{"id":3144,"date":"2017-08-19T02:52:41","date_gmt":"2017-08-19T02:52:41","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=3144"},"modified":"2024-09-04T14:25:05","modified_gmt":"2024-09-04T14:25:05","slug":"resolucoes-cinematica","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/vestibulares-recentes\/mecanica\/cinematica\/resolucoes-cinematica\/","title":{"rendered":"Resolu\u00e7\u00f5es – Cinem\u00e1tica"},"content":{"rendered":"

RESOLU\u00c7\u00d5ES<\/b><\/span><\/span><\/span><\/p>\n

CINEM\u00c1TICA<\/b><\/span><\/span><\/span><\/p>\n

(2015, 2016 e 2017)<\/b><\/span><\/span><\/span><\/p>\n

MOVIMENTO UNIFORME<\/b><\/span><\/span><\/span><\/p>\n

01<\/b><\/span><\/span>–\u00a0<\/b><\/span><\/span>Esse exerc\u00edcio \u00e9 de\u00a0<\/b><\/span><\/span>apenas velocidade m\u00e9dia<\/b><\/span><\/span>, s\u00f3 precisamos\u00a0<\/b><\/span><\/span>tomar cuidado com as transforma\u00e7\u00f5es<\/b><\/span><\/span>:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

02-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Considere:\u00a0<\/b><\/span><\/span>1 n\u00f3 = 1 milha mar\u00edtima\/hora<\/b><\/span><\/span>.<\/b><\/span><\/span><\/p>\n

Como um n\u00f3 equivale \u00e0 1 milha mar\u00edtima\/hora, a embarca\u00e7\u00e3o com velocidade de\u00a0<\/b><\/span><\/span>10 n\u00f3s, se movimenta em uma hora o equivalente \u00e0 10 milhas<\/b><\/span><\/span>. O mesmo racioc\u00ednio para a de\u00a0<\/b><\/span><\/span>30 n\u00f3s<\/b><\/span><\/span>, s\u00f3 que nesse caso\u00a0<\/b><\/span><\/span>s\u00e3o 30 milhas.<\/b><\/span><\/span><\/p>\n

Bom, se analisarmos que ambas est\u00e3o em\u00a0<\/b><\/span><\/span>dire\u00e7\u00f5es opostas<\/b><\/span><\/span>, a sua<\/b><\/span><\/span>velocidade relativa deve ser somada<\/b><\/span><\/span>, j\u00e1 que uma se move\u00a0<\/b><\/span><\/span>10 milhas\u00a0<\/b><\/span><\/span>e a outra\u00a0<\/b><\/span><\/span>30 para o mesmo ponto<\/b><\/span><\/span>, o de encontro.<\/b><\/span><\/span><\/p>\n

Sendo assim:<\/b><\/span><\/span><\/p>\n

Velocidade relativa das embarca\u00e7\u00f5es = 10 + 30 = 40 milhas\/hora<\/b><\/span><\/span><\/p>\n

Jogando na f\u00f3rmula da velocidade:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

Isolando o tempo:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

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

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

\"\"<\/p>\n

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

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

\"\"<\/p>\n

\"\"S<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span>\u00a0+\u00a0\"\"S<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span>\u00a0= 150 + 180\u00a0\"\"\u00a0\"\"<\/b><\/span><\/span><\/span>S<\/b><\/span><\/span><\/span>t<\/b><\/span><\/span>\u00a0= 330 m<\/b><\/span><\/span><\/span><\/p>\n

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

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

Vamos\u00a0<\/b><\/span><\/span>chamar<\/b><\/span><\/span>\u00a0de\u00a0<\/b><\/span><\/span>d\u00a0<\/b><\/span><\/span>a\u00a0<\/b><\/span><\/span>primeira metade do trajeto,<\/b><\/span><\/span>\u00a0tal que:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"\u00a0<\/span>Quando percorrida com V1\u00a0= 25 m\/s\u00a0<\/b><\/span><\/span>demora um\u00a0<\/b><\/span><\/span>intervalo de tempo \u2206t1<\/b><\/span><\/span>\u00a0<\/b><\/span>tal que\u00a0<\/b><\/span><\/span>V1\u00a0= d\/\u2206t1\u00a0\"\"\u00a0<\/b><\/span><\/span>25 = d\/\u2206t1\u00a0\"\"\u00a0<\/b><\/span><\/span>\u2206t1\u00a0=\u00a0\"\"\u00a0(I)<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"\u00a0<\/span>Quando percorrida com V2\u00a0= 15 m\/s\u00a0<\/b><\/span><\/span>demora um\u00a0<\/b><\/span><\/span>intervalo de tempo \u2206t2\u00a0= \u2206t1\u00a0+ 7200\u00a0<\/b><\/span><\/span>tal que\u00a0<\/b><\/span><\/span>V2\u00a0= d\/\u2206t2\u00a0\"\"\u00a015 = d\/(\u2206t1\u00a0+ 7200) (II).<\/b><\/span><\/span><\/span><\/span><\/p>\n

(I) em (II)\u00a0\"\"\u00a0<\/b><\/span><\/span>15 = d\/(\"\"\u00a0+ 7200)\u00a0\"\"\u00a015.\"\"\u00a0+ 15.7200 = d\u00a0\"\"\u00a00,6d \u2013 d = \u2013 108000\u00a0\"\"\u00a00,4d = 108000\u00a0\"\"\u00a0<\/b><\/span><\/span>d = 270 000 m = 270 km (III).<\/b><\/span><\/span><\/span><\/span><\/p>\n

(III) em (I)\u00a0\"\"\u00a0\u2206t1<\/b><\/span><\/span>\u00a0<\/b><\/span>=\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span>= 10800 s<\/b><\/span><\/span>\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span>\u2206t1\u00a0=\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span>= 3\u00a0<\/b><\/span><\/span>h\u00a0\"\"\u00a0<\/b><\/span><\/span>\u2206t2\u00a0= 3 + 2\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span>\u2206t2\u00a0= 5 h (tempo gasto na primeira metade do trajeto com velocidade de 15 m\/s).<\/b><\/span><\/span><\/span><\/span><\/p>\n

Como a<\/b><\/span><\/span>\u00a0primeira metade\u00a0<\/b><\/span><\/span>do trajeto\u00a0<\/b><\/span><\/span>tem 270 km, a\u00a0<\/b><\/span><\/span>segunda metade ter\u00e1<\/b><\/span><\/span>tamb\u00e9m 270 km e a\u00a0<\/b><\/span><\/span>dist\u00e2ncia<\/b><\/span><\/span><\/span><\/span><\/p>\n

entre as duas cidades<\/b><\/span><\/span>\u00a0ser\u00e1 de\u00a0<\/b><\/span><\/span>D\u00a0<\/b><\/span><\/span>= 270\u00d72\u00a0<\/b><\/span><\/span>= 540 km<\/b><\/span><\/span>.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Como, pelo\u00a0<\/b><\/span><\/span>enunciado,<\/b><\/span><\/span>\u00a0a\u00a0<\/b><\/span><\/span>segunda metade V2<\/b><\/span><\/span>\u00a0<\/b><\/span>desse trajeto \u00e9 sempre percorrida com\u00a0<\/b><\/span><\/span>velocidade constante e igual \u00e0 m\u00e9dia aritm\u00e9tica das duas velocidades anteriore<\/b><\/span><\/span>s\u00a0\"\"\u00a0<\/b><\/span><\/span>V2\u00a0=<\/b><\/span><\/span>\u00a0\"\"\u00a0= 20 m\/sx3,6\u00a0<\/b><\/span><\/span>= 72 km\/h.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Tempo \u2206t3<\/b><\/span><\/span>\u00a0<\/b><\/span>decorrido para percorrer a\u00a0<\/b><\/span><\/span>segunda metade do percurso de d = 270 km com V2\u00a0= 72 km\/h<\/b><\/span><\/span><\/span><\/span><\/p>\n

