{"id":993,"date":"2015-06-29T01:51:49","date_gmt":"2015-06-29T01:51:49","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=993"},"modified":"2024-08-23T12:23:38","modified_gmt":"2024-08-23T12:23:38","slug":"resolucao-comentada-dos-exercicios-sobre-muv","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/mecanica\/cinematica\/funcao-equacao-horaria-do-espaco-de-um-muv\/resolucao-comentada-dos-exercicios-sobre-muv\/","title":{"rendered":"Resolu\u00e7\u00e3o comentada MUV"},"content":{"rendered":"

Resolu\u00e7\u00e3o comentada dos exerc\u00edcios sobre MUV<\/span><\/span><\/span><\/p>\n

\u00a001- a)<\/p>\n

\"\"<\/p>\n

b) Quando ele muda o sentido se seu movimento ele p\u00e1ra (V=0) e, a partir desse instante, o movimento que era progressivo se torna retr\u00f3grado\u00a0 —\u00a0 V=-20 + 5t\u00a0 —\u00a0 0=-20 + 5t\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0t=4s<\/b><\/span><\/span>\u00a0(instante em que ele p\u00e1ra para inverter o sentido do movimento)<\/b><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

d) 2s \u2013 retr\u00f3grado retardado\u00a0 —\u00a0 10s – progressivo acelerado\u00a0 —\u00a0 Veja esquema acima)<\/b><\/span><\/span><\/p>\n

02- a) V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=-8m\/s\u00a0 —\u00a0 a=(8 \u2013 (-8))\/(4 \u2013 0)\u00a0 —\u00a0 a=16\/4\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0a=4m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

b) V= V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0+ at\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>V=-8 + 4t\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0inverte o sentido (p\u00e1ra) \u2013 V=0\u00a0 —\u00a0 0=-8 + 4t\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0t=2s<\/b><\/span><\/span>ou pelo gr\u00e1fico que corresponde ao ponto onde a reta intercepta o eixo t.<\/b><\/span><\/span><\/p>\n

c) entre 0 e 2s\u00a0 —\u00a0 retr\u00f3grado (V<0) e retardado (m\u00f3dulo de V est\u00e1 diminuindo)\u00a0 —\u00a0 ap\u00f3s 2s\u00a0 —\u00a0 progressivo (V>0) e acelerado (m\u00f3dulo de V est\u00e1 aumentando)<\/b><\/span><\/span><\/p>\n

d) \u0394S=\u00e1rea entre 0 e 4s, que corresponde \u00e0 soma das \u00e1reas hachuradas da figura abaixo<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u0394<\/b><\/span><\/span>S=b.h\/ + b.h\/2=2.(-8)\/2 + 8.2\/2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>\u0394<\/b><\/span><\/span>S=0<\/b><\/span><\/span><\/span><\/p>\n

03- V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub>=2m\/s\u00a0 —\u00a0 a=1ms<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup>\u00a0 —\u00a0 V=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub>\u00a0+ at\u00a0 —\u00a0 V=2 + 1.8\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0V=10m\/s<\/b><\/span><\/span><\/span><\/p>\n

04- \u0394S=\u00e1rea=(10 + 6).12,5\/2\u00a0 —\u00a0 \u0394S=100m\u00a0 —\u00a0 V<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=\u0394S\/\u0394t\u00a0 —\u00a0 V<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=10m\/s\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- B<\/b><\/span><\/span><\/p>\n

05- A \u2013 \u0394S<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=\u00e1rea do trap\u00e9zio=100m<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u0394S<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=(t<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0+ (t<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0\u2013 4)).11\/2\u00a0 —\u00a0 100=2t<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0\u2013 4).11\/2\u00a0 —\u00a0 t<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=244\/22\u00a0 —\u00a0 t<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=11,1s (treino A)<\/b><\/span><\/span><\/p>\n

