{"id":973,"date":"2015-06-29T01:27:15","date_gmt":"2015-06-29T01:27:15","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=973"},"modified":"2024-08-23T12:27:04","modified_gmt":"2024-08-23T12:27:04","slug":"resolucoes-dos-exercicios-sobre-graficos-do-muv","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/mecanica\/cinematica\/graficos-do-movimento-uniformemente-variado-muv\/resolucoes-dos-exercicios-sobre-graficos-do-muv\/","title":{"rendered":"Resolu\u00e7\u00f5es dos exerc\u00edcios gr\u00e1ficos do MUV"},"content":{"rendered":"

Resolu\u00e7\u00f5es dos exerc\u00edcios sobre gr\u00e1ficos do MUV<\/span><\/span><\/span><\/p>\n

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

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

a) S<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=8m (ponto em que a par\u00e1bola encosta na ordenada \u201ceixo S\u201d, ou seja, ponto quando t=0). Inverte o sentido de seu movimento no instante t<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>=3s (v\u00e9rtice da par\u00e1bola, onde V=0, onde o movimento passa de retr\u00f3grado para progressivo). Passa pela origem dos espa\u00e7os quando S=0, ou seja, nos instantes t=2s e t=4s.<\/b><\/span><\/span><\/span><\/p>\n

b) O movimento \u00e9 progressivo quando o m\u00f3vel se desloca no sentido dos marcos crescentes (V>0) o que ocorre ap\u00f3s t=3s e retr\u00f3grado\u00a0 quando o m\u00f3vel se desloca no sentido dos marcos decrescentes (V<0) o que ocorre entre 0 e 3s.<\/b><\/span><\/span><\/span><\/p>\n

c) O movimento \u00e9\u00a0acelerado\u00a0ap\u00f3s 3s, pois\u00a0a\u00a0e\u00a0V\u00a0tem mesmo sinal (a\u00a0\u00e9 positiva \u201ca concavidade da par\u00e1bola \u00e9 para cima\u201d e\u00a0V\u00a0\u00e9 positiva \u201cse desloca no sentido dos marcos crescentes\u201d).<\/b><\/span><\/span><\/span><\/p>\n

O movimento \u00e9\u00a0retardado\u00a0entre 0 e 3s 3s, pois\u00a0a\u00a0e\u00a0V\u00a0tem sinais contr\u00e1rios (a\u00a0\u00e9 positiva \u201ca concavidade da par\u00e1bola \u00e9 para cima\u201d e\u00a0V\u00a0\u00e9 negativa \u201cse desloca no sentido dos marcos decrescentes\u201d).<\/b><\/span><\/span><\/span><\/p>\n

d) S<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=8m\u00a0 —\u00a0 quando t=2s\u00a0 —\u00a0 S=0\u00a0 —\u00a0 S=S<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0+ V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.t + a.t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 0=8 + V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.2 + a.2<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 2.a + 2.V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0= -8\u00a0\u00a0I\u00a0 —\u00a0\u00a0 quando t=4s\u00a0 —\u00a0 S=0\u00a0 —\u00a0 S=S<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0+ V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.t + a.t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 0=8 + V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.4 + a.4<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 8.a + 4.V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0= -8\u00a0\u00a0II\u00a0\u00a0—\u00a0 Resolvendo o sistema composto por I e II\u00a0 —\u00a0 a=2m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0e V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=-6m\/s\u00a0 — S=S<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0+ V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.t + a.t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0\u00a0S= 8 – 6.t + t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span><\/p>\n

e) V=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0+ a.t\u00a0 —\u00a0\u00a0V= -6 + 2.t\u00a0 —\u00a0gr\u00e1fico VXt —\u00a0 V=0\u00a0 —\u00a0 0= -6 + 2.t\u00a0 —\u00a0 t=3s\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=-6m\/s\u00a0 —\u00a0 como \u00e9 uma fun\u00e7\u00e3o do primeiro grau seu gr\u00e1fico \u00e9 uma reta que pode ser definida por apenas dois pontos.\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

f) A acelera\u00e7\u00e3o \u00e9 constante e vale a=2m\/s<\/b><\/span><\/span><\/span>2\u00a0<\/b><\/span><\/span><\/sup><\/span>e seu gr\u00e1fico \u00e9 uma reta paralela ao eixo dos tempos<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

02- Trecho I \u2013 MUV (ramo de par\u00e1bola) com acelera\u00e7\u00e3o positiva \u201ca>0\u201d (concavidade para cima) e velocidade positiva \u201cV>0\u201d (se desloca no sentido dos marcos crescentes,movimento progressivo)\u00a0 —\u00a0 o movimento \u00e9\u00a0acelerado, pois a e V tem mesmo sinal.<\/b><\/span><\/span><\/span><\/p>\n