72 = 270\/\u2206t3\u00a0\"\"\u00a0\u2206t3\u00a0=\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span>= 3,75 h<\/b><\/span><\/span><\/span><\/span><\/p>\n

A\u00a0<\/b><\/span><\/span>velocidade m\u00e9dia (Vm) ao longo de todo o trajeto<\/b><\/span><\/span>\u00a0<\/b><\/span>\u00e9 fornecida por\u00a0\"\"\u00a0Vm=\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span>=<\/b><\/span><\/span>\u00a0\"\"\u00a0=\u00a0\"\"\u00a0\"\"\u00a0<\/b><\/span><\/span>Vm\u00a0= 80 km\/h<\/b><\/span><\/span><\/span><\/span><\/p>\n

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

 <\/p>\n

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

Ultrapassagem<\/b><\/span><\/span><\/span><\/span><\/p>\n

Trem de comprimento (x = 150 m)\u00a0<\/b><\/span><\/span>atravessando totalmente um\u00a0<\/b><\/span><\/span>t\u00fanel de comprimento (c=<\/b><\/span><\/span>?)\u00a0\"\"\u00a0observe que para\u00a0<\/b><\/span><\/span>atravessar completamente o t\u00fanel<\/b><\/span><\/span>, um\u00a0<\/b><\/span><\/span>ponto P fixo em qualquer parte do trem<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

deve\u00a0<\/b><\/span><\/span>percorrer uma dist\u00e2ncia \u0394S = 150 + c,<\/b><\/span><\/span>\u00a0com\u00a0<\/b><\/span><\/span>velocidade V num intervalo de tempo \u0394t, tal que\u00a0\"\"<\/b><\/span><\/span><\/span><\/span><\/p>\n

V =\u00a0\"\"\u00a0\"\"<\/b><\/span><\/span>\u00a016 =\u00a0\"\"\u00a0\"\"\u00a0150 + c = 800\u00a0\"\"\u00a0c = 800 \u2013 150\u00a0\"\"\u00a0<\/b><\/span><\/span>c = 650 m<\/b><\/span><\/span><\/span><\/span><\/p>\n

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

07-<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>VP\u00a0=\u00a0\"\"\u00a0\"\"<\/b><\/span><\/span>\u00a08 =\u00a0\"\"\u00a0\"\"\u00a0t\u2019 =\u00a0\"\"\u00a0\"\"\u00a0<\/b><\/span><\/span>t\u2019 = 150 s<\/b><\/span><\/span><\/span><\/span><\/p>\n

VS\u00a0=\u00a0\"\"<\/b><\/span><\/span>\u00a0\"\"\u00a05 =\u00a0\"\"\u00a0\"\"\u00a0t\u2019\u2019 =\u00a0\"\"\u00a0\"\"\u00a0<\/b><\/span><\/span>t\u2019\u2019 = 240 s<\/b><\/span><\/span><\/span><\/span><\/p>\n

\u2206<\/span>t =\u00a0<\/b><\/span><\/span>240 \u2013 150\u00a0\"\"\u00a0<\/b><\/span><\/span>\u2206t = 90 s<\/b><\/span><\/span><\/span><\/span><\/p>\n

08-<\/b><\/span><\/span><\/span>Intervalo de tempo (\u2206t\u2019)\u00a0<\/b><\/span><\/span>que demorou para percorrer a\u00a0<\/b><\/span><\/span>dist\u00e2ncia entre 0 km e 5 km com V\u2019 = 60 km\/h\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span>V\u2019 =\u00a0\"\"<\/b><\/span><\/span>\u00a0\"\"\u00a060 =\u00a0\"\"\u00a0\"\"\u00a0\u2206t\u2019 =\u00a0\"\"\u00a0\"\"\u00a0<\/b><\/span><\/span>\u2206t\u2019 =\"\"\u00a0h.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Intervalo de tempo (\u2206t\u2019\u2019)\u00a0<\/b><\/span><\/span>que demorou para percorrer a\u00a0<\/b><\/span><\/span>dist\u00e2ncia entre 5 km e 10 km com V\u2019\u2019 = 30 km\/h<\/b><\/span><\/span>\u00a0\"\"\u00a0<\/b><\/span><\/span>V\u2019\u2019 =\u00a0\"\"<\/b><\/span><\/span>\u00a0\"\"\u00a030 =\u00a0\"\"\u00a0\"\"\u00a0\u2206t\u2019\u2019 =\u00a0\"\"\u00a0\"\"<\/b><\/span><\/span>\u2206t\u2019 =\u00a0\"\"\u00a0h.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Velocidade m\u00e9dia total\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span>Vm\u00a0=\u00a0\"\"<\/b><\/span><\/span>\u00a0=\u00a0\"\"\u00a0=\u00a0\"\"\u00a0\"\"\u00a0Vm\u00a0=\u00a0\"\"\u00a0\"\"<\/b><\/span><\/span>Vm\u00a0= 40 km\/h<\/b><\/span><\/span><\/span><\/span><\/p>\n

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

09- <\/b><\/span><\/span><\/span>\u00a0Para esse exerc\u00edcio podemos dar a resposta logo de cara, ou realizar contas, vamos fazer ambos:<\/b><\/span><\/span><\/span><\/span><\/p>\n

Se analisarmos a descri\u00e7\u00e3o,\u00a0<\/b><\/span><\/span>a velocidade de um \u00e9 a metade da outra<\/b><\/span><\/span>, portanto ele\u00a0<\/b><\/span><\/span>ir\u00e1 percorrer metade da dist\u00e2ncia tamb\u00e9m<\/b><\/span><\/span>, sendo assim, <\/b><\/span><\/span>percorrer\u00e1\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span>do caminho total<\/b><\/span><\/span>, enquanto o outro carro percorrer\u00e1\u00a0\"\"<\/b><\/span><\/span>.<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

 <\/p>\n

\"\"<\/p>\n

10-<\/b><\/span><\/span><\/span>– <\/b><\/span><\/span><\/span>Velocidade<\/b><\/span><\/span><\/span> do <\/b><\/span><\/span><\/span>atleta 1 <\/b><\/span><\/span><\/span>que <\/b><\/span><\/span><\/span>partiu antes <\/b><\/span><\/span><\/span>e <\/b><\/span><\/span><\/span>percorreu \u2206S<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> = 42 km <\/b><\/span><\/span><\/span>em <\/b><\/span><\/span><\/span>\u2206t<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> = 2 h \"\"<\/b><\/span><\/span><\/span> V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> = \"\"<\/b><\/span><\/span><\/span> \"\" <\/b><\/span><\/span><\/span><\/p>\n

V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> = 21 km\/h.<\/b><\/span><\/span><\/span><\/p>\n

No <\/b><\/span><\/span><\/span>instante em que o atleta 2 partiu,<\/b><\/span><\/span><\/span> o <\/b><\/span><\/span><\/span>atleta 1 j\u00e1 havia percorrido, em 0,5 h a dist\u00e2ncia \u2206S<\/b><\/span><\/span><\/span> com <\/b><\/span><\/span><\/span>velocidade de V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> = 21 km\/h, <\/b><\/span><\/span><\/span>tal que \"\" <\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span> = \"\"<\/b><\/span><\/span><\/span> \"\" 21 = \"\" \"\" <\/b><\/span><\/span><\/span>\u2206S = 10,5 km.<\/b><\/span><\/span><\/span><\/p>\n

Para que eles <\/b><\/span><\/span><\/span>cheguem juntos no final dos 42 km <\/b><\/span><\/span><\/span>o <\/b><\/span><\/span><\/span>atleta 2 dever\u00e1 percorrer <\/b><\/span><\/span><\/span>essa dist\u00e2ncia <\/b><\/span><\/span><\/span>\u2206S<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> = 42 km em \u2206t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> <\/b>= 2,0 \u2013 0,5 <\/b><\/span><\/span><\/span>= 1,5 h com velocidade V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> tal que \"\" V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span> = \"\"<\/b><\/span><\/span><\/span> = 28 km\/h.<\/b><\/span><\/span><\/span><\/p>\n