B \u2013 \u0394S<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=\u00e1rea do trap\u00e9zio=100m<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u0394S<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=(t<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0+ (t<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0\u2013 3)).10\/2\u00a0 —\u00a0 100=10t<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0\u2013 15\u00a0 —\u00a0 t<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=115\/10\u00a0 —\u00a0 t<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=11,5s (treino B)<\/b><\/span><\/span><\/p>\n

\u0394<\/b><\/span><\/span><\/span>t=t<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0\u2013 t<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=0,4s\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- B<\/b><\/span><\/span><\/p>\n

06- a) V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=0\u00a0 —\u00a0 a=(V \u2013 V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>)\/(t \u2013 t<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>)=10\/5\u00a0 —\u00a0 a=2m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 V = V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0+ a.t\u00a0 —\u00a0 20=0 + 2t\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0t=10s<\/b><\/span><\/span><\/p>\n

b) o deslocamento \u00e9 fornecido pela \u00e1rea<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u0394S=b.h\/2=10.20\/2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>\u0394S=100m<\/b><\/span><\/span><\/p>\n

07- T=5s\u00a0 —\u00a0 V=50 \u2013 10.5\u00a0 —\u00a0 V=0\u00a0 —\u00a0 a=-10m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0(constante)\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- C<\/b><\/span><\/span><\/p>\n

08- V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=10m\/s\u00a0 —\u00a0 a=\u0394V\/\u0394t=(60m\/s)\/60s\u00a0 —\u00a0 a=1m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 V=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0+ at=10 + 1.30\u00a0 —<\/b><\/span><\/span>\u00a0V=40m\/s<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u0394S=(B + b).h\/2=(40 + 10).30\/2\u00a0 —\u00a0 \u0394S=750m\u00a0 —\u00a0 V<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>= \u0394S\/\u0394t=750\/30\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>V<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=25m\/s<\/b><\/span><\/span><\/p>\n

09- I \u2013 Falsa \u2013 \u00e9 no sentido contr\u00e1rio ao do movimento \u2013 a velocidade est\u00e1 diminuindo<\/b><\/span><\/span><\/p>\n

II \u2013<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u0394S=(15 + 11).2\/2=26m\u00a0 —\u00a0 Falsa<\/b><\/span><\/span><\/p>\n

III \u2013 a=(9 – 15)\/3=-2m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 Falsa<\/b><\/span><\/span><\/p>\n

iV \u2013 Correta \u2013 observe que \u00e0 medida que a velocidade diminui o tempo aumenta<\/b><\/span><\/span><\/p>\n

V \u2013 V= V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0+ at\u00a0 —\u00a0 0=15 \u2013 2t\u00a0 —\u00a0 t=7,5s\u00a0 —\u00a0 Correta<\/b><\/span><\/span><\/p>\n

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

10- entre 0 e 5s\u00a0 —\u00a0 V= V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0+ at=0 + 4.5\u00a0 —\u00a0 V<\/b><\/span><\/span>5<\/b><\/span><\/span><\/sub>=20m\/s\u00a0 —\u00a0 de 5s a 8s V vale 20m\/s\u00a0 —\u00a0 a partir de 8s \u2013 a=-2m\/s<\/b><\/span><\/span>2\u00a0<\/b><\/span><\/span><\/sup>at\u00e9 ele parar V=0\u00a0 — V=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0+ at\u00a0 —\u00a0 0=20 -2t\u00a0 —\u00a0 t=10s (de 8s at\u00e9 ele parar)\u00a0 —\u00a0 desde o inicio do movimento\u00a0 — t=8 + 10=18s\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- E<\/b><\/span><\/span><\/p>\n

11- \u0394S=\u00e1rea hachurada da figura abaixo<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u0394S=(12 + 4).7\/2 =\u00a0<\/b><\/span><\/span>56m<\/b><\/span><\/span><\/p>\n

