{"id":3244,"date":"2017-09-27T22:26:02","date_gmt":"2017-09-27T22:26:02","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=3244"},"modified":"2024-09-04T14:25:54","modified_gmt":"2024-09-04T14:25:54","slug":"resolucoes-cinematica-2013-2014","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/vestibulares-recentes\/mecanica\/cinematica-2013-2014\/resolucoes-cinematica-2013-2014\/","title":{"rendered":"Resolu\u00e7\u00f5es Cinem\u00e1tica – 2013 – 2014"},"content":{"rendered":"

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

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

01-\u00a0<\/b><\/span><\/span><\/span>01-\u00a0I.\u00a0Hor\u00e1rio do rush — carro — V<\/b><\/span><\/span><\/span>mc<\/b><\/span><\/span><\/sub><\/span>=12km\/h — \u0394S<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=12km — V<\/b><\/span><\/span><\/span>mc<\/b><\/span><\/span><\/sub><\/span>= \u0394S<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>\/\u0394t<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>\u00a0— 12=12\/\u0394t<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>\u00a0— \u0394t<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=1h — esse tempo \u00e9 o mesmo que o do metr\u00f4 fornecido pelo enunciado (1h) —\u00a0Falsa<\/b><\/span><\/span><\/span><\/p>\n

II.\u00a0Hor\u00e1rio do rush — carro — V<\/b><\/span><\/span><\/span>mc<\/b><\/span><\/span><\/sub><\/span>=12km\/h — \u0394S<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=12km — V<\/b><\/span><\/span><\/span>mc<\/b><\/span><\/span><\/sub><\/span>= \u0394S<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>\/\u0394t<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>\u00a0— 12=12\/\u0394t<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>\u00a0— \u0394t<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=1h —<\/b><\/span><\/span><\/span><\/p>\n

metr\u00f4 — V<\/b><\/span><\/span><\/span>mm<\/b><\/span><\/span><\/sub><\/span>=60km\/h — \u0394S<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>=20km — V<\/b><\/span><\/span><\/span>mm<\/b><\/span><\/span><\/sub><\/span>= \u0394S<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>\/\u0394t<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>\u00a0— 60=20\/\u0394t<\/b><\/span><\/span><\/span>m\u00a0<\/b><\/span><\/span><\/sub><\/span>— \u0394t<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>=1\/3h —\u00a0Correta<\/b><\/span><\/span><\/span><\/p>\n

III.\u00a0Fora do hor\u00e1rio de rush — carro — V<\/b><\/span><\/span><\/span>mc<\/b><\/span><\/span><\/sub><\/span>=42km\/h — \u0394S<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=12km — 42=12\/\u0394t<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>\u00a0— \u0394t<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=12\/42=0,29h — metr\u00f4 — V<\/b><\/span><\/span><\/span>mm<\/b><\/span><\/span><\/sub><\/span>=60km\/h — \u0394S<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>=20km — V<\/b><\/span><\/span><\/span>mm<\/b><\/span><\/span><\/sub><\/span>= \u0394S<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>\/\u0394t<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>\u00a0— 60=20\/\u0394t<\/b><\/span><\/span><\/span>m\u00a0<\/b><\/span><\/span><\/sub><\/span>— \u0394t<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>=1\/3h=0,33h —\u00a0Correta<\/b><\/span><\/span><\/span><\/p>\n

IV.\u00a0Fora do hor\u00e1rio de rush — carro — V<\/b><\/span><\/span><\/span>mc<\/b><\/span><\/span><\/sub><\/span>=42km\/h — \u0394S<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=12km — 42=12\/\u0394t<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>\u00a0— \u0394t<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=12\/42=0,29h — nova velocidade do metr\u00f4 — V<\/b><\/span><\/span><\/span>mm<\/b><\/span><\/span><\/sub><\/span>=70km\/h — \u0394S<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>=20km — V<\/b><\/span><\/span><\/span>mm<\/b><\/span><\/span><\/sub><\/span>= \u0394S<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>\/\u0394t<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>\u00a0— 70=20\/\u0394t<\/b><\/span><\/span><\/span>m\u00a0<\/b><\/span><\/span><\/sub><\/span>— \u0394t<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>=2\/7h=0,29h —\u00a0Correta<\/b><\/span><\/span><\/span><\/p>\n

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

02-\u00a002-\u00a0Dist\u00e2ncia total percorrida\u00a0—\u00a0\u0394S=(1380 + 730 + 490 + 640 + 1080)=4320m\u00a0—\u00a0tempo total\u00a0de viagem —\u00a0\u0394t=(2,5 + 2,0 + 1,0 + 1,5 + 2,0)=9minx60=540s.<\/b><\/span><\/span><\/span><\/p>\n

V<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>= \u0394S\/\u0394t=4320\/540=8m\/s.<\/b><\/span><\/span><\/span><\/p>\n

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

03-Tempo que o bondinho demora para percorrer o\u00a0primeiro trecho AB\u00a0da\u00a0Praia Vermelha ao Morro da Urca<\/b><\/span><\/span><\/span><\/p>\n

onde percorre\u00a0\u0394S<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=540m\u00a0com velocidade\u00a0V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=10,8\/3,6=3m\/s\u00a0— V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>= \u0394S<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/\u0394t<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0— 3=540\/\u0394t<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0— \u0394t<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=540\/3<\/b><\/span><\/span><\/span><\/p>\n

\u0394t<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=180s\/60 —\u00a0\u0394t<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=3 min\u00a0— o\u00a0segundo BC\u00a0trecho corresponde a uma\u00a0caminhada no Morro da Urca<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

cujo tempo gasto somado ao tempo de espera nas esta\u00e7\u00f5es \u00e9 de\u00a0\u0394t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=30 minutos\u00a0(dado fornecido) —<\/b><\/span><\/span><\/span><\/p>\n

no\u00a0terceiro trecho CD,\u00a0de bondinho, do\u00a0Morro da Urca ao P\u00e3o de A\u00e7\u00facar\u00a0ele percorreu\u00a0\u0394S<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>=720m\u00a0com velocidade\u00a0V<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>=14,4\/3,6=4m\/s\u00a0e demorou\u00a0\u0394t<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>\u00a0tal que — V<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>=\u0394S<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>\/\u00a0\u0394t<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>\u00a0— 4=720\/\u0394t<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>\u00a0—\u00a0\u0394t<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>=720\/4=190s —<\/b><\/span><\/span><\/span><\/p>\n