Assim, para que <\/b><\/span><\/span><\/span>eles terminem a prova empatados <\/b><\/span><\/span><\/span>o <\/b><\/span><\/span><\/span>atleta 1<\/b><\/span><\/span><\/span> dever\u00e1 ter <\/b><\/span><\/span><\/span>velocidade de V =<\/b><\/span><\/span><\/span> 28 \u2013 21 <\/b><\/span><\/span><\/span>= 7 km\/h a mais que a do atleta 2.<\/b><\/span><\/span><\/span><\/p>\n

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

 <\/p>\n

11-<\/b><\/span><\/span><\/span>Para analisar esse exerc\u00edcio precisamos primeiro entender que no final da corrida trem e homem estar\u00e3o lado a lado, na mesma posi\u00e7\u00e3o, ou seja,\u00a0<\/b><\/span><\/span>a posi\u00e7\u00e3o final de ambos \u00e9 igual<\/b><\/span><\/span>. Sabendo disso e que\u00a0<\/b><\/span><\/span>ambos est\u00e3o em velocidade constante:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

 <\/p>\n

12-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Em todo\u00a0<\/b><\/span><\/span>gr\u00e1fico VXt<\/b><\/span><\/span>\u00a0a\u00a0<\/b><\/span><\/span>\u00e1rea<\/b><\/span><\/span>\u00a0entre a\u00a0<\/b><\/span><\/span>reta representativa e o eixo dos tempos<\/b><\/span><\/span>\u00a0<\/b><\/span>\u00e9 numericamente<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

igual \u00e0<\/b><\/span><\/span>\u00a0varia\u00e7\u00e3o de espa\u00e7o \u0394S<\/b><\/span><\/span>, entre dois instantes quaisquer\u00a0<\/b><\/span><\/span>t1\u00a0e t2.<\/b><\/span><\/span><\/p>\n

Os corpos A e B voltam a se encontrar quando as\u00a0<\/b><\/span><\/span>\u00e1reas entre as curvas caracter\u00edsticas da velocidade<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

(v) e o tempo (t)\u00a0<\/b><\/span><\/span>forem iguais. Analisando o gr\u00e1fico as \u00e1reas s\u00e3o iguais no instante t4.<\/b><\/span><\/span><\/p>\n

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

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

\"\"<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

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

\"\"<\/p>\n

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

Veja a figura abaixo:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

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

16-<\/b><\/span><\/span><\/span>a)<\/b><\/span><\/span>\u00a0o\u00a0<\/b><\/span><\/span>m\u00f3dulo das velocidades escalares m\u00e9dias de A e de B, em m\/s, durante os 120 s<\/b><\/span><\/span>.<\/b><\/span><\/span><\/p>\n

Em\u00a0<\/b><\/span><\/span>todo gr\u00e1fico da velocidade escalar em fun\u00e7\u00e3o do tempo,<\/b><\/span><\/span>\u00a0a\u00a0<\/b><\/span><\/span>\u00e1rea A <\/b><\/span><\/span>entre a\u00a0<\/b><\/span><\/span>reta representativa<\/b><\/span><\/span>\u00a0(linha cheia) e o\u00a0<\/b><\/span><\/span>eixo dos tempos<\/b><\/span><\/span>\u00a0\u00e9 <\/b><\/span><\/span>numericamente igual ao deslocamento \u2206S\u00a0<\/b><\/span><\/span>efetuado pelo m\u00f3vel em <\/b><\/span><\/span>determinado intervalo de tempo \u2206t.<\/b><\/span><\/span><\/p>\n

Calculando o\u00a0<\/b><\/span><\/span>deslocamento do m\u00f3vel A<\/b><\/span><\/span>\u00a0entre\u00a0<\/b><\/span><\/span>to\u00a0= 0 e t = 120 s pela \u00e1rea AA<\/b><\/span><\/span>:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

Calculando o\u00a0<\/b><\/span><\/span>deslocamento do m\u00f3vel B<\/b><\/span><\/span>\u00a0entre\u00a0<\/b><\/span><\/span>to\u00a0= 0 e t = 120 s pela \u00e1rea AB<\/b><\/span><\/span>:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

b)<\/b><\/span><\/span>\u00a0a\u00a0<\/b><\/span><\/span>dist\u00e2ncia entre os ve\u00edculos, em metros<\/b><\/span><\/span>, no\u00a0<\/b><\/span><\/span>instante t = 60 s.<\/b><\/span><\/span><\/p>\n

Observe que, no\u00a0<\/b><\/span><\/span>instante t = 60 s eles possuem a mesma velocidade,<\/b><\/span><\/span>\u00a0que \u00e9 o\u00a0<\/b><\/span><\/span>ponto (P)<\/b><\/span><\/span>\u00a0onde as\u00a0<\/b><\/span><\/span>retas<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Acontece que\u00a0<\/b><\/span><\/span>no instante t = 60s, VB\u00a0= VA\u00a0= V\u2019<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>= 12 m\/s.<\/b><\/span><\/span><\/p>\n

Deslocamentos de A e de B entre 0 e 60 s pela \u00e1rea:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

MOVIMENTO UNIFORMEMENTE VARIADO<\/b><\/span><\/span><\/span><\/p>\n

 <\/p>\n

01-<\/b><\/span><\/span><\/span> Durante a\u00a0<\/b><\/span><\/span>primeira etapa\u00a0<\/b><\/span><\/span>entre a\u00a0<\/b><\/span><\/span>detec\u00e7\u00e3o do problema pelo motorista e o acionamento do freio,<\/b><\/span><\/span>\u00a0a\u00a0<\/b><\/span><\/span>velocidade\u00a0<\/b><\/span><\/span>do ve\u00edculo\u00a0<\/b><\/span><\/span>permanece constante <\/b><\/span><\/span>com ele em\u00a0<\/b><\/span><\/span>movimento uniforme (MU<\/b><\/span><\/span>) e o\u00a0<\/b><\/span><\/span>gr\u00e1fico \u00e9 uma reta horizontal paralela ao eixo da dist\u00e2ncia.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Durante a segunda etapa, ap\u00f3s o acionamento do freio,\u00a0<\/b><\/span><\/span>o ve\u00edculo sofre <\/b><\/span><\/span>desacelera\u00e7\u00e3o constante,\u00a0<\/b><\/span><\/span>e realiza um\u00a0<\/b><\/span><\/span>movimento uniformemente variado(retardado,<\/b><\/span><\/span>\u00a0pois est\u00e1\u00a0<\/b><\/span><\/span>freando).<\/b><\/span><\/span><\/span><\/span><\/p>\n

Como a\u00a0<\/b><\/span><\/span>velocidade V<\/b><\/span><\/span>\u00a0est\u00e1\u00a0<\/b><\/span><\/span>variando<\/b><\/span><\/span>\u00a0em fun\u00e7\u00e3o da\u00a0<\/b><\/span><\/span>dist\u00e2ncia d<\/b><\/span><\/span>\u00a0a equa\u00e7\u00e3o utilizada \u00e9\u00a0<\/b><\/span><\/span>a equa\u00e7\u00e3o de Torricelli\u00a0<\/b><\/span><\/span>(V<\/b><\/span><\/span>2<\/b><\/span><\/sup><\/span>\u00a0= Vo<\/b><\/span><\/span>2<\/b><\/span><\/sup><\/span>+ 2.(-a).d, sendo a acelera\u00e7\u00e3o negativa pois est\u00e1 freando)\u00a0<\/b><\/span><\/span>e, assim,\u00a0voc\u00ea pode<\/b><\/span><\/span>\u00a0concluir que o gr\u00e1fico da velocidade em fun\u00e7\u00e3o da dist\u00e2ncia \u00e9 um\u00a0<\/b><\/span><\/span>arco de par\u00e1bola<\/b><\/span><\/span>\u00a0com a intensidade da velocidade diminuindo.<\/b><\/span><\/span><\/span><\/span><\/p>\n