12- a) O \u00e2ngulo \u03b1 que a reta representativa da velocidade forma com um eixo horizontal \u00e9 tal que tg\u03b1=\u0394V\/\u0394t corresponde \u00e0 acelera\u00e7\u00e3o do<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

m\u00f3vel, pois a= \u0394V\/ \u0394t e \u00e9 denominada\u00a0<\/b><\/span><\/span>coeficiente angular<\/b><\/span><\/span>\u00a0da reta ou\u00a0<\/b><\/span><\/span>declividade da reta<\/b><\/span><\/span>. Observe que, se \u03b1 \u00e9 agudo, f(t) \u00e9 crescente e a>0 e se \u03b1 \u00e9 obtuso , f(t) \u00e9 decrescente e a<0.<\/b><\/span><\/span><\/p>\n

b) Em todo gr\u00e1fico VXt a \u00e1rea entre a reta representativa e o eixo dos tempos \u00e9 numericamente igual \u00e0 varia\u00e7\u00e3o de espa\u00e7o \u0394S, entre dois instantes quaisquer t<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0e t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/p>\n

\"\"<\/p>\n

 <\/p>\n

13- a) \u0394S=\u00e1rea total=b.h\/2 + (B + b).h\/2 + b.h\/2=3.8\/2 + (4 + 2).12\/2 + 2.12\/2=1260\u00a0 —\u00a0 12 + 36 + 12=60m\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>\u0394S=60m<\/b><\/span><\/span><\/p>\n

b) V<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=60\/16=3,75\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0V<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=3,75m<\/b><\/span><\/span><\/p>\n

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

15- Quando t=0 a dist\u00e2ncia entre eles \u00e9 de 32m\u00a0 —\u00a0 quando t=4s\u00a0 —\u00a0 \u0394S<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=\u00e1rea do trap\u00e9zio=(30 + 15).4\/2\u00a0\u00a0 —\u00a0 \u0394S<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=90m\u00a0 —\u00a0 \u0394S<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=\u00e1rea do ret\u00e2ngulo=4.15=60m\u00a0 —\u00a0 antes \u2013 \u0394S<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=32m\u00a0 —\u00a0 depois \u2013 \u0394S<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>=(90 \u2013 60)=30m\u00a0 —\u00a0 a dist\u00e2ncia entre eles no final da frenagem ser\u00e1 de d=32 \u2013 30=2m\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- B<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/p>\n

16- \u0394S<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=b.h\/2=6.20\/2=<\/b><\/span><\/span>60m<\/b><\/span><\/span>\u00a0 —\u00a0 \u0394S<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=B + b).h\/2=(8 + 2).10\/2=<\/b><\/span><\/span>50m\u00a0 —\u00a0<\/b><\/span><\/span>\u0394S<\/b><\/span><\/span>C<\/b><\/span><\/span><\/sub>=b.h\/2=4.20\/2=<\/b><\/span><\/span>40m\u00a0 —\u00a0 R- A<\/b><\/span><\/span><\/p>\n

17-\u00a0<\/b><\/span><\/span>\u0394<\/b><\/span><\/span><\/span>S=1.200=\u00e1rea do trap\u00e9zio=(B + b).h\/2=(90 + (90 – t)).20\/2\u00a0 —\u00a0 1.200=(180 \u2013 t).10\u00a0 —\u00a0 1200=1800 -10t\u00a0 —\u00a0 t=60s\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- C\u00a0<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/p>\n

18- Trata-se de um MUV em que a velocidade aumenta de 2,3m\/s em cada 1s e, assim, sua acelera\u00e7\u00e3o vale a=2,3m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=6,20m\/s\u00a0 —\u00a0 V= V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0+ at\u00a0 —\u00a0 V=6,20 + 2,3t\u00a0 —\u00a0 t=3,60s \u2013 V=6,20 + 2,3.3,60\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0V=14,48m\/s<\/b><\/span><\/span>\u00a0 —\u00a0 t=5,80s \u2013 V=6,20 + 2,3.5,80\u00a0 —\u00a0<\/b><\/span><\/span>V=19,54m\/s<\/b><\/span><\/span><\/p>\n