\u0394t<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>=3 min\u00a0— tempo total do passeio —\u00a0\u0394t=3 + 30 + 3=36 min.<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub><\/span>—\u00a0R- D<\/b><\/span><\/span><\/span><\/p>\n

04-\u00a0Tempo que o corredor mais r\u00e1pido (V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=32km\/h) demora para completar os d<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=100m=0,1km — V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=d<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/t<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0— 32=0,1\/t<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0— t<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=0,1\/32 h.<\/b><\/span><\/span><\/span><\/p>\n

Nesse tempo t<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=t=0,1\/32 h o corredor menos r\u00e1pido (V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=30km\/h) percorreu a dist\u00e2ncia d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0tal que —<\/b><\/span><\/span><\/span><\/p>\n

30=d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\/(0,1\/32) — d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=3\/32=0,09375km=93,375m.<\/b><\/span><\/span><\/span><\/p>\n

A\u00a0dist\u00e2ncia pedida entre eles vale\u00a0d\u2019=100 \u2013 93,375=6,25m<\/b><\/span><\/span><\/span><\/p>\n

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

05-\u00a0A)\u00a0Sim,\u00a0ambos no km 8, onde a reta representativa \u00e9 paralela ao eixo dos tempos, mas em intervalos de tempos<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

diferentes (veja gr\u00e1fico acima)<\/b><\/span><\/span><\/span><\/p>\n

B) (II)\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>=(d \u2013 d<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>)\/(t \u2013 t<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>)=(64 \u2013 8))\/(0,9 \u2013 0)=56\/0,9=62,2km\/h\u00a0 — \u00a0(I)\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>=(d \u2013 d<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>)\/(t \u2013 t<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>)=(64 \u2013 0)\/(0,95 \u2013 0)=67,4km\/h\u00a0 —\u00a0 a maior velocidade escalar m\u00e9dia \u00e9 a do ve\u00edculo I.<\/b><\/span><\/span><\/span><\/p>\n

C)\u00a0Sim\u00a0 —\u00a0 os dois\u00a0 —\u00a0 o ve\u00edculo I, por exemplo, entre 0,6h e 0,8h teve velocidade de V<\/b><\/span><\/span><\/span>I<\/b><\/span><\/span><\/sub><\/span>=(48 \u2013 24)\/(0,8 \u2013 0,6)=120km\/h\u00a0 —\u00a0 o ve\u00edculo II, por exemplo, entre 0,4h e 0,6h teve velocidade de V<\/b><\/span><\/span><\/span>I<\/b><\/span><\/span><\/sub><\/span>=(48 \u2013 24)\/(0,6 \u2013 0,4)=120km\/h .<\/b><\/span><\/span><\/span><\/p>\n

06-\u00a0C\u00e1lculo da velocidade real de Carlos na hora correta\u00a0 —\u00a0 se o rel\u00f3gio adianta \u2206t<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>=1min 40s=10 + 40s=100s, a hora real indicada ser\u00e1\u00a0 —\u00a0 \u2206t<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>=1h \u2013 100s=3600 \u2013 100=3500s\u00a0 —\u00a0 \u2206t<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>=3500s\u00a0\u00a0 —\u00a0 velocidade correta de Carlos\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>=\u2206S<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>\/\u2206t<\/b><\/span><\/span><\/span>C\u00a0<\/b><\/span><\/span><\/sub><\/span>\u00a0—\u00a0 V<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>=3500m\/3500s\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>=1m\/s (velocidade correta de Carlos)\u00a0 —\u00a0 c\u00e1lculo da velocidade real de Julieta na hora correta\u00a0 —\u00a0 se o rel\u00f3gio atrasa \u2206t<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>=1min 40s=10 + 40s=100s, a hora real indicada ser\u00e1\u00a0 —\u00a0 \u2206t<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>=1h + 100s=3600 + 100 =3700s\u00a0 —\u00a0 \u2206t<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>=3700s\u00a0\u00a0 —\u00a0 velocidade correta de Julieta\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>J<\/b><\/span><\/span><\/sub><\/span>=\u2206S<\/b><\/span><\/span><\/span>J<\/b><\/span><\/span><\/sub><\/span>\/\u2206t<\/b><\/span><\/span><\/span>J\u00a0<\/b><\/span><\/span><\/sub><\/span>\u00a0—\u00a0 V<\/b><\/span><\/span><\/span>J<\/b><\/span><\/span><\/sub><\/span>=3500m\/3700s\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>=0,9m\/s\u00a0 — (velocidade correta de Julieta)\u00a0 —\u00a0 em \u2206t=1h=3600s Carlos percorreu\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>=\u2206S<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>\/\u2206t<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 1=\u2206S<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>\/3600\u00a0 —\u00a0 \u2206S<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>=3600m\u00a0 —\u00a0 em \u2206t=1h=3600s Julieta percorreu\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>J<\/b><\/span><\/span><\/sub><\/span>=\u2206S<\/b><\/span><\/span><\/span>J<\/b><\/span><\/span><\/sub><\/span>\/\u2206t<\/b><\/span><\/span><\/span>J<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 0,9=\u2206S<\/b><\/span><\/span><\/span>J<\/b><\/span><\/span><\/sub><\/span>\/3600\u00a0 —\u00a0 \u2206S<\/b><\/span><\/span><\/span>J<\/b><\/span><\/span><\/sub><\/span>=3240m\u00a0 —\u00a0 \u2206S<\/b><\/span><\/span><\/span>pedido<\/b><\/span><\/span><\/sub><\/span>=\u2206S<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>\u00a0–\u00a0 \u2206S<\/b><\/span><\/span><\/span>J<\/b><\/span><\/span><\/sub><\/span>=3600 \u2013 3240=360m\u00a0 —\u00a0\u00a0R- A.<\/b><\/span><\/span><\/span><\/p>\n