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

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

\"\"<\/p>\n

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

\u00a0<\/span>As alternativas (A) e (B) est\u00e3o erradas, pois o gr\u00e1fico\u00a0<\/b><\/span><\/span>Sxt fornece apenas a posi\u00e7\u00e3o do m\u00f3vel em cada instante,\u00a0<\/b><\/span><\/span>independente da trajet\u00f3ria <\/b><\/span><\/span>do mesmo.<\/b><\/span><\/span>\"\"<\/p>\n

Pelo enunciado o\u00a0<\/b><\/span><\/span>trecho C n\u00e3o \u00e9 arco de par\u00e1bola (reta)\u00a0<\/b><\/span><\/span>o que significa que<\/b><\/span><\/span>nesse trecho o m\u00f3vel est\u00e1 em repouso numa posi\u00e7\u00e3o qualquer, por exemplo X.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Nos\u00a0<\/b><\/span><\/span>trechos A, B e D\u00a0<\/b><\/span><\/span>o movimento \u00e9 um\u00a0<\/b><\/span><\/span>MUV de equa\u00e7\u00e3o S = So\u00a0+ Vot +\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span>onde\u00a0<\/b><\/span><\/span>a<\/b><\/span><\/span>\u00a0\u00e9 a\u00a0<\/b><\/span><\/span>acelera\u00e7\u00e3o<\/b><\/span><\/span>\u00a0que \u00e9<\/b><\/span><\/span>positiva com a concavidade da par\u00e1bola para cima e negativa com a concavidade para baixo.<\/b><\/span><\/span><\/span><\/span><\/p>\n

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

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

\"\"<\/p>\n

 <\/p>\n

05-\u00a0<\/b><\/span><\/span><\/span><\/span><\/p>\n

Errado\u00a0\"\"\u00a0em t > 0 a\u00a0<\/b><\/span><\/span>trajet\u00f3ria<\/b><\/span><\/span>\u00a0\u00e9 uma\u00a0<\/b><\/span><\/span>exponencial<\/b><\/span><\/span>.<\/b><\/span><\/span><\/span><\/span><\/p>\n

b) Errado\u00a0\"\"\u00a0em\u00a0<\/b><\/span><\/span>t = 0\u00a0<\/b><\/span><\/span>o objeto est\u00e1 a 10 m de\u00a0<\/b><\/span><\/span>dist\u00e2ncia\u00a0<\/b><\/span><\/span>do observador.<\/b><\/span><\/span><\/span><\/span><\/p>\n

c) Correto\u00a0\"\"\u00a0<\/b><\/span><\/span>O movimento\u00a0<\/b><\/span><\/span>\u00e9 dado pela\u00a0<\/b><\/span><\/span>fun\u00e7\u00e3o<\/b><\/span><\/span>\u00a0x(t) = -10 + 4t +2t<\/b><\/span><\/span>2<\/b><\/span><\/sup><\/span>, dessa forma, o termo\u00a0<\/b><\/span><\/span>quadr\u00e1tico<\/b><\/span><\/span>\u00a0constante caracteriza um\u00a0<\/b><\/span><\/span>movimento retil\u00edneo<\/b><\/span><\/span>uniformemente variado.<\/b><\/span><\/span><\/span><\/span><\/p>\n

d) Errado\u00a0\"\"\u00a0em t = 0 a\u00a0<\/b><\/span><\/span>velocidade<\/b><\/span><\/span>\u00a0igual a velocidade inicial do\u00a0<\/b><\/span><\/span>objeto<\/b><\/span><\/span>, que vale 4 m\/s.<\/b><\/span><\/span><\/span><\/span><\/p>\n

e) Errado\u00a0\"\"\u00a0<\/b><\/span><\/span>quando t = 2 s\u00a0<\/b><\/span><\/span>o objeto est\u00e1 a 6 m do\u00a0<\/b><\/span><\/span>observador<\/b><\/span><\/span>\u00a0\"\"<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"\u00a0\"\"<\/span><\/p>\n

R-<\/b><\/span><\/span>\u00a0Portanto, a resposta correta \u00e9 a alternativa C.<\/b><\/span><\/span><\/span><\/span><\/p>\n

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

Fun\u00e7\u00e3o (equa\u00e7\u00e3o) hor\u00e1ria da velocidade do MUV<\/b><\/span><\/span><\/span><\/span><\/p>\n

\u00a0<\/span>O<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>m\u00f3vel<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>da figura abaixo tem\u00a0<\/b><\/span><\/span>velocidade Vo\u00a0no instante to<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>e<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>velocidade V num instante posterior t.<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Admitindo<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>to\u00a0= 0 como a origem dos tempos<\/b><\/span><\/span>\u00a0\"\"\u00a0\u00a0<\/b><\/span><\/span>a = (V \u2013 Vo)\/(t \u2013 to<\/b><\/span><\/span>)\u00a0\"\"\u00a0a = (V \u2013 Vo)\/t\u00a0\"\"\u00a0V \u2013 Vo\u00a0= a.t\u00a0\"\"\u00a0<\/b><\/span><\/span><\/span><\/span><\/p>\n

V = Vo\u00a0+ a.t, sendo\u00a0<\/b><\/span><\/span>a\u00a0<\/b><\/span><\/span>a\u00a0<\/b><\/span><\/span>acelera\u00e7\u00e3o<\/b><\/span><\/span>\u00a0do m\u00f3vel.<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Como Vo\u00a0= 0\u00a0\"\"\u00a0V = a.t.<\/b><\/span><\/span><\/span><\/span><\/p>\n

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

 <\/p>\n

07-<\/b><\/span><\/span><\/span>Correta\u00a0\"\"<\/b><\/span><\/span>\u00a0A\u00a0<\/b><\/span><\/span>crosta terrestre \u00e9 formada por v\u00e1rios blocos rochosos,<\/b><\/span><\/span>denominados\u00a0<\/b><\/span><\/span>placas litosf\u00e9ricas ou placas tect\u00f4nicas.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Esses\u00a0<\/b><\/span><\/span>blocos est\u00e3o em constante movimento,\u00a0<\/b><\/span><\/span>podendo se\u00a0<\/b><\/span><\/span>afastar uns dos outros,\u00a0<\/b><\/span><\/span>caracterizando uma zona de\u00a0<\/b><\/span><\/span>diverg\u00eancia ou se aproximar, <\/b><\/span><\/span>formando uma\u00a0<\/b><\/span><\/span>zona de converg\u00eancia.<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Nas\u00a0<\/b><\/span><\/span>zonas de converg\u00eancia pode ocorrer a colis\u00e3o ou a subduc\u00e7\u00e3o de placas,\u00a0<\/b><\/span><\/span>o que\u00a0<\/b><\/span><\/span>produz<\/b><\/span><\/span>\u00a0um\u00a0<\/b><\/span><\/span>ac\u00famulo de press\u00e3o e descarga de energia<\/b><\/span><\/span>, que se\u00a0<\/b><\/span><\/span>propaga em forma de ondas s\u00edsmicas, caracterizando o terremoto.<\/b><\/span><\/span><\/span><\/span><\/p>\n

08) Correta<\/b><\/span><\/span>\u00a0\"\"\u00a0Na\u00a0<\/b><\/span><\/span>figura abaixo\u00a0<\/b><\/span><\/span>podemos\u00a0<\/b><\/span><\/span>visualizar a divis\u00e3o das placas tect\u00f4nicas,\u00a0<\/b><\/span><\/span>entre as quais,<\/b><\/span><\/span><\/span><\/span><\/p>\n

01) Errada\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span>S(t) = -t + 15\u00a0\"\"\u00a0para qualquer t S(t) < 0 (retr\u00f3grado).<\/b><\/span><\/span><\/span><\/span><\/p>\n