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

19- Se, na ida ela tem velocidade de 50m\/s, na volta dever\u00e1 ter velocidade de -50m\/s\u00a0 —\u00a0 na ida, at\u00e9 parar (V=0) ela demorou\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

V=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0+ at\u00a0 —\u00a0 0=50 \u2013 0,2t\u00a0 —\u00a0 t=250s (na ida)\u00a0 —\u00a0 na volta\u00a0 —\u00a0 V<\/b><\/span><\/span>0<\/b><\/span><\/span><\/sub>=0 e V=-50m\/s\u00a0 —\u00a0 V=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0+ at\u00a0 —\u00a0 -50=0 -0,2t\u00a0 —\u00a0 t=250s (na volta)\u00a0 —\u00a0 t<\/b><\/span><\/span>pedido<\/b><\/span><\/span><\/sub>=t<\/b><\/span><\/span>ida<\/b><\/span><\/span><\/sub>\u00a0+ t<\/b><\/span><\/span>volta<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 t<\/b><\/span><\/span>pedido<\/b><\/span><\/span><\/sub>=250 + 250=500s\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>\u00a0t=500s\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- A<\/b><\/span><\/span><\/p>\n

20- Colocando no gr\u00e1fico\u00a0 —\u00a0 t=0 \u2013 V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=10m\/s\u00a0 —\u00a0 V=0\u00a0 —\u00a0\u00a0 0=10 -2t\u00a0 —\u00a0 t=5s<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u0394S=\u00e1rea=b.h\/2=5.10\/2\u00a0 —\u00a0 \u0394S=25m\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- D<\/b><\/span><\/span><\/p>\n

21- a=(20 \u2013 0)\/(10 \u2013 0)\u00a0 —\u00a0 a=2m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 d=\u00e1rea=5.10\/2\u00a0 —\u00a0 d=25m\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/p>\n

22- Considerando a \u00e1rea como sendo de um tri\u00e2ngulo<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u0394S=b.h\/=1,5.1,0\/2\u00a0 —\u00a0 \u0394S=0,75m\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- A<\/b><\/span><\/span><\/p>\n

23- \u0394S=\u00e1rea total=b.h + (B + b).h\/2 + b.h=10.5 + (15 + 5).10\/2 + 10.15=300m\u00a0 —\u00a0 \u0394S=300m\u00a0 —\u00a0 V<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>= \u0394S\/ \u0394t=300\/30\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>V<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=10m\/s<\/b><\/span><\/span><\/p>\n

24- a) entre 0 e 20s\u00a0 —\u00a0 a=2m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=0\u00a0 —\u00a0 V<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>= V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0+ at=0 + 2.20\u00a0 —\u00a0 V<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=40m\/s\u00a0 —\u00a0 entre 20s e 50s\u00a0 —\u00a0 V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=40m\/s\u00a0 —\u00a0 a=-1m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0+ at=40 \u2013 1.30\u00a0 —\u00a0 V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=10m\/s\u00a0 —\u00a0 gr\u00e1fico abaixo<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

 <\/p>\n

b) \u0394S=\u00e1rea=b.h\/2 + (B _ b).h\/2=20.40\/2 + (40 + 10).30\/2\u00a0 —\u00a0 A<\/b><\/span><\/span>\u00a0dist\u00e2ncia percorrida \u00e9 1150m.<\/b><\/span><\/span><\/p>\n

25- T=4s\u00a0 —\u00a0 V=5 \u2013 2.4\u00a0 —\u00a0 V=-3m\/s\u00a0 —\u00a0 velocidade inicial \u2013 movimento progressivo, velocidade de -3m\/s \u2013 movimento retr\u00f3grado\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- D<\/b><\/span><\/span><\/p>\n

