,0km\/h e o comprimento do rio (\u0394S=1,0km)\u00a0 —\u00a0 V<\/b><\/span><\/span>b<\/b><\/span><\/span><\/sub>=\u0394S\/\u0394t\u00a0 —\u00a0 3=1\/\u0394t\u00a0 —\u00a0 \u0394t=1\/3h\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>\u0394t=20min<\/b><\/span><\/span>\u00a0 —\u00a0 R- C<\/b><\/span><\/span><\/p>\n

21-<\/b><\/span><\/span>\u00a0A favor do vento\u00a0 —\u00a0 V<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>\u00a0+ V<\/b><\/span><\/span>v<\/b><\/span><\/span><\/sub>=180 I\u00a0 —\u00a0 contra o vento\u00a0 —\u00a0 V<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>\u00a0\u2013 V<\/b><\/span><\/span>v<\/b><\/span><\/span><\/sub>=150 II\u00a0 —\u00a0 resolvendo o sistema composto por I e II\u00a0 —\u00a0 V<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=165km\/h e V<\/b><\/span><\/span>v<\/b><\/span><\/span><\/sub>=15km\/h\u00a0 —<\/b><\/span><\/span>\u00a0 R- A<\/b><\/span><\/span><\/p>\n

22-<\/b><\/span><\/span>\u00a0V<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=240 + 260\u00a0 —\u00a0 V<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=500m\/s\u00a0 —\u00a0 lembre-se que a velocidade do som independe da velocidade da fonte\u00a0 —\u00a0 V<\/b><\/span><\/span>s<\/b><\/span><\/span><\/sub>=300m\/s\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/p>\n

R- D<\/b><\/span><\/span><\/p>\n

23-<\/b><\/span><\/span>\u00a0A velocidade do ponto superior do cilindro que est\u00e1 em contato com a prancha vale 2V<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>e \u00e9 igual \u00e0 velocidade da prancha V<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>, ou seja, V<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>=2V<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/p>\n

\"\"<\/p>\n

Portanto\u00a0 —\u00a0 V<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>\/V<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>=2\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- A<\/b><\/span><\/span><\/p>\n

24-<\/b><\/span><\/span>\u00a0O tempo de travessia depende apenas da largura do rio (L) e da velocidade do barco em rela\u00e7\u00e3o \u00e0s margens (u)\u00a0 —\u00a0 u=L\/t\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>t=L\/u\u00a0 —\u00a0<\/b><\/span><\/span>quanto maior a velocidade das \u00e1guas, maior ser\u00e1 o deslocamento do barco para a direita\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- B<\/b><\/span><\/span><\/p>\n

25-<\/b><\/span><\/span>\u00a0Subindo o rio\u00a0 —\u00a0 V=\u0394S\/\u0394t\u00a0 —\u00a0 V<\/b><\/span><\/span>b<\/b><\/span><\/span><\/sub>\u00a0\u2013 V<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=d\/10 (I)\u00a0 —\u00a0 descendo o rio\u00a0 —\u00a0 V= \u0394S\/\u0394t\u00a0 —\u00a0 V<\/b><\/span><\/span>b<\/b><\/span><\/span><\/sub>\u00a0+ V<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=d\/4 (II)\u00a0 —\u00a0 fazendo (II) \u2013 (I)\u00a0 —\u00a0 (V<\/b><\/span><\/span>b<\/b><\/span><\/span><\/sub>\u00a0+ V<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>) \u2013 (V<\/b><\/span><\/span>b<\/b><\/span><\/span><\/sub>\u00a0\u2013 V<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>)=4\/4 \u2013 d\/10\u00a0 —\u00a0 V<\/b><\/span><\/span>b<\/b><\/span><\/span><\/sub>\u00a0+ V<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>\u00a0\u2013 V<\/b><\/span><\/span>b<\/b><\/span><\/span><\/sub>\u00a0+ V<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>\u00a0= (5d – 2d)\/20\u00a0 —\u00a0 2V<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=3d\/20 (III)\u00a0 —\u00a0 descendo o rio com o motor desligado o barco percorre\u00a0 dist\u00e2ncia d com velocidade que \u00e9 a mesma que a da \u00e1gua V<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

V<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=d\/t\u00a0 —\u00a0 d=V<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>t (IV)\u00a0 —\u00a0 substituindo (IV) em (III)\u00a0 —\u00a0 2V<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=3.V<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>t\/20 \u00a0—\u00a0 t=40\/3=13h + 1\/3h\u00a0 —\u00a0 t=13h e 20min\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- B<\/b><\/span><\/span><\/p>\n

26-\u00a0<\/b><\/span><\/span>Dados: v<\/b><\/span><\/span>carro<\/b><\/span><\/span><\/sub>\u00a0= 80 km\/h; sen\u00a0q\u00a0= 0,8 e cos\u00a0q\u00a0= 0,6\u00a0 —\u00a0 a figura abaixo mostra o autom\u00f3vel e as velocidades do autom\u00f3vel<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