02) Errada\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span>S(t) n\u00e3o \u00e9 injetora (a\u00a0<\/b><\/span><\/span>\uf0b9<\/b><\/span><\/span><\/span>\u00a00),\u00a0<\/b><\/span><\/span>portanto, tamb\u00e9m\u00a0<\/b><\/span><\/span>n\u00e3o \u00e9 bijetora (bijetora = injetora + sobrejetora)<\/b><\/span><\/span><\/span><\/span><\/p>\n

04) Correta\u00a0\"\"\u00a0S(t) = t<\/b><\/span><\/span>2<\/b><\/span><\/sup><\/span>+2t+3\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span>S(2) = 2<\/b><\/span><\/span>2<\/b><\/span><\/sup><\/span>+2.2+3 = 11\u00a0\"\"<\/b><\/span><\/span>\u00a0S(6) = 62+2.6+3 = 51\u00a0\"\"\u00a0<\/b><\/span><\/span>VM\u00a0= \u2206S\/\u2206t\u00a0\"\"<\/b><\/span><\/span>\u00a0VM=(51-11)\/(6-2) = 40\/4\u00a0\"\"\u00a0<\/b><\/span><\/span>VM= 10 m\/s.<\/b><\/span><\/span><\/span><\/span><\/p>\n

08) Errada\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span>at2\u00a0\u2013 bt + c (MUV), s\u00f3 ser\u00e1 UM se a\u00a0<\/b><\/span><\/span>\uf0b9<\/b><\/span><\/span><\/span>\u00a00.<\/b><\/span><\/span><\/span><\/span><\/p>\n

16) Correta\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span>S(0) = c = 3\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span>S(1) = a + b + c = 5\u00a0\"\"\u00a0S(2) = 4\u00aa + 2b+c = 7\u00a0\"\"\u00a0a + b = 2, 4a + 2b = 4, 2a + b = 2\u00a0\"\"\u00a02a +(2-a) = 2\u00a0\"\"\u00a0a=0, b=2\u00a0\"\"<\/b><\/span><\/span>S(t) = 2t + 3.<\/b><\/span><\/span><\/span><\/span><\/p>\n

R- 04 e 16.<\/b><\/span><\/span><\/span><\/span><\/p>\n

 <\/p>\n

08-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Observando o enunciado percebemos<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>que<\/b><\/span><\/span>\u00a0o trem realiza um Movimento Uniformemente Variado (<\/b><\/span><\/span>MUV<\/b><\/span><\/span>) com acelera\u00e7\u00e3o m\u00e1xima a = 0,09g = 0,09\u00d710 = 0,9 m\/s<\/b><\/span><\/span>2<\/b><\/span>\u00a0e que n\u00e3o foi fornecido\u00a0<\/b><\/span><\/span>nenhuma informa\u00e7\u00e3o de<\/b><\/span><\/span>tempo\u00a0<\/b><\/span><\/span>para determinarmos<\/b><\/span><\/span>\u00a0a dist\u00e2ncia m\u00ednima ( \u2206S) percorrida.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

 <\/p>\n

MOVIMENTOS VERTICAIS<\/b><\/span><\/span><\/span><\/p>\n

01-<\/b><\/span><\/span><\/span>Observe<\/b><\/span><\/span>\u00a0que, pelo\u00a0<\/b><\/span><\/span>enunciado, \u00e9 pedida a velocidade do parafuso em rela\u00e7\u00e3o ao ch\u00e3o do elevador<\/b><\/span><\/span>\u00a0e,\u00a0<\/b><\/span><\/span>antes de se desprender<\/b><\/span><\/span>\u00a0do teto\u00a0<\/b><\/span><\/span>(instante t<\/b><\/span><\/span>o<\/b><\/span>),<\/b><\/span><\/span>\u00a0o\u00a0<\/b><\/span><\/span>parafuso encontra-se em repouso em rela\u00e7\u00e3o ao ch\u00e3o do elevador, <\/b><\/span><\/span>pois a\u00a0<\/b><\/span><\/span>dist\u00e2ncia entre eles n\u00e3o est\u00e1 variando.<\/b><\/span><\/span><\/p>\n

Assim, as\u00a0<\/b><\/span><\/span>alternativas a, b e c n\u00e3o satisfazem.<\/b><\/span><\/span><\/p>\n

Ap\u00f3s desprender-se do teto<\/b><\/span><\/span>\u00a0(instante t<\/b><\/span><\/span>o<\/b><\/span>) o\u00a0<\/b><\/span><\/span>parafuso efetua uma queda livre em rela\u00e7\u00e3o ao ch\u00e3o do elevador<\/b><\/span><\/span>\u00a0com a\u00a0<\/b><\/span><\/span>velocidade<\/b><\/span><\/span>\u00a0variando com o <\/b><\/span><\/span>tempo<\/b><\/span><\/span>\u00a0conforme a\u00a0<\/b><\/span><\/span>equa\u00e7\u00e3o V = V<\/b><\/span><\/span>o<\/b><\/sub><\/span>\u00a0+<\/b><\/span><\/span>\u00a0gt<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>= 0\u00a0+<\/b><\/span><\/span>\u00a0gt\u00a0\"\"<\/b><\/span><\/span><\/span><\/p>\n

V =\u00a0<\/b><\/span><\/span>gt<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>(o sinal\u00a0de gt depende da\u00a0<\/b><\/span><\/span>orienta\u00e7\u00e3o da trajet\u00f3ria<\/b><\/span><\/span>).<\/b><\/span><\/span><\/p>\n

Como o\u00a0<\/b><\/span><\/span>enunciado<\/b><\/span><\/span>\u00a0pede o\u00a0<\/b><\/span><\/span>m\u00f3dulo da velocidade<\/b><\/span><\/span>\u00a0voc\u00ea ter\u00e1\u00a0<\/b><\/span><\/span>\u0399V\u0399 = gt<\/b><\/span><\/span>\u00a0e, essa <\/b><\/span><\/span>fun\u00e7\u00e3o<\/b><\/span><\/span>\u00a0\u00e9 do\u00a0<\/b><\/span><\/span>primeiro grau<\/b><\/span><\/span>\u00a0representada por uma\u00a0<\/b><\/span><\/span>reta inclinada.<\/b><\/span><\/span><\/p>\n

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

02-<\/b><\/span><\/span><\/span>Como a\u00a0<\/b><\/span><\/span>bola sai da mesa com velocidade inicial (horizontal Vo)<\/b><\/span><\/span>\u00a0e\u00a0<\/b><\/span><\/span>cai<\/b><\/span><\/span>sujeita \u00e0\u00a0<\/b><\/span><\/span>acelera\u00e7\u00e3o da gravidade g, trata-se de um lan\u00e7amento horizontal.<\/b><\/span><\/span><\/p>\n

Olhando de cima\u00a0<\/b><\/span><\/span>voc\u00ea est\u00e1 vendo o\u00a0<\/b><\/span><\/span>movimento horizontal da bola<\/b><\/span><\/span>\u00a0e<\/b><\/span><\/span>olhando de frente, o movimento vertical da mesma.<\/b><\/span><\/span><\/p>\n

Veja como a\u00a0<\/b><\/span><\/span>situa\u00e7\u00e3o apresentada pode ser esquematizada\u00a0<\/b><\/span><\/span>(lan\u00e7amento horizontal) com as\u00a0<\/b><\/span><\/span>seguintes caracter\u00edsticas:<\/b><\/span><\/span><\/p>\n

Colocando-se\u00a0<\/b><\/span><\/span>a\u00a0<\/b><\/span><\/span>origem do sistema de refer\u00eancia<\/b><\/span><\/span>\u00a0no\u00a0<\/b><\/span><\/span>ponto de lan\u00e7am<\/b><\/span><\/span>ento, orienta-se, por exemplo,\u00a0<\/b><\/span><\/span>o\u00a0eixo X para a direita e o eixo Y para baixo.<\/b><\/span><\/span><\/p>\n