26-\u00a0<\/b><\/span><\/span>A dist\u00e2ncia percorrida corresponde \u00e0 \u00e1rea compreendida entre a reta representativa\u00a0 e o eixo do tempo, entre 0 e 3s, ou seja, \u00e0 \u00e1rea de um trap\u00e9zio\u00a0 —\u00a0 \u0394S=\u00e1rea=(B + b).h\/2=(80\/3,6 + 40\/3,6).3\/2= (22,2 + 11,1).1,5\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0\u0394S=49,95\u224850m\u00a0<\/b><\/span><\/span><\/p>\n

27-\u00a0<\/b><\/span><\/span>Observe atentamente que o \u00fanico gr\u00e1fico coerente com o enunciado \u00e9 o da alternativa E\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- E<\/b><\/span><\/span><\/p>\n

28-\u00a0<\/b><\/span><\/span>Quando t=0 eles est\u00e3o na mesma posi\u00e7\u00e3o (dado do exerc\u00edcio) —\u00a0 quando t=4s, o deslocamento de cada carro \u00e9 fornecido pela \u00e1rea<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0entre 0 e 4s\u00a0 —\u00a0 carro A, a \u00e1rea \u00e9 de um tri\u00e2ngulo\u00a0 —\u00a0 \u0394S<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=b.h\/2=4.20\/2\u00a0 —\u00a0 \u0394S<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=40m\u00a0 —\u00a0 carro B, a \u00e1rea \u00e9 de um ret\u00e2ngulo\u00a0 —\u00a0 \u0394S<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=b.h=4.10\u00a0 —\u00a0 \u0394S<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=40m\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- A<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/p>\n

29- \u00a0<\/b><\/span><\/span>Calculando o espa\u00e7o percorrido pela \u00e1rea\u00a0 —\u00a0 \u0394S<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=(B + b).h\/2=(45 + 30).10\/2\u00a0 —\u00a0 \u0394S<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=375m\u00a0 —\u00a0 \u0394S<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=(-10 \u2013 30).10\/2\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

\u0394S<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=-200m\u00a0 —\u00a0 d=375 + 200\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>d=575m<\/b><\/span><\/span>\u00a0 —\u00a0 ou acelera\u00e7\u00e3o escalar de cada m\u00f3vel, lembrando que\u00a0 —\u00a0 a<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=(45 \u2013 30)\/(10 \u2013 0)\u00a0 —\u00a0 a<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=1,5m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 a<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=(-30 \u2013 (-10)\/(10 \u2013 0)\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

a<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>= -2m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 — S<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=S<\/b><\/span><\/span>oA<\/b><\/span><\/span><\/sub>\u00a0+ 30t + 0,75t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 S<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=S<\/b><\/span><\/span>oB<\/b><\/span><\/span><\/sub>\u00a0\u2013 10t \u2013 t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 supondo S<\/b><\/span><\/span>oA<\/b><\/span><\/span><\/sub>=0 e fazendo t=10s no encontro onde voc\u00ea iguala as equa\u00e7\u00f5es\u00a0 —\u00a0 30(10) + 0,75(10)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0= S<\/b><\/span><\/span>oB<\/b><\/span><\/span><\/sub>\u00a0\u2013 10(10) \u2013 (10)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 375 = S<\/b><\/span><\/span>oB<\/b><\/span><\/span><\/sub>\u00a0\u2013 200\u00a0 —\u00a0 S<\/b><\/span><\/span>oB<\/b><\/span><\/span><\/sub>\u00a0= 575 m, que \u00e9 a dist\u00e2ncia inicial entre os m\u00f3veis, pois supusemos o m\u00f3vel A partindo da origem.<\/b><\/span><\/span><\/p>\n

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

30-\u00a0<\/b><\/span><\/span>Propriedade do gr\u00e1fico v = f(t)\u00a0 —\u00a0\u00a0 a \u00e1rea entre a linha do gr\u00e1fico e o eixo t representa o espa\u00e7o percorrido pelo m\u00f3vel (DS)\u00a0 —\u00a0 como n\u00e3o h\u00e1 mudan\u00e7a de sentido, o espa\u00e7o percorrido \u00e9 igual \u00e0 dist\u00e2ncia percorrida\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- D\u00a0<\/b><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