Trecho II \u2013 MUV (ramo de par\u00e1bola) com acelera\u00e7\u00e3o negativa \u201ca<0\u201d (concavidade para baixo) e velocidade positiva \u201cV>0\u201d (se desloca no sentido dos marcos crescentes,movimento progressivo)\u00a0 —\u00a0 o movimento \u00e9\u00a0retardado, pois a e V sinais contr\u00e1rios.<\/b><\/span><\/span><\/span><\/p>\n

Trecho III \u2013 MU (reta obl\u00edqua) com\u00a0V constante\u00a0e negativa (se desloca no sentido dos marcos decrescentes, movimento\u00a0retr\u00f3grado).<\/b><\/span><\/span><\/span><\/p>\n

Trecho IV \u2013 MUV (ramo de par\u00e1bola) com acelera\u00e7\u00e3o positiva \u201ca>0\u201d (concavidade para cima) e velocidade negativa \u201cV<0\u201d (se desloca no sentido dos marcos decrescentes,\u00a0movimento retr\u00f3grado)\u00a0 —\u00a0 o movimento \u00e9\u00a0retardado, pois a e V sinais contr\u00e1rios.<\/b><\/span><\/span><\/span><\/p>\n

Trecho IV \u2013 MUV (ramo de par\u00e1bola) com acelera\u00e7\u00e3o positiva \u201ca>0\u201d (concavidade para cima) e velocidade negativa \u201cV<0\u201d (se desloca no sentido dos marcos decrescentes,\u00a0movimento retr\u00f3grado)\u00a0 —\u00a0 o movimento \u00e9\u00a0retardado, pois a e V sinais contr\u00e1rios.<\/b><\/span><\/span><\/span><\/p>\n

Trecho V \u2013 MUV (ramo de par\u00e1bola) com acelera\u00e7\u00e3o positiva \u201ca>0\u201d (concavidade para cima) e velocidade positiva \u201cV>0\u201d (se desloca no sentido dos marcos crescentes,\u00a0movimento progressivo)\u00a0 —\u00a0 o movimento \u00e9\u00a0acelerado, pois a e V sinais contr\u00e1rios.<\/b><\/span><\/span><\/span><\/p>\n

03- A acelera\u00e7\u00e3o \u00e9 m\u00e1xima onde a curva tem maior inclina\u00e7\u00e3o em rela\u00e7\u00e3o \u00e0 horizontal\u00a0 —\u00a0R- A<\/b><\/span><\/span><\/span><\/p>\n

04- V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=4m\/s\u00a0 —\u00a0 a=(V \u2013 V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>)\/(t \u2013 t<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>)=(12 \u2013 4)\/(4 \u2013 0)\u00a0 —\u00a0 a=2m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 V= V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0+ a.t\u00a0 —\u00a0 V=4 + 2.7\u00a0 —\u00a0 V=18m\/s (velocidade no instante t=7s)\u00a0 —\u00a0 a dist\u00e2ncia percorrida corresponde \u00e0 \u00e1rea hachurada do gr\u00e1fico abaixo<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u0394S=(B + b).h\/2=(18 + 4).7\/2\u00a0 —\u00a0\u00a0\u0394S=77m<\/b><\/span><\/span><\/span><\/p>\n

05-\u00a0<\/b><\/span><\/span><\/span>\u0394<\/b><\/span><\/span><\/span><\/span>S=S \u2013 S<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=9 \u2013 0=9m\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=0\u00a0 —\u00a0 V=6m\/s\u00a0 —\u00a0 Torricelli\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0= V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ 2.a.<\/b><\/span><\/span><\/span>\u0394<\/b><\/span><\/span><\/span><\/span>S\u00a0 —\u00a0 36=0 + 2.a.9\u00a0 —\u00a0 a=2m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0\u00a0R- B<\/b><\/span><\/span><\/span><\/p>\n

06- a) Na vertical\u00a0 —\u00a0 \u00a0S<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=0\u00a0 —\u00a0 t=0,3s\u00a0 —\u00a0 S=1,2m\u00a0 —\u00a0 S=S<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0+ V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.t + a.t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0\u00a0 —\u00a0 1,2=0 + V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.0,3 + a,(0,3)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 0,3.V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0+ 0,045.a=1,2\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0+ 0,15.a=4\u00a0\u00a0I\u00a0 —\u00a0 t=1,1s\u00a0 —\u00a0 S=0\u00a0 —\u00a0 S=S<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0+ V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.t + a.t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 0=0 + V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.1,1 + a.(1,1)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 1,1V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0+ a.0,605=0\u00a0\u00a0II\u00a0\u00a0—\u00a0 resolvendo o sistema composto por I e II\u00a0 —\u00a0 a=-10m\/s<\/b><\/span><\/span><\/span>2\u00a0<\/b><\/span><\/span><\/sup><\/span>\u00a0—\u00a0 V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=5,5m\/s\u00a0 —\u00a0 a altura m\u00e1xima ocorre quando V=0 (ela p\u00e1ra para come\u00e7ar a descer) que vale 0,55s pelo gr\u00e1fico\u00a0 — S=S<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0+ V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.t + a.t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 ===\u00a0 S<\/b><\/span><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub><\/span>=0 + 5,5.(0,55) \u2013 10.(0,55)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 S<\/b><\/span><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub><\/span>=3,025 \u2013 1,5125\u00a0 —\u00a0\u00a0S<\/b><\/span><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub><\/span>=1,5125m<\/b><\/span><\/span><\/span><\/p>\n