Decomp\u00f5e-se o movimento em\u00a0duas parcelas:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"\u00a0<\/span>Segundo o eixo X \u00a0\"\"<\/b><\/span><\/span>\u00a0 trata-se de um\u00a0<\/b><\/span><\/span>movimento horizontal uniforme\u00a0<\/b><\/span><\/span>com<\/b><\/span><\/span>velocidade constante\u00a0<\/b><\/span><\/span>de\u00a0<\/b><\/span><\/span>intensidade\u00a0Vo\u00a0,\u00a0<\/b><\/span><\/span>que \u00e9 a\u00a0<\/b><\/span><\/span>velocidade de lan\u00e7amento\u00a0\u00a0\"\"<\/b><\/span><\/span>\u00a0 S = So\u00a0+ V.t\u00a0\"\"\u00a0X = 0 + Vo.t\u00a0\u00a0\"\"\u00a0\u00a0\u00a0<\/b><\/span><\/span>X = Vo.t<\/b><\/span><\/span><\/p>\n

\"\"\u00a0<\/span>Segundo o eixo Y \u00a0\"\"<\/b><\/span><\/span>\u00a0 trata-se de um\u00a0<\/b><\/span><\/span>movimento uniformemente variado<\/b><\/span><\/span>com\u00a0<\/b><\/span><\/span>velocidade inicial<\/b><\/span><\/span><\/p>\n

Vo\u00a0= 0,\u00a0<\/b><\/span><\/span>ou seja, \u00e9 uma\u00a0<\/b><\/span><\/span>queda livre com o corpo abandonado da origem,\u00a0sujeito\u00a0apenas \u00e0 acelera\u00e7\u00e3o da\u00a0gravidade, de intensidade\u00a0g, dire\u00e7\u00e3o vertical e sentido para baixo.\u00a0<\/b><\/span><\/span><\/p>\n

Equa\u00e7\u00f5es:<\/b><\/span><\/span><\/span><\/p>\n

S = So\u00a0+ Vo.t + at<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 \u00a0\"\"<\/b><\/span><\/span><\/span>\u00a0 Y = 0 + 0.t + gt<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0\"\"\u00a0<\/b><\/span><\/span><\/span>Y = g.t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2<\/b><\/span><\/span><\/span><\/p>\n

Vy\u00a0= Voy\u00a0+ a.t \u00a0\"\"<\/b><\/span><\/span><\/span>\u00a0 Vy= 0 + g.t\u00a0\"\"\u00a0\u00a0<\/b><\/span><\/span><\/span>Vy\u00a0= g.t<\/b><\/span><\/span><\/span><\/p>\n

V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0= Vo<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ 2.a.\u0394S \u00a0\"\"<\/b><\/span><\/span><\/span>\u00a0 Vy<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0\u00a0= Voy<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ 2.g.\u0394h \u00a0\"\"\u00a0 Vy2\u00a0\u00a0= 02\u00a0+ 2.g.\u0394h\u00a0\u00a0\"\"\u00a0\u00a0<\/b><\/span><\/span><\/span>Vy<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0= 2.g.\u0394h<\/b><\/span><\/span><\/span><\/p>\n

No caso do exerc\u00edcio:<\/b><\/span><\/span><\/p>\n

C\u00e1lculo do tempo t que a bola demora para atingir o solo onde y = 1,25 m\u00a0\"\"\u00a0<\/b><\/span><\/span>Y = g.t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0\"\"\u00a01,25 =<\/b><\/span><\/span><\/span><\/p>\n

\"\"\u00a0\"\"\u00a0<\/span>t =\u00a0\"\"\u00a0\"\"\u00a0t = 0,5 s.<\/b><\/span><\/span><\/p>\n

Esse tempo de queda na vertical \u00e9 o mesmo que a bola demora para percorrer X = 2,4 m na horizontal com velocidade Vo\u00a0\"\"\u00a0X = Vo.t\u00a0\"\"\u00a02,4 = Vo.0,5\u00a0\"\"\u00a0Vo\u00a0= 4,8 m\/s.<\/b><\/span><\/span><\/p>\n

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

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

\"\"<\/p>\n

\"\"<\/p>\n

04-<\/b><\/span><\/span><\/span> <\/b>Corpos\u00a0<\/b><\/span><\/span>abandonados com Vo\u00a0= 0 ou lan\u00e7ados verticalmente para baixo com Vo\u00a0\u2260 0<\/b><\/span><\/span>\u00a0de um\u00a0<\/b><\/span><\/span>mesmo local e da mesma altura<\/b><\/span><\/span>\u00a0(desprezando-se os efeitos do ar)<\/b><\/span><\/span>demoram o mesmo tempo para chegar ao solo, independente de suas massas.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

A\u00a0<\/b><\/span><\/span>velocidade\u00a0<\/b><\/span><\/span>com que eles chegam ao solo\u00a0<\/b><\/span><\/span>tamb\u00e9m \u00e9 a mesma.<\/b><\/span><\/span><\/p>\n

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

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

\u00a0\"\"<\/span><\/p>\n

\"\"<\/p>\n

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

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

\"\"<\/p>\n

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

LAN\u00c7AMENTO OBL\u00cdQUO<\/b><\/span><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

02-\u00a0<\/b><\/span><\/span>Para calcularmos a altura m\u00e1xima precisaremos\u00a0<\/b><\/span><\/span>decompor esse exerc\u00edcio em x e y<\/b><\/span><\/span>, para obtermos todas as informa\u00e7\u00f5es poss\u00edveis de y, inicialmente\u00a0<\/b><\/span><\/span>decompondo a velocidade V0\u00a0em V0x\u00a0e V0y<\/b><\/span><\/span>:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"\"\"\"\"<\/p>\n

A seta vermelha representa nosso V0y\u00a0e a azul nosso V0x\u00ad. Como fornecido no exerc\u00edcio\u00a0<\/b><\/span><\/span>o \u00e2ngulo entre x e V0\u00a0\u00e9 de 30\u00b0, o mesmo para x e V0x<\/b><\/span><\/span>.\u00a0Vamos calcular V0y:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"\"\"\"\"<\/p>\n

Passando V0y\u00a0para o lado direito<\/b><\/span><\/span>, podemos observar que\u00a0<\/b><\/span><\/span>ele \u00e9 o cateto oposto de\u00a0\"\"<\/b><\/span><\/span>, sendo assim:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

Substituindo os valores:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Multiplicando em cruz:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Isolando V0y:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Sabendo disso, podemos\u00a0<\/b><\/span><\/span>calcular a altura por Torricelli<\/b><\/span><\/span>:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Onde: Vfy\u00a0\u00e9 a velocidade final em y<\/b><\/span><\/span><\/span><\/span><\/p>\n

V0y\u00a0\u00e9 a velocidade inicial em y<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"\u00a0<\/span>\u00e9 o deslocamento em y<\/b><\/span><\/span><\/span><\/span><\/p>\n

a \u00e9 a acelera\u00e7\u00e3o em y<\/b><\/span><\/span><\/span><\/span><\/p>\n

Vamos substituir os valores:<\/b><\/span><\/span><\/span><\/span><\/p>\n

A altura m\u00e1xima\u00a0<\/b><\/span><\/span>equivale ao ponto aonde\u00a0<\/b><\/span><\/span>a velocidade \u00e9 0<\/b><\/span><\/span>, portanto nosso\u00a0<\/b><\/span><\/span>Vfy\u00a0deve ser 0<\/b><\/span><\/span>.<\/b><\/span><\/span><\/span><\/span><\/p>\n

O\u00a0\"\"<\/b><\/span><\/span>,<\/b><\/span><\/span>\u00a0portanto, ser\u00e1\u00a0<\/b><\/span><\/span>nossa altura<\/b><\/span><\/span>.<\/b><\/span><\/span><\/span><\/span><\/p>\n