07-\u00a0C\u00e1lculo da dist\u00e2ncia d<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0indicada na figura (tri\u00e2ngulo hachurado) —\u00a0 tg30<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>=12\/d<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 \u221a3\/3=d<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/12\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

d<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=4\u221a3\u00a0 —\u00a0d<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=4.1,7=6,8km=6800m\u00a0 —\u00a0 observe que voc\u00ea obt\u00e9m um tri\u00e2ngulo ret\u00e2ngulo e que a dist\u00e2ncia AC=12km\u00a0 —\u00a0 assim, d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=12000 \u2013 6800=5200m\u00a0 —\u00a0 V=1872km\/h\/3,6=520m\/s\u00a0 —\u00a0 V=d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\/t\u00a0 —\u00a0520=5200\/t\u00a0 —\u00a0 t=10s\u00a0 —\u00a0\u00a0R-C.<\/b><\/span><\/span><\/span><\/p>\n

08-\u00a0\u00a0V<\/b><\/span><\/span><\/span><\/span>m<\/b><\/span><\/span><\/span><\/sub><\/span>=\u2206S\/\u2206t=(S \u2013 S<\/b><\/span><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub><\/span>)\/(t \u2013 t<\/b><\/span><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub><\/span>)=(50 \u2013 0)\/(40 \u2013 0)\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span><\/span>m<\/b><\/span><\/span><\/span><\/sub><\/span>=1,25m\/s\u00a0 —\u00a0\u00a0R- B.<\/b><\/span><\/span><\/span><\/span><\/p>\n

09-\u00a0–\u00a0Primeiro trecho\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>m1<\/b><\/span><\/span><\/sub><\/span>=\u2206S<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/\u2206t<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 80=80\/\u2206t<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 \u2206t<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=1h\u00a0 (intervalo de tempo que demora para<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

percorrer o primeiro trecho)\u00a0 —\u00a0 segundo trecho\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>m2<\/b><\/span><\/span><\/sub><\/span>=\u2206S<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\/\u2206t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 60=120\/\u2206t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 \u2206t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=60\/120\u00a0 —\u00a0 \u2206t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=0,5h\u00a0 ((intervalo de tempo que demora para percorrer o segundo trecho)\u00a0 —\u00a0 tempo necess\u00e1rio para efetuar a entrega\u00a0 —\u00a0 \u2206t=\u2206t<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0+ \u2206t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=1 + 0,5\u00a0 —\u00a0 \u2206t=1,5h\u00a0 —\u00a0\u00a0R- C.<\/b><\/span><\/span><\/span><\/p>\n

10-\u00a0Primeiro trecho: Trata-se de um movimento uniformemente acelerado, onde a locomotiva parte do repouso, com acelera\u00e7\u00e3o constante, sendo que esse movimento \u00e9 expresso por uma fun\u00e7\u00e3o do segundo grau 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, onde o gr\u00e1fico posi\u00e7\u00e3oxtempo \u00e9 um arco de par\u00e1bola com concavidade para cima (acelera\u00e7\u00e3o positiva).<\/b><\/span><\/span><\/span><\/p>\n

Segundo trecho: Mant\u00e9m velocidade constante e a fun\u00e7\u00e3o hor\u00e1ria \u00e9 do primeiro grau S=S<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0+ Vt, onde o gr\u00e1fico posi\u00e7\u00e3oxtempo \u00e9 uma reta inclinada ascendente.<\/b><\/span><\/span><\/span><\/p>\n

Terceiro trecho: A composi\u00e7\u00e3o freia at\u00e9 parar\u00a0 —\u00a0 trata-se de um movimento uniformemente retardado, com acelera\u00e7\u00e3o constante, sendo que esse movimento \u00e9 expresso por uma fun\u00e7\u00e3o do segundo grau 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, onde o gr\u00e1fico posi\u00e7\u00e3oxtempo \u00e9 um arco de par\u00e1bola com concavidade para baixo (acelera\u00e7\u00e3o negativa)\u00a0 —\u00a0 no trecho de parada enquanto a composi\u00e7\u00e3o permanece em repouso, o gr\u00e1fico \u00e9 uma reta paralela ao eixo dos tempos.<\/b><\/span><\/span><\/span><\/p>\n

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

11-\u00a0Correndo sem vento, o velocista demora \u2206t=10s para percorrer \u2206S=100m\u00a0 —\u00a0 correndo com o vento a favor, para percorrer os mesmos \u2206S=100m ele demora \u2206t\u2019=10,0 \u2013 0,1=9,9s\u00a0 —\u00a0 com vento a favor\u00a0 —\u00a0 V=\u2206S\/\u2206t\u2019=100\/9,9\u00a0 —<\/b><\/span><\/span><\/span><\/p>\n

V=10,101m\/s\u00a0 —\u00a0\u00a0R- C.<\/b><\/span><\/span><\/span><\/p>\n

Movimento Uniformemente variado (MUV)<\/b><\/span><\/span><\/span><\/p>\n

01-\u00a0C\u00e1lculo da acelera\u00e7\u00e3o a\u00a0 — a=\u2206V\/\u2206t=(28 \u2013 4)\/(7 \u2013 1)=24\/6\u00a0 —\u00a0 a=4m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 quando t=7s, V=28m\/s\u00a0 —\u00a0 V=V<\/b><\/span><\/span><\/span>0<\/b><\/span><\/span><\/sub><\/span>\u00a0+ at\u00a0 —\u00a0 28= V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0+ 4.7\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=1m\/s\u00a0 —\u00a0 o deslocamento \u2206S entre 3s e 5s \u00e9 fornecido pela \u00e1rea do trap\u00e9zio\u00a0 —\u00a0\u00a0 \u2206S=\u00e1rea=(B +<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

b).h\/2\u00a0 —\u00a0\u00a0 \u2206S=(20 + 12).2\/2\u00a0 —\u00a0\u00a0 \u2206S=32m\u00a0 —\u00a0\u00a0R- C.<\/b><\/span><\/span><\/span><\/p>\n