b) Na horizontal a velocidade \u00e9 constante e vale V<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>=\u0394S\/\u0394t=1,3\/1,1\u00a0 —\u00a0\u00a0V<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>=1,18m\/s<\/b><\/span><\/span><\/span><\/p>\n

c)\u00a0V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=5,5m\/s\u00a0(veja item a)<\/b><\/span><\/span><\/span><\/p>\n

07- V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=0 (v\u00e9rtice da par\u00e1bola \u201cp\u00e1ra para come\u00e7ar a inverter o sentido de seu movimento\u201d)\u00a0 —\u00a0 quanto mais inclinada a curva em rela\u00e7\u00e3o \u00e0 horizontal, maior ser\u00e1 a velocidade, ou seja, V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>>V<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0\u00a0R- B<\/b><\/span><\/span><\/span><\/p>\n

08- a) Ficou parado entre 1h e 1,8h, ou seja, 48minutos, no km100 (veja gr\u00e1fico)<\/b><\/span><\/span><\/span><\/p>\n

b) V<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>=\u0394S\/\u0394t=(120 \u2013 0)\/(3 \u2013 0)\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>=40km\/h (lembre-se de que o tempo de parada est\u00e1 sempre inclu\u00eddo)<\/b><\/span><\/span><\/span><\/p>\n

09- Os gr\u00e1ficos n\u00e3o podem se referir ao mesmo movimento; se a acelera\u00e7\u00e3o \u00e9 uma constante negativa, a velocidade \u00e9 uma reta com inclina\u00e7\u00e3o negativa, ou seja, est\u00e1 diminuindo. Logo, a fun\u00e7\u00e3o\u00a0posi\u00e7\u00e3o x(t)\u00a0s\u00f3 pode ser representada por uma\u00a0par\u00e1bola com concavidade para baixo, ao contr\u00e1rio do que est\u00e1 mostrado.<\/b><\/span><\/span><\/span><\/p>\n

10- a)Os deslocamentos de cada carro s\u00e3o fornecidos pelas \u00e1reas hachuradas nos gr\u00e1ficos abaixo:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u0394<\/b><\/span><\/span><\/span><\/span>S<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=b.h=15.30=450m\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>\u0394<\/b><\/span><\/span><\/span><\/span>S<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=b.h\/2=10.40\/2=200m\u00a0 —\u00a0 d=450 \u2013 200\u00a0 —\u00a0\u00a0d=250m<\/b><\/span><\/span><\/span><\/p>\n

b) Se encontram no instante t em que efetuaram o mesmo deslocamento, ou seja, a \u00e1rea em cada gr\u00e1fico deve ser a mesma:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

\u0394S<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=b.h=t.30=30.t\u00a0 —\u00a0 \u0394S<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=(B + b).h\/2=(t \u2013 5 + t \u2013 15).40\/2=40t \u2013 400\u00a0 —\u00a0 no encontro, percorreram a mesma dist\u00e2ncia\u00a0 —\u00a0 \u0394S<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=\u0394S<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 30t=40t \u2013 400\u00a0 —\u00a0\u00a0t=40s<\/b><\/span><\/span><\/span><\/p>\n

11- a) \u0394S=soma das \u00e1reas=3.8\/2 + (4 + 2).12\/2 + 2.12\/2\u00a0 —\u00a0\u00a0\u0394S=60m<\/b><\/span><\/span><\/span><\/p>\n

b) V<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>= \u0394S\/ \u0394t=60\/15\u00a0 —\u00a0\u00a0V<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>=4m\/s<\/b><\/span><\/span><\/span><\/p>\n

12- Entre 0 e t<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 velocidade aumentando, acelera\u00e7\u00e3o positiva e concavidade da par\u00e1bola para cima\u00a0 —\u00a0 entre t<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0e t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 velocidade constante, acelera\u00e7\u00e3o nula, movimento uniforme com reta ascendente e movimento progressivo\u00a0 —\u00a0 entre t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0e t<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 velocidade diminuindo, acelera\u00e7\u00e3o negativa e par\u00e1bola com concavidade para baixo\u00a0 —\u00a0\u00a0R- D<\/b><\/span><\/span><\/span><\/p>\n