(\"\") e da chuva (\"\"), para a pessoa parada na beira da estrada. O diagrama vetorial mostra a composi\u00e7\u00e3o dessas velocidades para o estudante\u00a0 —\u00a0 tg\u00a0q\u00a0=V<\/b><\/span><\/span>carro<\/b><\/span><\/span><\/sub>\/V\u00a0 —\u00a0 senq\/cosq=V<\/b><\/span><\/span>carro<\/b><\/span><\/span><\/sub>\/V\u00a0 —\u00a0 0,8\/0.6=80\/V\u00a0 —\u00a0 V=60km\/h\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- B<\/b><\/span><\/span><\/p>\n

 <\/p>\n

27-\u00a0<\/b><\/span><\/span>Dados: v<\/b><\/span><\/span>b<\/b><\/span><\/span><\/sub>\u00a0= 11 km\/h; v<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>\u00a0= 0,83 m\/s = (0,83\u00a0\u00b4\u00a03,6) = 3 km\/h\u00a0 —\u00a0 na descida\u00a0 —\u00a0 \u00a0v = v<\/b><\/span><\/span>b<\/b><\/span><\/span><\/sub>\u00a0+ v<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>= 11 + 3 = 14 km\/ h\u00a0 —\u00a0 na subida\u00a0<\/b><\/span><\/span><\/p>\n

—\u00a0 v = v<\/b><\/span><\/span>b<\/b><\/span><\/span><\/sub>\u00a0\u2013 v<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>\u00a0= 11 \u2013 3 = 8 km\/ h\u00a0 —\u00a0 R- A<\/b><\/span><\/span><\/p>\n

28-\u00a0<\/b><\/span><\/span>Sejam v<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>\u00a0a velocidade da correnteza de v<\/b><\/span><\/span>b<\/b><\/span><\/span><\/sub>\u00a0a velocidade pr\u00f3pria do barco\u00a0 —\u00a0 na descida\u00a0 —\u00a0 v<\/b><\/span><\/span>b<\/b><\/span><\/span><\/sub>\u00a0+ v<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>\u00a0= 20\u00a0 (I)\u00a0 —\u00a0 na subida\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

V<\/b><\/span><\/span>b<\/b><\/span><\/span><\/sub>\u00a0\u2013 v<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>\u00a0= 12\u00a0 (II)\u00a0 —\u00a0 somando as duas express\u00f5es\u00a0 —\u00a0 (I) + (II)\u00a0\u00de\u00a0(v<\/b><\/span><\/span>b<\/b><\/span><\/span><\/sub>\u00a0+ v<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>) + (v<\/b><\/span><\/span>b<\/b><\/span><\/span><\/sub>\u00a0\u2013 v<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>) = 32\u00a0 —\u00a0 2 v<\/b><\/span><\/span>b<\/b><\/span><\/span><\/sub>\u00a0= 32\u00a0 —\u00a0 v<\/b><\/span><\/span>b<\/b><\/span><\/span><\/sub>\u00a0= 16 km\/h\u00a0 —<\/b><\/span><\/span><\/p>\n

Substituindo em (I)\u00a0 —\u00a0 16 + v<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>\u00a0= 20\u00a0 —\u00a0 v<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>\u00a0= 4 km\/h \u00a0—\u00a0<\/b><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/p>\n

29-\u00a0<\/b><\/span><\/span>Como o transatl\u00e2ntico se move em linha reta com velocidade constante ele est\u00e1 em equil\u00edbrio din\u00e2mico e comporta-se como se estivesse em repouso (equil\u00edbrio est\u00e1tico) , n\u00e3o afetando o movimento da bola\u00a0 —<\/b><\/span><\/span>\u00a0 R- D<\/b><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

30-\u00a0<\/b><\/span><\/span>Observe a figura abaixo\u00a0 —\u00a0\u00a0aplicando o Teorema de Pit\u00e1goras\u00a0 —\u00a0 130<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0= 50<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0+ x<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0\u00a0\u00a0—\u00a0 x =120 m\u00a0 —\u00a0 da express\u00e3o<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

fornecida\u00a0 —\u00a0V<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>\u00a0= V<\/b><\/span><\/span>r<\/b><\/span><\/span><\/sub>\u00a0<\/b><\/span><\/span>\u22c5<\/b><\/span><\/span>\u00a0cos(\u03b1)\u00a0 —\u00a0 72=V<\/b><\/span><\/span>r<\/b><\/span><\/span><\/sub>.120\/130\u00a0 —\u00a0 V<\/b><\/span><\/span>r<\/b><\/span><\/span><\/sub>=78km\/h\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/p>\n