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

32-\u00a0<\/b><\/span><\/span>a) V=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0+ at\u00a0 —\u00a0 80=0 + 2.t\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0t=40s<\/b><\/span><\/span><\/p>\n

b) Na situa\u00e7\u00e3o em que a pista tem o comprimento m\u00ednimo (d<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>), o avi\u00e3o perde o contato com a pista exatamente em seu final\u00a0 —\u00a0\u00a0\u00a0 equa\u00e7\u00e3o de Torricelli\u00a0 —\u00a0 v<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0\u2013 v<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0+ 2.a.d<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 80<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=0<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0+ 2.2.d<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 6.400=4d<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>\u00a0 —<\/b><\/span><\/span>\u00a0 d<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=1.600m<\/b><\/span><\/span><\/p>\n

33-\u00a0<\/b><\/span><\/span>Observe a figura abaixo\u00a0 —\u00a0 determinando a dist\u00e2ncia d por Pit\u00e1goras\u00a0 —\u00a0 (130)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=d<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0+ (50)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 16.900=d<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0+ 2.500\u00a0 —\u00a0 d=120m\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

—\u00a0 V<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=V<\/b><\/span><\/span>r<\/b><\/span><\/span><\/sub>.cos\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\u00a0<\/b><\/span><\/span>R- C<\/b><\/span><\/span><\/p>\n

 <\/p>\n

34-<\/b><\/span><\/span>\u00a0Observe na figura abaixo onde a origem da trajet\u00f3ria foi colocada no ponto de partida do m\u00f3vel A S<\/b><\/span><\/span>oA<\/b><\/span><\/span><\/sub>=0\u00a0 —\u00a0 a<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

trajet\u00f3ria foi orientada para a direita S<\/b><\/span><\/span>oB<\/b><\/span><\/span><\/sub>=19200m\u00a0 —\u00a0 ambos os m\u00f3veis partiram do repouso, V<\/b><\/span><\/span>oA<\/b><\/span><\/span><\/sub>=V<\/b><\/span><\/span>oB<\/b><\/span><\/span><\/sub>=0\u00a0 —\u00a0 como o m\u00f3vel B est\u00e1 em movimento retr\u00f3grado e acelerado sua velocidade e sua acelera\u00e7\u00e3o s\u00e3o negativas (veja fisicaevestibular.com.br \u2013 mec\u00e2nica \u2013 cinem\u00e1tica \u2013 gr\u00e1ficos do MUV)\u00a0 —\u00a0 dedu\u00e7\u00e3o da fun\u00e7\u00e3o hor\u00e1ria de cada m\u00f3vel\u00a0 —\u00a0 S<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=S<\/b><\/span><\/span>oA<\/b><\/span><\/span><\/sub>\u00a0+ V<\/b><\/span><\/span>oA<\/b><\/span><\/span><\/sub>t + a<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2=0 + 0.t + 2t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 S<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 S<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=S<\/b><\/span><\/span>oB<\/b><\/span><\/span><\/sub>\u00a0+ V<\/b><\/span><\/span>oB<\/b><\/span><\/span><\/sub>t + a<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2=19200 \u2013 0.t \u2013 4.t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 S<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=19200 \u2013 2t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 no encontro AS=S<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=19200 \u2013 2t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=6400\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0t=80s<\/b><\/span><\/span>.<\/b><\/span><\/span><\/p>\n