02-\u00a0V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=30m\/s — V=? — a=-5m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0(freando) — \u0394S=50m — equa\u00e7\u00e3o de Torricelli — V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ 2.a.\u0394S — V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=30<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ 2.(-5).50=900 \u2013 500 — V=\u221a400 — V=20m\/sx3.6=72km\/h —\u00a0R- E<\/b><\/span><\/span><\/span><\/p>\n

03-\u00a0\"\"<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

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

\"\"<\/p>\n

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

05-\u00a0Em todo gr\u00e1fico Vxt, o deslocamento \u00e9 numericamente igual \u00e0 \u00e1rea entre a reta representativa e o eixo t — se o motorista n\u00e3o tivesse reduzido a velocidade do ve\u00edculo e mantido a velocidade constante de<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

V=90km\/h=25m\/s, a dist\u00e2ncia percorrida seria a \u00e1rea \u0394S<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0da figura abaixo — \u0394S<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=b.h=80×25=2000m —\u00a0\u0394S<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=2000m<\/b><\/span><\/span><\/span><\/p>\n

A dist\u00e2ncia\u00a0(\u0394S<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>)\u00a0que ele realmente percorreu devido \u00e0s obras corresponde \u00e0 soma das \u00e1reas da figura abaixo — A<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=b.h=10×25=250m — A<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=(B + b).h\/2=(25 + 15).10\/2=200m — A<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>=b.h=20×15=300m —<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

A<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sub><\/span>=(B + b).h\/2=(25 + 15).20\/2=400m — A<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sub><\/span>=b.h=20×25=500m — \u0394S<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=250 + 200 + 300 + 400 + 500= <\/b><\/span><\/span><\/span>1650m —\u00a0<\/b><\/span><\/span><\/span>\u0394S<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=1650m\u00a0<\/b><\/span><\/span><\/span>— se ele n\u00e3o tivesse reduzido a velocidade devido \u00e0s obras a dist\u00e2ncia adicional percorrida seria de d=2000 \u2013 1650=350m —<\/b><\/span><\/span><\/span>d=350m\u00a0<\/b><\/span><\/span><\/span>—\u00a0<\/b><\/span><\/span><\/span>R- D<\/b><\/span><\/span><\/span><\/p>\n

Observa\u00e7\u00e3o: Voc\u00ea poderia tamb\u00e9m subtrair as \u00e1reas, conforme as figuras abaixo e, nesse caso, a dist\u00e2ncia<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

d\u00a0pedida seria a \u00e1rea do trap\u00e9zio — d=(B +b).h\/2=(50 + 20).10\/2=350m —<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>d=350m<\/b><\/span><\/span><\/span><\/p>\n

06-\u00a0Se voc\u00ea fixar um ponto P no in\u00edcio da locomotiva , quando P come\u00e7ar a passar diante do observador (marco inicial 0), voc\u00ea ter\u00e1 velocidade inicial V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0e t<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=0,<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>e quando a locomotiva terminar de passar pelo observador P estar\u00e1 a 24m de 0 no instante t=4s (figura I).<\/b><\/span><\/span><\/span><\/p>\n

Pelo enunciado, quando o\u00a0<\/b><\/span><\/span><\/span>primeiro vag\u00e3o terminar de passar pelo observador o ponto P estar\u00e1 a 48m de 0 e o instante ser\u00e1 t=4 + 2 = 6s (figura II)<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

Na\u00a0<\/b><\/span><\/span><\/span>figura I, quando t=4s o ponto P, com velocidade inicial V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0e acelera\u00e7\u00e3o a, percorreu \u0394S=24m\u00a0<\/b><\/span><\/span><\/span>—<\/b><\/span><\/span><\/span><\/p>\n

\u0394S= 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<\/b><\/span><\/span><\/span>— 24 = 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 — 24 = 4V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0+ 8a — 6 = V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0+ 2a —\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>= 6 \u2013 2a (1).<\/b><\/span><\/span><\/span><\/p>\n

Na\u00a0<\/b><\/span><\/span><\/span>figura II, quando t=6s o ponto P, com velocidade inicial V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0e acelera\u00e7\u00e3o a, percorreu \u0394S=48m\u00a0<\/b><\/span><\/span><\/span>—<\/b><\/span><\/span><\/span><\/p>\n

\u0394<\/b><\/span><\/span><\/span>S= V<\/b><\/span><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub><\/span>t + at<\/b><\/span><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup><\/span>\/2\u00a0<\/b><\/span><\/span><\/span><\/span>— 48 = V<\/b><\/span><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub><\/span>.6 + a.6<\/b><\/span><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup><\/span>\/2 — 48 = 6V<\/b><\/span><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub><\/span>\u00a0+ 18a —\u00a0<\/b><\/span><\/span><\/span><\/span>8 = V<\/b><\/span><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub><\/span>\u00a0+ 3a (2).<\/b><\/span><\/span><\/span><\/span><\/p>\n

(1) em (2)\u00a0<\/b><\/span><\/span><\/span>—\u00a0<\/b><\/span><\/span><\/span>8=6 \u2013 2a + 3a —\u00a0<\/b><\/span><\/span><\/span>a=2m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0(acelera\u00e7\u00e3o da composi\u00e7\u00e3o).<\/b><\/span><\/span><\/span><\/p>\n

V\u00ad<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>= 6 \u2013 2.2\u00a0<\/b><\/span><\/span><\/span>—\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=2m\/s (velocidade inicial do ponto P).<\/b><\/span><\/span><\/span><\/p>\n

Quando o\u00a0<\/b><\/span><\/span><\/span>primeiro vag\u00e3o come\u00e7a a passar diante do observador o ponto P est\u00e1 na posi\u00e7\u00e3o 24m e o instante \u00e9 t=4s\u00a0<\/b><\/span><\/span><\/span>—\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0+ at=2 + 2.4=10m\/sx3,6<\/b><\/span><\/span><\/span>=36km\/h.<\/b><\/span><\/span><\/span><\/p>\n