A\u00a0<\/b><\/span><\/span>acelera\u00e7\u00e3o deve ser negativa<\/b><\/span><\/span>, pois o movimento est\u00e1 subindo e\u00a0<\/b><\/span><\/span>a gravidade exerce uma for\u00e7a contr\u00e1ria<\/b><\/span><\/span>, para baixo.<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Passando o h para o outro lado:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Calculando o h:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

N\u00e3o podemos esquecer que\u00a0<\/b><\/span><\/span>a catapulta j\u00e1 estava na altura de 1m<\/b><\/span><\/span>, portanto h = 20 + 1 = 21m.<\/b><\/span><\/span><\/span><\/span><\/p>\n

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

 <\/p>\n

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

 <\/p>\n

\"\"<\/p>\n

d) Errada\u00a0\"\"\u00a0<\/b><\/span><\/span>Podemos calcular\u00a0<\/b><\/span><\/span>o tempo em que o objeto atinge o ponto mais alto e\u00a0<\/b><\/span><\/span>chega ao solo\u00a0<\/b><\/span><\/span>novamente, atrav\u00e9s da\u00a0<\/b><\/span><\/span>equa\u00e7\u00e3o dada\u00a0<\/b><\/span><\/span>no enunciado:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

Resolvendo essa\u00a0<\/b><\/span><\/span>equa\u00e7\u00e3o<\/b><\/span><\/span>\u00a0de segundo grau encontramos que o\u00a0<\/b><\/span><\/span>tempo de subida\u00a0<\/b><\/span><\/span>\u00e9 de 1,5 segundos. Como o\u00a0<\/b><\/span><\/span>tempo de descida\u00a0<\/b><\/span><\/span>\u00e9 igual ao tempo de subida, o tempo total \u00e9 3\u00a0<\/b><\/span><\/span>segundos<\/b><\/span><\/span>.<\/b><\/span><\/span><\/span><\/span><\/p>\n

e) Correto\u00a0\"\"\u00a0Podemos\u00a0<\/b><\/span><\/span>calcular<\/b><\/span><\/span>\u00a0a dist\u00e2ncia horizontal atrav\u00e9s da equa\u00e7\u00e3o em x, dada no enunciado,\u00a0<\/b><\/span><\/span>utilizando o tempo\u00a0<\/b><\/span><\/span>calculado no item d, j\u00e1 que \u00e9 o mesmo.<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

R-<\/b><\/span><\/span>\u00a0Portanto, a resposta correta \u00e9 a alternativa E.<\/b><\/span><\/span><\/span><\/span><\/p>\n

 <\/p>\n

04-<\/b><\/span><\/span><\/span> Observe que s\u00e3o desprezados os atritos e o\u00a0<\/b><\/span><\/span>helic\u00f3ptero<\/b><\/span><\/span>\u00a0est\u00e1 em <\/b><\/span><\/span>movimento retil\u00edneo\u00a0<\/b><\/span><\/span>e, como\u00a0<\/b><\/span><\/span>cada pacote atinge o solo em um ponto exatamente embaixo do helic\u00f3ptero<\/b><\/span><\/span>, ele (helic\u00f3ptero) est\u00e1 com <\/b><\/span><\/span>velocidade constante (acelera\u00e7\u00e3o nula).<\/b><\/span><\/span><\/span><\/span><\/p>\n

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

05-<\/b><\/span><\/span><\/span> Trata-se de um lan\u00e7amento obl\u00edquo onde,na vertical (Y) voc\u00ea tem as fun\u00e7\u00f5es(equa\u00e7\u00f5es), com os respectivos gr\u00e1ficos<\/b><\/span><\/span><\/span><\/span><\/p>\n

Equa\u00e7\u00e3o da posi\u00e7\u00e3o (dos espa\u00e7os) ou da altura Y<\/b><\/span><\/span><\/span><\/span><\/p>\n

S=So\u00a0+ Vo.t + a.t2\/2\u00a0 \u2014\u00a0 Y=Yo\u00a0+ Voy.t \u2013 gt2\/2\u00a0 \u2014\u00a0 Y=0 + Voy.t \u2013 gt2\/2\u00a0 \u2014 \u00a0Y= Voy.t \u2013 gt2\/2, com Voy=Vo.sen\u03b1\u00a0<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Equa\u00e7\u00e3o da velocidade Vy<\/b><\/span><\/span><\/span><\/span><\/p>\n

V=Vo\u00a0+ a.t\u00a0 \u2014\u00a0\u00a0 Vy=Voy\u00a0\u2013 g.t, com Voy=Vo.sen\u03b1\u00a0<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

Movimento parcial na dire\u00e7\u00e3o horizontal (eixo X)<\/b><\/span><\/span><\/span><\/span><\/p>\n

Sendo o movimento apenas na horizontal, a\u00a0<\/b><\/span><\/span>proje\u00e7\u00e3o do vetor acelera\u00e7\u00e3o da gravidade (\"\") no eixo X<\/b><\/span><\/span>\u00a0\u00e9\u00a0<\/b><\/span><\/span>nula<\/b><\/span><\/span>\u00a0e, nesse caso o movimento \u00e9 uniforme (MU) com<\/b><\/span><\/span>\u00a0velocidade constante Vox,\u00a0<\/b><\/span><\/span>de equa\u00e7\u00e3o<\/b><\/span><\/span>\u00a0X = Voxt e cujo gr\u00e1fico da velocidade Vx\u00a0, constante, \u00e9 uma reta paralela ao eixo dos tempos:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

MOVIMENTO CIRCULAR<\/b><\/span><\/span><\/span><\/p>\n

01-<\/b><\/span><\/span><\/span>A seguir, um<\/b><\/span><\/span>\u00a0breve resumo te\u00f3rico:<\/b><\/span><\/span><\/p>\n

Acoplamento de polias e engrenagens<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

Como\u00a0<\/b><\/span><\/span>todos os dentes de cada engrenagem<\/b><\/span><\/span>\u00a0se\u00a0<\/b><\/span><\/span>encaixam perfeitamente<\/b><\/span><\/span>, os\u00a0<\/b><\/span><\/span>raios\u00a0<\/b><\/span><\/span>das mesmas s\u00e3o\u00a0<\/b><\/span><\/span>diretamente proporcionais ao n\u00famero de dentes\"\"RA<\/b><\/span><\/span>= 24;\u00a0<\/b><\/span><\/span>RB<\/b><\/span><\/span>\u00a0= 72;\u00a0<\/b><\/span><\/span>RC<\/b><\/span><\/span>= 36 e\u00a0<\/b><\/span><\/span>RD<\/b><\/span><\/span>\u00a0= 108.<\/b><\/span><\/span><\/p>\n

Observe na<\/b><\/span><\/span>figura\u00a0<\/b><\/span><\/span>que a\u00a0<\/b><\/span><\/span>engrenagem A est\u00e1 presa ao motor de freq\u00fc\u00eancia 18 rpm,<\/b><\/span><\/span>\u00a0ent\u00e3o\u00a0<\/b><\/span><\/span>fA\u00a0= 18 rpm.<\/b><\/span><\/span><\/p>\n

Engrenagens A e B\"\"fA.RA\u00a0= fB.RB\"\"<\/b><\/span><\/span>\u00a018.24 = fB.72\u00a0\"\"fB\u00a0= 432\/72\u00a0\"\"<\/b><\/span><\/span>fB\u00a0= 6 rpm (freq\u00fc\u00eancia da engrenagem B).<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

02<\/b><\/span><\/span>\u00a0\u2013\u00a0<\/b><\/span><\/span>Para resolver esse exerc\u00edcio primeiramente precisamos\u00a0<\/b><\/span><\/span>passar as unidades para segundos<\/b><\/span><\/span>, como est\u00e1 em minutos, ent\u00e3o\u00a0<\/b><\/span><\/span>vamos dividir por 60<\/b><\/span><\/span>:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