13- Trata-se de uma queda livre com a=-g (acelera\u00e7\u00e3o da gravidade) com concavidade para baixo\u00a0 e a velocidade a partir de zero e aumentando em m\u00f3dulo (movimento retr\u00f3grado acelerado)\u00a0 —\u00a0\u00a0R- B<\/b><\/span><\/span><\/span><\/p>\n

14- Entre 0 e 4s\u00a0 —\u00a0 a=4m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 V=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0+ at\u00a0 —\u00a0 V=0 =4.4\u00a0 —\u00a0 V=16m\/s\u00a0 —\u00a0 entre 4s (onde V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=16m\/s) e 8s\u00a0 —\u00a0 V=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0+ at\u00a0 —\u00a0 V=16 -2.4\u00a0 —\u00a0\u00a0V=8ms<\/b><\/span><\/span><\/span><\/p>\n

15- Durante 1s ele se move com velocidade constante de 20m\/s e depois, sua velocidade diminui at\u00e9 zero segundo uma reta, pois se trata de um MUV com desacelera\u00e7\u00e3o constante\u00a0 —\u00a0\u00a0R- D<\/b><\/span><\/span><\/span><\/p>\n

16- Parte do repouso\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=0\u00a0 —\u00a0 entre 0 e 2s a acelera\u00e7\u00e3o \u00e9 positiva e vale 0,5m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 V=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0+ at=0 + 0,5.2=1m\/s e sua velocidade variade 0 a 1m\/s\u00a0 — ntre 2s e 4s, a acelera\u00e7\u00e3o \u00e9 nula e ele segue em movimento uniforme com velocidade constante de 1m\/s\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

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

17- entre 0 e 10s\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=0\u00a0 —\u00a0 V=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0+ at=0 + 1.10=10m\/s\u00a0 —\u00a0 entre 10s e 20s\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=10m\/s\u00a0 —\u00a0 V=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0+ at=10 + 2.10=30m\/s\u00a0 —\u00a0 entre 20s e 50s\u00a0 —\u00a0 a acelera\u00e7\u00e3o \u00e9 nula e a velocidade constante de 30m\/s\u00a0 —\u00a0 entre 50s e t ele freia e p\u00e1ra com sua velocidade variando de 30m\/s para 0\u00a0 —\u00a0 V=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0+ at\u00a0 —\u00a0 0=30 -1.t —\u00a0 t=30s\u00a0 —\u00a0\u00a0t<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>=50 + 30=80s<\/b><\/span><\/span><\/span><\/p>\n

\u00a0Construindo o gr\u00e1fico Vxt<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u0394S<\/b><\/span><\/span><\/span>total<\/b><\/span><\/span><\/sub><\/span>=soma das \u00e1reas=10.10\/2 + (30 + 10).10\/2 + 30.30 + 30.30\/2\u00a0 —\u00a0\u0394S<\/b><\/span><\/span><\/span>total<\/b><\/span><\/span><\/sub><\/span>=1.600m\u00a0 —\u00a0 R- A<\/b><\/span><\/span><\/span><\/p>\n

18- Colocando a origem da trajet\u00f3ria vertical no ponto de lan\u00e7amento e orientando-a para baixo, na subida o movimento \u00e9 retr\u00f3grado retardado e na descida progressivo acelerado (observe atentamente figuras abaixo)<\/b><\/span><\/span><\/span><\/p>\n

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

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

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

19- Veja explica\u00e7\u00e3o baseada nos gr\u00e1ficos abaixo:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

20-\u00a0<\/b><\/span><\/span>R- B\u00a0 —\u00a0 a acelera\u00e7\u00e3o do m\u00f3vel A \u00e9 positiva e diferente de zero e o m\u00f3vel B est\u00e1 em movimento uniforme com velocidade constante e acelera\u00e7\u00e3o nula.<\/b><\/span><\/span><\/span><\/p>\n

21-\u00a0<\/b><\/span><\/span>O v\u00e9rtice da par\u00e1bola \u00e9 o ponto onde ocorre a invers\u00e3o do sentido do movimento. Observe que no gr\u00e1fico existem tr\u00eas v\u00e9rtices\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

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

22-\u00a0<\/b><\/span><\/span>1 \u2013 V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>>0\u00a0 —\u00a0 movimento progressivo (se move no sentido dos marcos crescentes) e a<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span><0 (par\u00e1bola com concavidade para baixo)<\/b><\/span><\/span><\/span><\/p>\n

2 \u2013 V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span><0\u00a0 —\u00a0 movimento retr\u00f3grado (se move no sentido dos marcos decrescentes) e a<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span><0 (par\u00e1bola com concavidade para baixo)<\/b><\/span><\/span><\/span><\/p>\n

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

23-\u00a0<\/b><\/span><\/span>Veja pela resolu\u00e7\u00e3o anterior que, nesse intervalo de tempo, V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00e9 sempre positiva e V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>sempre negativa e, assim, elas n\u00e3o se igualam\u00a0 —\u00a0\u00a0R- E<\/b><\/span><\/span><\/span><\/p>\n