 <\/p>\n

31-<\/b><\/span><\/span><\/p>\n

O tempo m\u00ednimo de travessia s\u00f3 ocorre quando o bote atravessa o rio com sua velocidade (V\u00ad<\/b><\/span><\/span>b<\/b><\/span><\/span><\/sub>=10m\/s) sempre perpendicular \u00e0s margens e, consequentemente perpendicular \u00e0 velocidade da \u00e1gua (correnteza) (veja figura)\u00a0 —\u00a0 observe que se voc\u00ea utilizar apenas a<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

componente V<\/b><\/span><\/span>b<\/b><\/span><\/span><\/sub>\u00a0da velocidade, a dist\u00e2ncia percorrida ser\u00e1 a largura do rio, que \u00e9 de 140m\u00a0 —\u00a0 V<\/b><\/span><\/span>b<\/b><\/span><\/span><\/sub>=\u2206S\/\u2206t\u00a0 —\u00a0 10=140\/\u2206t\u00a0 —\u00a0 \u2206t=14s\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- D<\/b><\/span><\/span>. Observa\u00e7\u00e3o: Esse tempo n\u00e3o depende da velocidade da correnteza e, dependendo dela ele apenas chegar\u00e1 \u00e0 outra margem mais pr\u00f3ximo ou mais afastado do ponto de partida, mas o tempo de travessia ser\u00e1 o mesmo.<\/b><\/span><\/span><\/p>\n

32-<\/b><\/span><\/span><\/p>\n

Trata-se de deslocamentos vetoriais cuja representa\u00e7\u00e3o esquem\u00e1tica est\u00e1 na figura\u00a0\u00a0 —\u00a0 observe que o tri\u00e2ngulo \u00e9<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

ret\u00e2ngulo\u00a0 —\u00a0 Pit\u00e1goras —\u00a0 d2=120<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0+ 160<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 d=200km\u00a0 —\u00a0 V<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=d\/\u2206t=200\/(1\/4)\u00a0 —\u00a0 V<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=800km\/h\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- E<\/b><\/span><\/span><\/p>\n

 <\/p>\n

 <\/p>\n

33-<\/b><\/span><\/span><\/p>\n

-A velocidade vetorial (\"\") do barco em rela\u00e7\u00e3o \u00e0 margem (velocidade com que uma pessoa parada na margem do rio veria o barco passar por ela), quando ele sobe o rio \u00e9 fornecida por\u00a0\"\"\u00a0\u00a0—\u00a0 observe na soma vetorial abaixo que, em m\u00f3dulo (intensidade)<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

V=Vb\u00a0\u2013 Vc\u00a0sendo, Vb\u00a0\u2013 m\u00f3dulo da velocidade do barco, Vc\u00a0\u2013 m\u00f3dulo da velocidade da correnteza e V \u2013 m\u00f3dulo da velocidade do barco em rela\u00e7\u00e3o \u00e0 margem\u00a0 —\u00a0 assim, no exerc\u00edcio V= 13 \u2013 5\u00a0 —\u00a0 V=8m\/s\u00a0 —\u00a0 sendo a dist\u00e2ncia entre as duas cidades \u2206S=40km=40.103m\u00a0 —\u00a0 V=\u2206S\/\u2206t\u00a0 —\u00a0 8=40.103\/\u2206t\u00a0 —\u00a0 \u2206t=40.103\/8=5.103\u00a0s\u00a0 —\u00a0 \u2206t = 1h 23min 20s\u00a0 —\u00a0\u00a0R- B<\/b><\/span><\/span><\/p>\n

 <\/p>\n

Voltar para os Exerc\u00edcios<\/span><\/a><\/h3>\n","protected":false},"excerpt":{"rendered":"

Resolu\u00e7\u00e3o comentada dos exerc\u00edcios de vestibulares sobre COMPOSI\u00c7\u00c3O DE MOVIMENTOS   01-\u00a0R- A\u00a0 (veja teoria) 02-\u00a0Veja a figura abaixo: Sen\u03b2=50\/250=0,2\u00a0 —\u00a0 \u03b2=aproximadamente 12o\u00a0 —\u00a0\u00a0R-D 03-\u00a0Aplicando Pit\u00e1goras no tri\u00e2ngulo abaixo: VT2=(120)2\u00a0+ (90)2\u00a0 —\u00a0 VT=150km\/h\u00a0 —\u00a0\u00a0R- A 04-\u00a0Ida do ninho para a \u00e1rvore (contra o vento)—-\u00a0 VR=5m\/s\u00a0 —\u00a0 VR=\u0394S\/\u0394t\u00a0 —\u00a0 5=75\/\u0394t\u00a0 —\u00a0 \u0394t=15s\u00a0 —\u00a0 volta da \u00e1rvore para o ninho (a favor<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":1105,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"om_disable_all_campaigns":false,"_monsterinsights_skip_tracking":false,"_monsterinsights_sitenote_active":false,"_monsterinsights_sitenote_note":"","_monsterinsights_sitenote_category":0,"footnotes":""},"class_list":["post-1109","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1109","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/comments?post=1109"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1109\/revisions"}],"predecessor-version":[{"id":10788,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1109\/revisions\/10788"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1105"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=1109"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}