 <\/p>\n

 <\/p>\n

35-<\/b><\/span><\/span>Colocando a origem das posi\u00e7\u00f5es no instante inicial (t=0 e S<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=0) e deduzindo a equa\u00e7\u00e3o de cada carro\u00a0 —\u00a0 S<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>t + at<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 S<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=2V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>t +(a\/2).t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 S<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=2V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>t + at<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/4\u00a0 —\u00a0 no encontro\u00a0 —\u00a0 \u00a0S<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=S<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>t + at<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2 = 2V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>t + at<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/4\u00a0 —\u00a0 V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>t \u2013 2V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>t + at<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2 \u2013 at<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/4 = 0\u00a0\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

-4V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub>t + at<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup>\u00a0= 0\u00a0 —\u00a0 t(at \u2013 4V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub>)=0\u00a0 —\u00a0 at \u2013 4V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub>=0\u00a0 —<\/b><\/span><\/span><\/span>\u00a0 t=4V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub>\/a.<\/b><\/span><\/span><\/span><\/p>\n

 <\/p>\n

36-<\/b><\/span><\/span>Observe na express\u00e3o D=K.V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>, onde K \u00e9 constante e, nela voc\u00ea observa que a velocidade V \u00e9 diretamente proporcional ao quadrado da velocidade\u00a0 —\u00a0 assim, se a velocidade dobra passando de 80kmh para 160kmh, a dist\u00e2ncia percorrida pelo carro at\u00e9 parar fica 2<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=4 vezes maior\u00a0 —<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/p>\n

37-\u00a0<\/b><\/span><\/span>a) Falsa\u00a0 —\u00a0 a acelera\u00e7\u00e3o \u00e9 positiva (concavidade para cima)\u00a0 —\u00a0 Entre 0 e t\u00a0\u2013 o espa\u00e7o decresce (movimento retr\u00f3grado, V<0)<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

e o movimento \u00e9 retardado, pois a e V tem sinais contr\u00e1rios (a>0 e V<0)\u00a0 —\u00a0 ap\u00f3s t\u00a0\u2013 o espa\u00e7o cresce (movimento progressivo, V>0) e o movimento \u00e9 acelerado, pois a e V tem mesmo sinal (a>0 e V>0)<\/b><\/span><\/span><\/p>\n

b) Correta\u00a0 —\u00a0 como o movimento \u00e9 retr\u00f3grado, a velocidade \u00e9 negativa e, para que o movimento seja acelerado, a acelera\u00e7\u00e3o tamb\u00e9m tem que ser negativa, o que \u00e9 o caso.<\/b><\/span><\/span><\/p>\n

c) Falsa\u00a0 —\u00a0 o movimento \u00e9 acelerado mas \u00e9 progressivo (V>0).<\/b><\/span><\/span><\/p>\n

d)\u00a0 Falsa\u00a0 —\u00a0 se existe acelera\u00e7\u00e3o, o gr\u00e1fico Sxt tem que ser uma par\u00e1bola (equa\u00e7\u00e3o do segundo grau).<\/b><\/span><\/span><\/p>\n

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

 <\/p>\n

38-\u00a0<\/b><\/span>\u00a0<\/b><\/span>A<\/b><\/span>\u00a0<\/b><\/span>\u00e1rea compreendida entre a reta representativa e o eixo dos tempos corresponde ao deslocamento \u2206S do m\u00f3vel no intervalo de tempo considerado\u00a0 —\u00a0 A e B (0 e t+Em todo gr\u00e1fico da velocidade em fun\u00e7\u00e3o do tempo1)\u00a0 —\u00a0 \u2206SAB=\u00e1rea do ret\u00e2ngulo=b.h\u00a0 —\u00a0 \u2206SAB=Vot1\u00a0 —\u00a0 B e C (t1<\/b><\/span>\u00a0<\/b><\/span>e t2)\u00a0 —\u00a0 \u2206SBC=\u00e1rea do tri\u00e2ngulo=b.h\/2\u00a0 — \u2206SBC=(t2<\/b><\/span>\u00a0<\/b><\/span>\u2013 t1).(Vo<\/b><\/span>\u00a0<\/b><\/span>\u2013 0)\/2\u00a0 —\u00a0 \u2206SBC=Vo.(t2<\/b><\/span>\u00a0<\/b><\/span>\u2013 t1)\/2\u00a0 —\u00a0 \u2206SBC\/\u2206SAB= [Vo.(t2<\/b><\/span>\u00a0<\/b><\/span>\u2013 t1)\/2]\/ Vot1\u00a0 —\u00a0 \u2206SBC\/\u2206SAB=(t2<\/b><\/span>\u00a0<\/b><\/span>\u2013 t1)\/2t1\u00a0 —\u00a0<\/b><\/span>\u00a0<\/b><\/span>R- C<\/b><\/span><\/p>\n