No caso<\/b><\/span><\/span>\u00a0hertz representa rota\u00e7\u00f5es por segundo<\/b><\/span><\/span>.<\/b><\/span><\/span><\/p>\n

Bom como\u00a0<\/b><\/span><\/span>cada rota\u00e7\u00e3o equivale \u00e0 2\u03c0<\/b><\/span><\/span>, afinal temos um c\u00edrculo:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Agora que temos\u00a0<\/b><\/span><\/span>ambas as velocidades angulares<\/b><\/span><\/span>, precisamos apenas <\/b><\/span><\/span>subtrair uma da outra para calcularmos a varia\u00e7\u00e3o<\/b><\/span><\/span>:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

R \u2013 B<\/b><\/span><\/span><\/p>\n

03-<\/b><\/span><\/span> a)<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>Se a\u00a0<\/b><\/span><\/span>freq\u00fc\u00eancia da h\u00e9lice \u00e9 de f = 4 H<\/b><\/span><\/span>z<\/b><\/span>, o per\u00edodo vale\u00a0<\/b><\/span><\/span>T =\u00a0\"\"<\/b><\/span><\/span>\u00a0=<\/b><\/span><\/span>\u00a0\"\"\u00a0\"\"\u00a0<\/b><\/span><\/span>T = 0,25 s (tempo que demora para efetuar uma volta completa).<\/b><\/span><\/span><\/p>\n

Se,\u00a0<\/b><\/span><\/span>para efetuar uma volta completa a h\u00e9lice demora T = 0,25 s,<\/b><\/span><\/span>\u00a0para <\/b><\/span><\/span>efetuar 12 voltas completas<\/b><\/span><\/span>\u00a0ela demorar\u00e1\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>= 12\u00d70,25\u00a0<\/b><\/span><\/span>= 3 s.<\/b><\/span><\/span><\/p>\n

Como o avi\u00e3o se desloca\u00a0\"\"\u00a0com velocidade escalar constante V = 2 m\/s em\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>= 3 s\u00a0\"\"\u00a0V =\u00a0\"\"<\/b><\/span><\/span>\u00a0\"\"<\/b><\/span><\/span><\/p>\n

2 =<\/b><\/span><\/span>\u00a0\"\"<\/b><\/span><\/span>\u00a0\"\"\u00a0<\/b><\/span><\/span>\u2206S = 6 m.<\/b><\/span><\/span><\/p>\n

b)<\/b><\/span><\/span><\/p>\n

Velocidade angular (W) de um MCU<\/b><\/span><\/span><\/p>\n

\u00a0<\/span><\/p>\n

\u00a0<\/span>Considere um<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>m\u00f3vel em MCU<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>de modo que no<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>instante t<\/b><\/span><\/span>o<\/b><\/span>\u00a0<\/b><\/span><\/span>sua posi\u00e7\u00e3o seja determinada pelo<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>\u00e2ngulo<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>de fase \u03c6<\/b><\/span><\/span>o<\/b><\/span>\u00a0e num\u00a0outro instante t\u00a0(t > t<\/b><\/span><\/span>o<\/b><\/span>)\u00a0pelo \u00e2ngulo de fase \u03c6.<\/b><\/span><\/span><\/p>\n

Assim, observa-se que,<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>no intervalo de tempo\u00a0<\/b><\/span><\/span>\u0394t = t \u2013 t<\/b><\/span><\/span>o<\/b><\/span>,\u00a0o<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>raio<\/b><\/span><\/span>\u00a0que\u00a0<\/b><\/span><\/span>acompanha\u00a0<\/b><\/span><\/span>o m\u00f3vel em seu movimento\u00a0descreveu \u201cvarreu\u201d um \u00e2ngulo \u0394\u03c6 = \u03c6 \u2013 \u03c6<\/b><\/span><\/span>o<\/b><\/span>.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

C\u00e1lculo da\u00a0<\/b><\/span><\/span>velocidade V\u2019 de rota\u00e7\u00e3o\u00a0<\/b><\/span><\/span>de um ponto da\u00a0<\/b><\/span><\/span>extremidade R = 1m de uma das p\u00e1s\u00a0<\/b><\/span><\/span>da h\u00e9lice\u00a0\"\"\u00a0<\/b><\/span><\/span>W =\u00a0\"\"<\/b><\/span><\/span>\u00a0\"\"\u00a0<\/b><\/span><\/span>24 =\u00a0\"\"<\/b><\/span><\/span>\u00a0\"\"\u00a0<\/b><\/span><\/span>V\u2019 = 24 m\/s<\/b><\/span><\/span>\"\"<\/p>\n

Acontece que a\u00a0<\/b><\/span><\/span>velocidade vetorial, em rela\u00e7\u00e3o ao solo, \u00e9 a\u00a0<\/b><\/span><\/span>soma vetorial<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>da <\/b><\/span><\/span>velocidade\u00a0<\/b><\/span><\/span>devida \u00e0\u00a0<\/b><\/span><\/span>rota\u00e7\u00e3o\u00a0<\/b><\/span><\/span>na extremidade da h\u00e9lice\u00a0<\/b><\/span><\/span>(V\u2019 = 24 m\/s)\u00a0<\/b><\/span><\/span>com a\u00a0<\/b><\/span><\/span>velocidade de transla\u00e7\u00e3o\u00a0<\/b><\/span><\/span>do avi\u00e3o\u00a0<\/b><\/span><\/span>(V = 2 m\/s), pois ambas s\u00e3o perpendiculares.<\/b><\/span><\/span><\/p>\n

V<\/b><\/span><\/span>R2<\/b><\/span>\u00a0= V<\/b><\/span><\/span>2<\/b><\/span>\u00a0+ V\u2019<\/b><\/span><\/span>2<\/b><\/span>\u00a0<\/b><\/span><\/span>= 2<\/b><\/span><\/span>2<\/b><\/span>\u00a0+ 24<\/b><\/span><\/span>2<\/b><\/span>\u00a0= 4 + 576 = 580\u00a0\"\"\u00a0V<\/b><\/span><\/span>R<\/b><\/span>=\u00a0\"\"\u00a0\"\"\u00a0<\/b><\/span><\/span>V<\/b><\/span><\/span>R<\/b><\/span>\u00a0\"\"\u00a024,1 m\/s<\/b><\/span><\/span>.<\/b><\/span><\/span><\/p>\n

04-<\/b><\/span><\/span><\/span>Pode-se<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>interligar duas ou mais\u00a0 polias atrav\u00e9s de uma correia (figura 1) ou acoplar duas ou mais engrenagens (figura 2)<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

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

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

\"\"<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

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

\"\"<\/p>\n

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

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

\"\"<\/p>\n

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

\"\"<\/p>\n

\"\"<\/p>\n

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

\"\"<\/p>\n

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

RESOLU\u00c7\u00d5ES CINEM\u00c1TICA (2015, 2016 e 2017) MOVIMENTO UNIFORME 01–\u00a0Esse exerc\u00edcio \u00e9 de\u00a0apenas velocidade m\u00e9dia, s\u00f3 precisamos\u00a0tomar cuidado com as transforma\u00e7\u00f5es: 02-\u00a0Considere:\u00a01 n\u00f3 = 1 milha mar\u00edtima\/hora. Como um n\u00f3 equivale \u00e0 1 milha mar\u00edtima\/hora, a embarca\u00e7\u00e3o com velocidade de\u00a010 n\u00f3s, se movimenta em uma hora o equivalente \u00e0 10 milhas. O mesmo racioc\u00ednio para a de\u00a030 n\u00f3s, s\u00f3 que nesse<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":3142,"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-3144","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3144","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=3144"}],"version-history":[{"count":2,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3144\/revisions"}],"predecessor-version":[{"id":11080,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3144\/revisions\/11080"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3142"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=3144"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}