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

07-\u00a0Colocando a origem da trajet\u00f3ria no ponto de partida do carro1 e orientando-a para a direita, quando 1 parte do\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

repouso do marco zero, 2 parte tamb\u00e9m do repouso do marco 50m\u00a0 —\u00a0 equa\u00e7\u00e3o hor\u00e1ria do carro 1\u00a0 —\u00a0 S<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=S<\/b><\/span><\/span><\/span>o1<\/b><\/span><\/span><\/sub><\/span>\u00a0+ V<\/b><\/span><\/span><\/span>o1<\/b><\/span><\/span><\/sub><\/span>t + a<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=<\/b><\/span><\/span><\/span><\/p>\n

0 + 0 + a<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 S<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=0,5.a<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0\u00a0 S<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=S<\/b><\/span><\/span><\/span>o2<\/b><\/span><\/span><\/sub><\/span>\u00a0+ V<\/b><\/span><\/span><\/span>o2<\/b><\/span><\/span><\/sub><\/span>t + 2a<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=50 + 0 + a<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 S<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=50+ a<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 no encontro S<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0= S<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —<\/b><\/span><\/span><\/span><\/p>\n

0,5.a<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>= 50 +a<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 0,5a<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=50\u00a0 —\u00a0 a<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=100\u00a0 —\u00a0 substituindo a<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=100 em S<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 S<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=0,5.100=50m\u00a0 —\u00a0 substituindo a<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=100 em S<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 S<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=50 + 100=150m\u00a0 —\u00a0 assim,\u00a0 o carro 2 alcan\u00e7a o carro 1 ap\u00f3s percorrer \u2206S=150 \u2013 50=100m\u00a0 —<\/b><\/span><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/span>R- C.<\/b><\/span><\/span><\/span><\/p>\n

08-\u00a0<\/b><\/span><\/span><\/span>Como, durante a frenagem eles percorrem a mesma dist\u00e2ncia, a dist\u00e2ncia m\u00ednima, d, entre eles para que n\u00e3o ocorra colis\u00e3o deve ser a dist\u00e2ncia percorrida pelo carro II durante o tempo de rea\u00e7\u00e3o (t=0,6s) com velocidade de V=20m\/s\u00a0 — V=d\/t\u00a0 —\u00a0 20=d\/0,6\u00a0 —\u00a0 d=12m\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- D<\/b><\/span><\/span><\/span><\/p>\n

09-\u00a0Colocando a origem da trajet\u00f3ria orientada para a direita no ponto onde a viatura parte com V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=0 e localizando o carro nesse instante, ele estar\u00e1 no marco S-V<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>.t=20.5=100m, voc\u00ea ter\u00e1 a situa\u00e7\u00e3o esquematizada na figura abaixo\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Equa\u00e7\u00e3o da viatura\u00a0 —\u00a0 S<\/b><\/span><\/span><\/span>v<\/b><\/span><\/span><\/sub><\/span>=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 S<\/b><\/span><\/span><\/span>v<\/b><\/span><\/span><\/sub><\/span>=0 + 0.t + at<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 S<\/b><\/span><\/span><\/span>v<\/b><\/span><\/span><\/sub><\/span>=at<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 (I)\u00a0 —\u00a0 equa\u00e7\u00e3o do carro\u00a0 —\u00a0 S<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=S<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0+ V<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>t\u00a0 —\u00a0 S<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=100 + 20t (II)\u00a0 —\u00a0 no encontro, pelo enunciado, o carro est\u00e1 no marco S<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=2100m, que substitu\u00eddo em (II) fornece o instante do encontro\u00a0 —\u00a0 2100=100 + 20t\u00a0 —\u00a0 t=2000\/20=100s\u00a0 —\u00a0 a viatura nesse instante t=100s tamb\u00e9m est\u00e1 no marco S<\/b><\/span><\/span><\/span>v<\/b><\/span><\/span><\/sub><\/span>=2100m, que substitu\u00eddo em (I) fornece sua acelera\u00e7\u00e3o\u00a0 —\u00a0 2100=a.100<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 a=2100\/5000\u00a0 — a=0,42m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 velocidade da viatura no encontro quando t=100s\u00a0 —\u00a0 V=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0+ at=0 + 0,42.100=42m\/s\u00a0 —\u00a0\u00a0R- E.<\/b><\/span><\/span><\/span><\/p>\n

10-\u00a0Em 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><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0e t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>, no caso entre 0 e 15s\u00a0 —\u00a0 c\u00e1lculo do deslocamento do carro A pela<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00e1reahachurada do trap\u00e9zio da figura\u00a0 —\u00a0 \u2206S<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=(B + b)xh\/2=(15 + 10)x10\/2\u00a0 —\u00a0 \u2206S<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=25×5=125m\u00a0\u00a0(deslocamento do carro A em 15s – \u2206S<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=125m)\u00a0 — c\u00e1lculo do deslocamento \u2206S<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0do carro A entre 0 e t (veja figura abaixo)\u00a0 —\u00a0 \u2206S<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=(B + b).h\/2=[t + (t \u2013 5].10\/2\u00a0 —\u00a0 \u2206S<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=(2t \u2013 5).5\u00a0 —\u00a0 \u2206S<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=10t \u2013 25\u00a0 —\u00a0 equa\u00e7\u00e3o do carro A\u00a0 —\u00a0 S<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>= S<\/b><\/span><\/span><\/span>oA<\/b><\/span><\/span><\/sub><\/span>\u00a0+ \u2206S<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0= 0 + 10t \u2013 25\u00a0 —\u00a0 S<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=10t \u2013 25 (I)\u00a0 —\u00a0 c\u00e1lculo do deslocamento \u2206S<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0do carro B entre 0 e t (veja figura abaixo)\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u2206<\/span>S<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=(B + b).h\/2=[t + (t \u2013 8].(-10)\/2\u00a0 —\u00a0 \u2206S<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=(2t \u2013 8).(-5)\u00a0 —\u00a0 \u2206S<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=-10t + 40\u00a0 —\u00a0 equa\u00e7\u00e3o do carro B\u00a0 —\u00a0 S<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>= S<\/b><\/span><\/span><\/span>oB<\/b><\/span><\/span><\/sub><\/span>\u00a0+ \u2206S<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0= 3 – 10t + 40\u00a0 —\u00a0 S<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=-10t + 43 (II)\u00a0 —\u00a0 nesse instante t, a dist\u00e2ncia entre os carros A e B \u00e9 fornecida <\/b><\/span><\/span><\/span>e vale d<\/b><\/span><\/span><\/span>AB<\/b><\/span><\/span><\/sub><\/span>=332m\u00a0 —\u00a0 d<\/b><\/span><\/span><\/span>AB<\/b><\/span><\/span><\/sub><\/span>=S<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 S<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 332=10t \u2013 25 \u2013 (- 10t + 43)\u00a0 —\u00a0 332=20t \u2013 68\u00a0 —\u00a0 t=400\/20\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>t=20s.<\/b><\/span><\/span><\/span><\/p>\n