24-\u00a0<\/b><\/span><\/span>S<\/b><\/span><\/span><\/span>oII<\/b><\/span><\/span><\/sub><\/span>=0 e supondo V<\/b><\/span><\/span><\/span>oII<\/b><\/span><\/span><\/sub><\/span>=0 a equa\u00e7\u00e3o do m\u00f3vel II, que \u00e9 um MUV \u00e9 S=S<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0+ V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.t + at<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 t=15s\u00a0 —\u00a0 S=225m\u00a0 —\u00a0 225=0 + 0 + a.15<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 a=2m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 V=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0+ at=0 +2.15\u00a0 — V=30m\/s\u00a0 —\u00a0\u00a0R- D<\/b><\/span><\/span><\/span><\/p>\n

25-\u00a0Entre 0 e 2s\u00a0 —\u00a0 a=(V \u2013 V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>)\/(t \u2013 t<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>)=(14 \u2013 26)\/(2 \u2013 0)\u00a0 —\u00a0 a=-6m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 ap\u00f3s 2s, V \u00e9 constante e assim, a=0\u00a0 —\u00a0\u00a0R- A<\/b><\/span><\/span><\/span><\/p>\n

26-\u00a0V=V<\/b><\/span><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub><\/span>\u00a0+ at\u00a0 —\u00a0 20=0 + a.10\u00a0 —\u00a0 a=2,0m\/s<\/b><\/span><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span><\/span>\u0394<\/b><\/span><\/span><\/span>S=area=b.h\/2=5.10\/2=25m\u00a0 —\u00a0\u00a0R- C<\/b><\/span><\/span><\/span><\/span><\/p>\n

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

a) O gr\u00e1fico solicitado entre t = 0 e t = 10 s.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

b) Se x = 5 + 16.t – 2.t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0\u00a0 v = 16 – 4.t\u00a0 —\u00a0 v = 16 – 4.4 = 16 – 16 = 0 m\/s<\/b><\/span><\/span><\/span><\/p>\n

c) t=o\u00a0 —\u00a0 S<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=5m\u00a0 —\u00a0 t=5s\u00a0 —\u00a0 S=5 + 16.5 -2.25\u00a0 —\u00a0 S=35m\u00a0 —\u00a0 deslocamento\u00a0 —\u00a0 \u0394S=S \u2013 S<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=35 \u2013 5=30m\u00a0 —\u00a0 dist\u00e2ncia percorrida d\u00a0 —\u00a0 at\u00e9 parar ele demora\u00a0 — V=Vo +a.t\u00a0 —\u00a0 0=16 \u2013 4t\u00a0 —\u00a0 t=4s e percorre se desloca at\u00e9 o marco\u00a0 —\u00a0 S=5 +16.4 -2.16=37m\u00a0 —\u00a0 entre 4s e 5s ele retorna ao marco S=35m<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

d ida=(37 \u2013 5)=32m + 2 (volta)\u00a0 —\u00a0\u00a0d=34m<\/b><\/span><\/span><\/span><\/p>\n

28- a) Parte do repouso\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=0\u00a0 —\u00a0 entre 0 e 20s\u00a0 —\u00a0 V=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0+ at\u00a0 —\u00a0 V=0 + 2.20\u00a0 —\u00a0 V=40m\/s\u00a0 —\u00a0 entre 20s e 50s\u00a0 —\u00a0 V=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0+ at=40 \u2013 1.30\u00a0 —\u00a0 V=10m\/s\u00a0 —\u00a0 veja gr\u00e1fico abaixo<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

b) A dist\u00e2ncia percorrida \u00e9 fornecida pela \u00e1rea hachurada da figura abaixo<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

d=b.h\/2 + (B + b).h\/2=20.40\/2 + (40 + 10).30\/2\u00a0 —\u00a0\u00a0d=1.150m<\/b><\/span><\/span><\/span><\/p>\n

29-\u00a0Sendo o movimento acelerado o gr\u00e1fico posi\u00e7\u00e3o x tempo \u00e9 um arco de par\u00e1bola com concavidade para cima e a velocidade aumenta de modo uniforme, assim o gr\u00e1fico velocidade x tempo \u00e9 uma reta com inclina\u00e7\u00e3o ascendente\u00a0 —\u00a0\u00a0R- D<\/b><\/span><\/span><\/span><\/p>\n

30-\u00a0Entre 0 e 6s\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>I<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0+ a<\/b><\/span><\/span><\/span>I<\/b><\/span><\/span><\/sub><\/span>t<\/b><\/span><\/span><\/span>I<\/b><\/span><\/span><\/sub><\/span>=2 + 4.6\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>=26 cm\/s\u00a0 — \u00a0entre 6s e 10s\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>I<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=26cm\/s\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>II<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0+ a<\/b><\/span><\/span><\/span>II<\/b><\/span><\/span><\/sub><\/span>t<\/b><\/span><\/span><\/span>II<\/b><\/span><\/span><\/sub><\/span>=26 + (-3).4\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