 <\/p>\n

39<\/b><\/span>–<\/b><\/span>\u00a0<\/b><\/span>Primeira situa\u00e7\u00e3o\u00a0 —\u00a0 c\u00e1lculo da acelera\u00e7\u00e3o do carro com a pista seca\u00a0 —\u00a0 V2=Vo2<\/b><\/span>\u00a0<\/b><\/span>+ 2.a.\u2206S\u00a0 —\u00a0 02<\/b><\/span>\u00a0<\/b><\/span>= 102<\/b><\/span>\u00a0<\/b><\/span>+ 2.a.5\u00a0 —\u00a0 a=-10ms2\u00a0 —\u00a0 c\u00e1lculo da acelera\u00e7\u00e3o do carro com a pista molhada\u00a0 —\u00a0 V2=Vo2<\/b><\/span>\u00a0<\/b><\/span>+ 2.a.\u2206S\u00a0 —\u00a0 02<\/b><\/span>\u00a0<\/b><\/span>= 102<\/b><\/span>\u00a0<\/b><\/span>+ 2.a.6\u00a0 —\u00a0 a=-100\/12=-25\/3 ms2\u00a0 —\u00a0 Segunda situa\u00e7\u00e3o\u00a0 —\u00a0 c\u00e1lculo da dist\u00e2ncia percorrida com a pista seca\u00a0 —\u00a0 V2=Vo2<\/b><\/span>\u00a0<\/b><\/span>+ 2.a.\u2206S\u00a0 —\u00a0 02<\/b><\/span>\u00a0<\/b><\/span>= 302<\/b><\/span>\u00a0<\/b><\/span>+ 2.(-10). \u2206S\u00a0 —\u00a0 \u2206S =45m\u00a0 —\u00a0 c\u00e1lculo da dist\u00e2ncia percorrida com a pista molhada\u00a0 —\u00a0 V2=Vo2+ 2.a.\u2206S \u00a0—\u00a0 02<\/b><\/span>\u00a0<\/b><\/span>= 302<\/b><\/span>\u00a0<\/b><\/span>+ 2.(-25\/3).\u2206S\u00a0 —\u00a0 \u2206S=54m\u00a0 —\u00a0 dist\u00e2ncia a mais\u00a0 —\u00a0 d=54 \u2013 45=9m\u00a0 —<\/b><\/span>\u00a0<\/b><\/span>\u00a0R- D<\/b><\/span><\/p>\n

 <\/p>\n

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

Resolu\u00e7\u00e3o comentada dos exerc\u00edcios sobre MUV \u00a001- a) b) Quando ele muda o sentido se seu movimento ele p\u00e1ra (V=0) e, a partir desse instante, o movimento que era progressivo se torna retr\u00f3grado\u00a0 —\u00a0 V=-20 + 5t\u00a0 —\u00a0 0=-20 + 5t\u00a0 —\u00a0\u00a0t=4s\u00a0(instante em que ele p\u00e1ra para inverter o sentido do movimento) c) d) 2s \u2013 retr\u00f3grado retardado\u00a0 —\u00a0 10s<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":988,"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-993","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/993","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=993"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/993\/revisions"}],"predecessor-version":[{"id":10791,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/993\/revisions\/10791"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/988"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=993"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}