Lan\u00e7amento obliquo<\/b><\/span><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

b) Pelo enunciado, o\u00a0<\/b><\/span><\/span><\/span>melhor \u00e2ngulo poss\u00edvel\u00a0<\/b><\/span><\/span><\/span>ocorre quando o\u00a0<\/b><\/span><\/span><\/span>alcance \u00e9 m\u00e1ximo\u00a0<\/b><\/span><\/span><\/span>e, o\u00a0<\/b><\/span><\/span><\/span>maior valor para sen (2\u03b1) ocorre quando \u03b1=45<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>, pois sen(2\u03b1)=sen(2.45<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>)=<\/b><\/span><\/span><\/span>sen90<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>=1\u00a0<\/b><\/span><\/span><\/span>— 98=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>.1 10 — V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=\u221a980\u224831msx3,6 —<\/b><\/span><\/span><\/span><\/p>\n

V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2248 111km\/h<\/b><\/span><\/span><\/span><\/p>\n

02-\u00a0<\/b><\/span><\/span><\/span>C\u00e1lculo da intensidade velocidade inicial da bola na vertical V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>0<\/b><\/span><\/span><\/sub><\/span>.sen45<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>,\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>=(\u221a2\/2).V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0<\/b><\/span><\/span><\/span>e na<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

horizontal,\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>ox<\/b><\/span><\/span><\/sub><\/span>=(\u221a2\/2).V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

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

03-\u00a0<\/b><\/span><\/span><\/span>Trata-se de um lan\u00e7amento obliquo com\u00a0<\/b><\/span><\/span><\/span>\u00e2ngulo de 45<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\u00a0com a horizontal e velocidade de V=20m\/s<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

Tempo<\/b><\/span><\/span><\/span>\u00a0que a bola demora para\u00a0<\/b><\/span><\/span><\/span>subir e descer na vertical\u00a0<\/b><\/span><\/span><\/span>com\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>=20.\u221a2\/2=10\u221a2m\/s\u00a0<\/b><\/span><\/span><\/span>e quando chega novamente ao solo\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>= – 10\u221a2m\/s\u00a0<\/b><\/span><\/span><\/span>— -10\u221a2 = 10\u221a2 \u2013 gt — -20\u221a2 = -10t —\u00a0<\/b><\/span><\/span><\/span>t=2\u221a2s<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Esse tempo\u00a0<\/b><\/span><\/span><\/span>t=2\u221a2s\u00a0<\/b><\/span><\/span><\/span>\u00e9 o\u00a0<\/b><\/span><\/span><\/span>mesmo tempo\u00a0<\/b><\/span><\/span><\/span>que a bola demora para percorrer a\u00a0<\/b><\/span><\/span><\/span>dist\u00e2ncia horizontal x com velocidade constante\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>ox<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>cos45<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>=20. \u221a2=10\u221a2m\/<\/b><\/span><\/span><\/span>s —\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>ox<\/b><\/span><\/span><\/sub><\/span>=x\/t\u00a0<\/b><\/span><\/span><\/span>— 10\u221a2=x\/2\u221a2 —\u00a0<\/b><\/span><\/span><\/span>x=40m<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

Nesse mesmo tempo t=2\u221a2s o jogador\u00a0<\/b><\/span><\/span><\/span>deve percorrer d=57 \u2013 40=17m com velocidade V\u00a0<\/b><\/span><\/span><\/span>tal que —<\/b><\/span><\/span><\/span>V=d\/t<\/b><\/span><\/span><\/span>=17\/2\u221a2=17\/2.1,4=17\/2,8 —\u00a0<\/b><\/span><\/span><\/span>V=6,07m<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

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

04-\u00a0<\/b><\/span><\/span><\/span>O\u00a0<\/b><\/span><\/span><\/span>observador A, dentro do vag\u00e3o<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>observa a trajet\u00f3ria da pedra como um\u00a0<\/b><\/span><\/span><\/span>la\u00e7amento vertical para<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

cima<\/b><\/span><\/span><\/span>\u00a0e,\u00a0<\/b><\/span><\/span><\/span>no ponto mais alto da trajet\u00f3ria a velocidade da pedra \u00e9\u00a0<\/b><\/span><\/span><\/span>zero,<\/b><\/span><\/span><\/span>\u00a0pois, nesse instante ela\u00a0<\/b><\/span><\/span><\/span>para e em seguida come\u00e7a a retornar (figura1).<\/b><\/span><\/span><\/span><\/p>\n

O\u00a0<\/b><\/span><\/span><\/span>observador B\u00a0<\/b><\/span><\/span><\/span>do lado de fora do vag\u00e3o e em repouso\u00a0<\/b><\/span><\/span><\/span>em rela\u00e7\u00e3o \u00e0 Terra, observa o vag\u00e3o passar\u00a0<\/b><\/span><\/span><\/span>e\u00a0<\/b><\/span><\/span><\/span>v\u00ea a trajet\u00f3ria da bola como um arco de par\u00e1bola (lan\u00e7amento obl\u00edquo)\u00a0<\/b><\/span><\/span><\/span>e, no ponto\u00a0<\/b><\/span><\/span><\/span>mais alto da trajet\u00f3ria a velocidade vertical \u00e9 nula mas a horizontal \u00e9 constante e vale V=5m\/s (figura 2).<\/b><\/span><\/span><\/span><\/p>\n