V<\/b><\/span><\/span><\/span>II<\/b><\/span><\/span><\/sub><\/span>=14 cm\/s\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>II<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=14 cm\/s\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>III<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0+ a<\/b><\/span><\/span><\/span>III<\/b><\/span><\/span><\/sub><\/span>.t<\/b><\/span><\/span><\/span>III<\/b><\/span><\/span><\/sub><\/span>=14 + 4.6\u00a0 —\u00a0\u00a0V<\/b><\/span><\/span><\/span>III<\/b><\/span><\/span><\/sub><\/span>=38 cm\/s<\/b><\/span><\/span><\/span><\/p>\n

31-\u00a001) Correta\u00a0 —\u00a0 no gr\u00e1fico representado a seguir, seja k o coeficiente angular (declividade) da reta secante \u00e0 curva entre os
\n\"\"<\/b><\/span><\/span><\/span><\/p>\n

instantes 0 e t\u2019\u00a0 —\u00a0 k=tg\u03b1=(X<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 X<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>)\/(t\u2019 \u2013 0)=V<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>=\u0394X\/\u0394t.
\n02) Falsa\u00a0 —\u00a0 no diagrama D, no intervalo considerado, a declividade da reta tangente \u00e0 curva dada est\u00e1 aumentando (em m\u00f3dulo) logo o m\u00f3dulo da velocidade est\u00e1 aumentando\u00a0 —\u00a0 portanto, o movimento \u00e9 acelerado.<\/b><\/span><\/span><\/span><\/p>\n

04) Falsa\u00a0 —\u00a0 no instante considerado, o m\u00f3vel est\u00e1 numa posi\u00e7\u00e3o negativa (antes da origem).<\/b><\/span><\/span><\/span><\/p>\n

08) Falsa\u00a0 —\u00a0 no intervalo considerado, o corpo est\u00e1 se deslocando para posi\u00e7\u00f5es cada vez mais negativas, portanto afastando-se da origem (em movimento retr\u00f3grado retardado).<\/b><\/span><\/span><\/span><\/p>\n

16) Correta\u00a0 —\u00a0 nesse instante, a reta tangente \u00e0 curva \u00e9 horizontal, tendo declividade (coeficiente anular) nula.<\/b><\/span><\/span><\/span><\/p>\n

32) Falsa\u00a0 —\u00a0 nesse intervalo, em m\u00f3dulo, o coeficiente angular da reta tangente est\u00e1 diminuindo, logo o movimento \u00e9 uniformemente retardado.<\/b><\/span><\/span><\/span><\/p>\n

64) Correta\u00a0 —\u00a0 pode corresponder a um lan\u00e7amento vertical para cima com trajet\u00f3ria orientada para baixo (v<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0< 0), sendo o ponto de lan\u00e7amento acima do plano de refer\u00eancia (x<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0< 0), como indicado no esquema a seguir.
\n\"\"
\nR- (01 + 16 + 64) = 81<\/b><\/span><\/span><\/span><\/p>\n

32-\u00a0Trata-se de um gr\u00e1fico de acelera\u00e7\u00e3o\u00a0\u00b4\u00a0tempo\u00a0 —\u00a0 analisando-o voc\u00ea pode afirmar que a acelera\u00e7\u00e3o \u00e9 constante e n\u00e3o nula nos intervalos C e G e nula no intervalo E, onde a velocidade \u00e9 constante\u00a0 —\u00a0\u00a0R- A<\/b><\/span><\/span><\/span><\/p>\n

33-\u00a0Sendo a trajet\u00f3ria \u00e9 retil\u00ednea, a acelera\u00e7\u00e3o restringe-se \u00e0 componente tangencial (\"\"), que, em m\u00f3dulo, \u00e9 igual a acelera\u00e7\u00e3o escalar (a), dada pela taxa de varia\u00e7\u00e3o da velocidade (Dv) em rela\u00e7\u00e3o ao tempo (Dt)\u00a0 —\u00a0 a=\u0394V\/\u0394t:<\/b><\/span><\/span><\/span><\/p>\n

I. a<\/b><\/span><\/span><\/span>I<\/b><\/span><\/span><\/sub><\/span>\u00a0=\u00a0\"\"\u00a0\u00a0\u00a0\u00de\u00a0\u00a0 a<\/b><\/span><\/span><\/span>I<\/b><\/span><\/span><\/sub><\/span>\u00a0= 10 m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

II. a<\/b><\/span><\/span><\/span>II<\/b><\/span><\/span><\/sub><\/span>\u00a0= 0 (n\u00e3o houve varia\u00e7<\/b><\/span><\/span>\u0645<\/span><\/span><\/span><\/span><\/span>o da velocidade)<\/b><\/span><\/span><\/span><\/p>\n

III. a<\/b><\/span><\/span><\/span>III<\/b><\/span><\/span><\/sub><\/span>\u00a0=\u00a0\"\"\u00a0\u00a0\u00de\u00a0 a<\/b><\/span><\/span><\/span>III<\/b><\/span><\/span><\/sub><\/span>\u00a0= \u2013<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub><\/span>5 m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>.\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\u00a0R- C<\/b><\/span><\/span><\/span><\/p>\n

34-\u00a0Nos intervalos de tempo em que a velocidade escalar \u00e9 constante (1 s a 2 s; 3 s a 4 s e 5 s a 6 s) a acelera\u00e7\u00e3o escalar \u00e9 nula\u00a0 —\u00a0 nos intervalos 0 a 1 s; 2 s a 3 s; 4 s a 5 s e 6 s a 7 s, a velocidade varia linearmente com o tempo, sendo, ent\u00e3o, a acelera\u00e7\u00e3o escalar \u00e9 constante.<\/b><\/span><\/span><\/span><\/p>\n

De 0 a 1 s: a =\u00a0\"\"\u00a0\u00a0—\u00a0 de 2 s a 3 s: a =\u00a0\"\"m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>;\u00a0 —\u00a0 de 4 s a 5 s: a =\u00a0\"\"m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>;\u00a0 —\u00a0 De 6 s a 7 s:<\/b><\/span><\/span><\/span><\/p>\n

\u00a0a =\u00a0\"\"m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

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

35-\u00a0Analisando cada intervalo:<\/b><\/span><\/span><\/span><\/p>\n

\u2013 De 0 a 3 s: o movimento \u00e9 uniformemente acelerado; a acelera\u00e7\u00e3o escalar \u00e9\u00a0 —\u00a0 a<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0= \u0394V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/\u0394t<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=8\/3=2,7m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 o espa\u00e7o percorrido \u00e9 calculado pela \u201c\u00e1rea\u201d de 0 a 3 s\u00a0 —\u00a0 \u0394S<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=3.8\/2\u00a0 —\u00a0\u0394S<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=12m.<\/b><\/span><\/span><\/span><\/p>\n

\u2013 De 3 s a 5 s: o movimento \u00e9 uniforme, com velocidade escalar\u00a0v<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0= 8 m\/s\u00a0 —\u00a0 o espa\u00e7o percorrido \u00e9\u00a0 —\u00a0\u00a0\u00a0DS<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0= v<\/b><\/span><\/span><\/span>2\u00a0<\/b><\/span><\/span><\/sub><\/span>Dt<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=8<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub><\/span>\u00b4<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/span><\/p>\n

DS<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=16 m.<\/b><\/span><\/span><\/span><\/p>\n

\u2013 De 5 s s 7 s: o movimento \u00e9 uniformemente retardado; a acelera\u00e7\u00e3o escalar \u00e9\u00a0 —\u00a0 a<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>=(0 \u2013 8)\/(7 \u2013 5)\u00a0 —\u00a0 a<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>=-4m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0\u00a0\u0394S<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>=2.8\/2\u00a0 —\u00a0\u0394S<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>=8m<\/b><\/span><\/span><\/span><\/p>\n

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

36-\u00a0I. C\u00e1lculo do deslocamento entre 0 e 8s pela \u00e1rea do tri\u00e2ngulo\u00a0 —\u00a0 \u0394S<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=b.h\/2=8.80\/2\u00a0 —\u00a0 \u0394S<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=320m\u00a0 —\u00a0 n\u00e3o completou a volta, pois \u0394S<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0< 400m\u00a0 —\u00a0 Falsa.<\/b><\/span><\/span><\/span><\/p>\n

II. C\u00e1lculo do deslocamento entre 0 e 9s pela soma das \u00e1reas do tri\u00e2ngulo com a do ret\u00e2ngulo\u00a0 —\u00a0 \u0394S<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=320 + b.h=320 + 1.80\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\u0394S<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=400m\u00a0 —\u00a0 em 9s o piloto completou uma volta\u00a0 —\u00a0 Correta.<\/b><\/span><\/span><\/span><\/p>\n

III. Entre 8s e 10s, o movimento \u00e9 retil\u00edneo e uniforme com velocidade constante de 80m\/s e consequentemente a for\u00e7a resultante \u00e9 nula\u00a0 —\u00a0 Correta.<\/b><\/span><\/span><\/span><\/p>\n

IV. A componente da for\u00e7a resultante na dire\u00e7\u00e3o do movimento \u00e9 a tangencial de intensidade\u00a0 —\u00a0 FR=ma=m\u0394V\/\u0394t\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=m.60\/2\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=30m\u00a0 —\u00a0 P=mg=m10\u00a0 —\u00a0 P=10m\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>\/P=30m\/10m\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=3p\u00a0 —\u00a0 Correta.<\/b><\/span><\/span><\/span><\/p>\n