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

05-\u00a0<\/b><\/span><\/span><\/span>C\u00e1lculo de\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>\u00a0quando a bola atinge a altura\u00a0<\/b><\/span><\/span><\/span>m\u00e1xima h<\/b><\/span><\/span><\/span>max<\/b><\/span><\/span><\/sub><\/span>=10m, onde V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>=0\u00a0<\/b><\/span><\/span><\/span>utilizando a equa\u00e7\u00e3o de Torricelli —\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0= V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0\u2013 2.g.h<\/b><\/span><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub><\/span>\u00a0<\/b><\/span><\/span><\/span>— 0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0= V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0\u2013 2.10.10 —\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>\u00a0= \u221a(200) m\/s.<\/b><\/span><\/span><\/span><\/p>\n

C\u00e1lculo de\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>ox<\/b><\/span><\/span><\/sub><\/span>\u00a0com\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0fornecida e igual a 20m\/s\u00a0<\/b><\/span><\/span><\/span>—\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0= V<\/b><\/span><\/span><\/span>ox<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0<\/b><\/span><\/span><\/span>— 20<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0= V<\/b><\/span><\/span><\/span>ox<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ {\u221a(200)}<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0—<\/b><\/span><\/span><\/span><\/p>\n

V<\/b><\/span><\/span><\/span>ox<\/b><\/span><\/span><\/sub><\/span>=\u221a(200) m\/s.<\/b><\/span><\/span><\/span><\/p>\n

Na horizontal o movimento \u00e9 uniforme com V<\/b><\/span><\/span><\/span>ox<\/b><\/span><\/span><\/sub><\/span>=\u221a(200)m\/s (constante)\u00a0<\/b><\/span><\/span><\/span>e a bola\u00a0<\/b><\/span><\/span><\/span>demora 2s\u00a0<\/b><\/span><\/span><\/span>para atingir a parede —<\/b><\/span><\/span><\/span>x=V<\/b><\/span><\/span><\/span>ox<\/b><\/span><\/span><\/sub><\/span>t<\/b><\/span><\/span><\/span>=\u221a(200).t=14.2 —\u00a0<\/b><\/span><\/span><\/span>x\u224828m<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

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

06-\u00a0<\/b><\/span><\/span><\/span>Segundo a vertical, a componente da velocidade inicial vale V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.sen30<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>=20.1\/2\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>=10m\/s e a acelera\u00e7\u00e3o \u00e9 a da gravidade g=10m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 estudando o movimento na vertical voc\u00ea verifica que trata-se de um lan\u00e7amento vertical para cima\u00a0 —\u00a0 colocando a origem no ponto de lan\u00e7amento, orientando a trajet\u00f3ria para cima e aplicando a equa\u00e7\u00e3o de<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Torricelli lembrando que na altura m\u00e1xima h\u2019 a velocidade vertical \u00e9 nula, V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>=0\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0= V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0\u2013 2.g.h\u2019\u00a0 —\u00a0 0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0= 10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0\u2013<\/b><\/span><\/span><\/span><\/p>\n

2.10.h\u2019\u00a0 —\u00a0 h\u2019=100\/20=5m\u00a0 —\u00a0 como a altura pedida \u00e9 em rela\u00e7\u00e3o ao solo\u00a0 —\u00a0 h=5 + 5=10m\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- B.<\/b><\/span><\/span><\/span><\/p>\n

07-\u00a0a) Em todo lan\u00e7amento obl\u00edquo o movimento na dire\u00e7\u00e3o horizontal \u00e9 retil\u00edneo e uniforme, com velocidade\u00a0 horizontal , valendo\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>ox<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>cos45<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>ox<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.\u221a2\/2\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>ox<\/b><\/span><\/span><\/sub><\/span>=1,4V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\/2\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>ox<\/b><\/span><\/span><\/sub><\/span>=(\u221a2\/2)V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 equa\u00e7\u00e3o do movimento na horizontal\u00a0 —\u00a0 x=x<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0+ V<\/b><\/span><\/span><\/span>ox<\/b><\/span><\/span><\/sub><\/span>t\u00a0 — \u00a0para percorrer os x=80m na horizontal ele demora certo tempo t\u00a0 —\u00a0 80=0 + (\u221a2\/2)V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>t\u00a0 —\u00a0 t= 80\u221a2\/V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0(I)\u00a0 —\u00a0 na dire\u00e7\u00e3o vertical quando a bola chega ao solo, no instante t, ela ocupa a<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

ordenada y=-0,5m sendo lan\u00e7ada com velocidade inicial vertical V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>sen45<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>=(\u221a2\/2).V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 equa\u00e7\u00e3o de um lan\u00e7amento vertical\u00a0 — y=y<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0+ V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>t \u2013 gt<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 – 0,5 = 0 + (\u221a2\/2).V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.80\u221a2\/V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 5.t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 -0,5 = 80 – 5t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=80,5\/5=16,1\u00a0 —\u00a0 t\u22484s (tempo que demora para percorrer 8om na horizontal, que \u00e9 o mesmo tempo que demora para subir, descer e chegar ao solo)\u00a0 —\u00a0 substituindo t=4s em (I)\u00a0 —\u00a0\u00a0 t= 80\u221a2\/V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 4= 80\u221a2\/V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0= 80\u221a2\/4=28m\/s\u00a0 —\u00a0\u00a0V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=28m\/s ou V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=20\u221a2m\/s.<\/b><\/span><\/span><\/span><\/p>\n