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

37-\u00a0a=(V – V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>)\/(t \u2013 t<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>)=(-20 \u2013 20)\/(2 \u2013 0)\u00a0 —\u00a0 a=-20m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0\u00a0R- A<\/b><\/span><\/span><\/span><\/p>\n

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

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

\u00a0-1. Correta\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>=\u2206S\/\u2206t=(600 \u2013 200)\/4\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>=400\/4=100m\/min=0,1km\/(1\/60)h\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>=0,1×60=6km\/h.<\/b><\/span><\/span><\/span><\/p>\n

2. Correta\u00a0 —\u00a0 a dist\u00e2ncia n\u00e3o variou entre 6min e 8min, ou seja, durante 2 minutos.<\/b><\/span><\/span><\/span><\/p>\n

3. Correta\u00a0 —\u00a0 no eixo da dist\u00e2ncia (vertical)\u00a0 —\u00a0 d=(1400 \u2013 200)= 1200m<\/b><\/span><\/span><\/span><\/p>\n

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

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

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

39-Primeira etapa\u00a0 —\u00a0 queda livre no v\u00e1cuo (durante 5,0s) com velocidade variando de V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0at\u00e9 V , com a=g=10m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 V=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0+ a.t\u00a0 —\u00a0 V=0 + 10.5\u00a0 —\u00a0 V=50m\/s\u00a0 —\u00a0 durante esse tempo ele caiu uma altura h=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>t + at<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=0 + 10.25\/2\u00a0 —\u00a0 h=125m\u00a0 —\u00a0 observe no gr\u00e1fico que no instante t=5,0s o p\u00e1ra-quedas abriu sua velocidade caiu instantaneamente de 50m\/s para 10m\/s\u00a0 —\u00a0 no intervalo de tempo (t \u2013 5)s ele percorreu, com velocidade constante de V=10m\/s a a altura h\u2019=325 \u2013 125=200m\u00a0 —\u00a0 V=h\u00b4\/(t \u2013 5)\u00a0 —\u00a0 10= 200\/(t \u2013 5)\u00a0 —\u00a0 10t \u2013 50=200\u00a0 —\u00a0 t=250\/10\u00a0 —\u00a0 t=25s\u00a0 —\u00a0\u00a0R- B\u00a0\u00a0—\u00a0 voc\u00ea poderia tamb\u00e9m resolver pela \u00e1rea\u00a0 —\u00a0 h<\/b><\/span><\/span><\/span>total<\/b><\/span><\/span><\/sub><\/span>=325=\u00e1rea do tri\u00e2ngulo + \u00e1rea do ret\u00e2ngulo=5.50\/2 +<\/b><\/span><\/span><\/span><\/p>\n

(t \u2013 5).10\u00a0 —\u00a0 t=25s.<\/b><\/span><\/span><\/span><\/p>\n

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

40-\u00a0\u00a0Como partem juntos (instante t<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>), eles se encontrar\u00e3o ap\u00f3s sofrerem o mesmo deslocamento \u2206S\u00a0 —\u00a0 em todo gr\u00e1fico Vxt o deslocamento \u00e9 numericamente igual \u00e0 \u00e1rea entre a reta representativa e o tempo\u00a0\u00a0 —\u00a0 observe que quando o tempo \u00e9 t<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sub><\/span>\u00a0as \u00e1reas<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

s\u00e3o iguais\u00a0 —\u00a0 carro B\u00a0 —\u00a0 \u00e1rea=\u2206S<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=b.h=4.2=8 unidades\u00a0 —\u00a0 carro P\u00a0 —\u00a0 \u00e1rea=\u2206S<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/sub><\/span>=b.h\/2=4.4\/2=8 unidades\u00a0 —\u00a0\u00a0R- D<\/b><\/span><\/span><\/span><\/p>\n

 <\/p>\n

Voltar para os exerc\u00edcios<\/span><\/a><\/h2>\n","protected":false},"excerpt":{"rendered":"

Resolu\u00e7\u00f5es dos exerc\u00edcios sobre gr\u00e1ficos do MUV \u00a0 01- a) So=8m (ponto em que a par\u00e1bola encosta na ordenada \u201ceixo S\u201d, ou seja, ponto quando t=0). Inverte o sentido de seu movimento no instante ti=3s (v\u00e9rtice da par\u00e1bola, onde V=0, onde o movimento passa de retr\u00f3grado para progressivo). Passa pela origem dos espa\u00e7os quando S=0, ou seja, nos instantes t=2s<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":969,"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-973","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/973","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=973"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/973\/revisions"}],"predecessor-version":[{"id":10794,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/973\/revisions\/10794"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/969"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=973"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}