b) Na altura m\u00e1xima a componente vertical da velocidade \u00e9 nula (V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>=0)\u00a0 —\u00a0 equa\u00e7\u00e3o de Torricelli\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0= V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0\u2013 2.g.h<\/b><\/span><\/span><\/span>max<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0= [V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>. (\u221a2\/2)]<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0\u2013 2.10.h<\/b><\/span><\/span><\/span>max<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0\u00a0 0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0= [20\u221a2.(\u221a2\/2)]<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0\u2013 2.10.h<\/b><\/span><\/span><\/span>max<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=400 \u2013 20h<\/b><\/span><\/span><\/span>max<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 h<\/b><\/span><\/span><\/span>max<\/b><\/span><\/span><\/sub><\/span>=20m (em rela\u00e7\u00e3o ao ponto de lan\u00e7amento)\u00a0 —\u00a0 em rela\u00e7\u00e3o ao solo a altura m\u00e1xima atingida ser\u00e1\u00a0 —\u00a0 H<\/b><\/span><\/span><\/span>max<\/b><\/span><\/span><\/sub><\/span>=20,0 + 0,5\u00a0 —<\/b><\/span><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/span>H<\/b><\/span><\/span><\/span>max<\/b><\/span><\/span><\/sub><\/span>=20,5m.<\/b><\/span><\/span><\/span><\/p>\n

08-\u00a0Trata-se de um lan\u00e7amento obl\u00edquo e, se voc\u00ea n\u00e3o domina a teoria voc\u00ea a encontrar\u00e1 em fisicaevestibular.com.br \u2013 mec\u00e2nica \u2013 cinem\u00e1tica \u2013 lan\u00e7amento obliquo\u00a0 —\u00a0 em todo lan\u00e7amento obl\u00edquo, no ponto mais alto da trajet\u00f3ria a<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

componente vertical da velocidade (\"\") \u00e9 nula existindo apenas a componente horizontal (\"\") que \u00e9 constante e de intensidade\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>ox<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.cos60<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>=30.0,5=15m\/s\u00a0 —\u00a0\u00a0R- B.<\/b><\/span><\/span><\/span><\/p>\n

09-\u00a0Trata-se de um lan\u00e7amento horizontal cuja teoria \u00e9 fornecida a seguir:<\/b><\/span><\/span><\/span><\/p>\n

Colocando-se a origem do sistema de refer\u00eancia no ponto de lan\u00e7amento, orienta-se, por exemplo, o eixo X para a direita e o eixo Y para baixo.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

\"\"<\/p>\n

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

\"\"Segundo o eixo Y\u00a0 —\u00a0 trata-se de um movimento uniformemente variado com velocidade inicial V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=0, ou seja, \u00e9 uma\u00a0queda livre com o corpo abandonado da origem, sujeito apenas \u00e0 acelera\u00e7\u00e3o da gravidade, de intensidade\u00a0g, dire\u00e7\u00e3o vertical e sentido para baixo.\u00a0<\/b><\/span><\/span><\/span>Equa\u00e7\u00f5es:<\/b><\/span><\/span><\/span><\/span><\/p>\n

S = S<\/b><\/span><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub><\/span>=\u00a0+ V<\/b><\/span><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub><\/span>.t + at<\/b><\/span><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 Y= 0 + 0.t + gt<\/b><\/span><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0\u00a0Y=g.t<\/b><\/span><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup><\/span>\/2<\/b><\/span><\/span><\/span><\/span><\/p>\n

V<\/b><\/span><\/span><\/span><\/span>y<\/b><\/span><\/span><\/span><\/sub><\/span>\u00a0= V<\/b><\/span><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/span><\/sub><\/span>\u00a0+ a.t\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span><\/span>y<\/b><\/span><\/span><\/span><\/sub><\/span>= 0 + g.t\u00a0 —\u00a0\u00a0V<\/b><\/span><\/span><\/span><\/span>y<\/b><\/span><\/span><\/span><\/sub><\/span>=g.t<\/b><\/span><\/span><\/span><\/span><\/p>\n

V<\/b><\/span><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup><\/span>=V<\/b><\/span><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/span><\/sup><\/span>\u00a0+ 2.a.<\/b><\/span><\/span><\/span><\/span>\u0394<\/b><\/span><\/span><\/span>S\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span><\/span>y<\/b><\/span><\/span><\/span><\/sub><\/span>2\u00a0= V<\/b><\/span><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/span><\/sup><\/span>\u00a0+ 2.g.<\/b><\/span><\/span><\/span><\/span>\u0394<\/b><\/span><\/span><\/span>h\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span><\/span>y<\/b><\/span><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/span><\/sup><\/span>\u00a0= 0<\/b><\/span><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup><\/span>\u00a0+ 2.g.<\/b><\/span><\/span><\/span><\/span>\u0394<\/b><\/span><\/span><\/span>h\u00a0 —\u00a0\u00a0V<\/b><\/span><\/span><\/span><\/span>y<\/b><\/span><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/span><\/sup><\/span>\u00a0= 2.g.<\/b><\/span><\/span><\/span><\/span>\u0394<\/b><\/span><\/span><\/span>h<\/b><\/span><\/span><\/span><\/span><\/p>\n

O tempo de movimento na horizontal \u00e9 o mesmo de queda e vale t=0,6s\u00a0 —\u00a0 \u00a0na horizontal, Segundo o eixo X\u00a0 —\u00a0 X=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>t=8.0,6=4,8m\u00a0 —\u00a0 na vertical, segundo o eixo Y\u00a0 —\u00a0 Y=g.t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=10.(0,6)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 Y=1,6m\u00a0 —\u00a0 no instante em que as pedras chegam ao vag\u00e3o elas possuem duas velocidades, a horizontal de valor V<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>=8m\/e e a vertical de valor\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>=gt =10.0,6=6m\/s\u00a0 —\u00a0 a velocidade com que as pedras chegam ao vag\u00e3o corresponde \u00e0 soma vetorial das velocidades V<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>\u00a0e<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=V<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>=8<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ 6<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 V=10m\/s\u00a0 —\u00a0\u00a0R- D.<\/b><\/span><\/span><\/span><\/p>\n

10-\u00a0Abaixo est\u00e3o as equa\u00e7\u00f5es de um lan\u00e7amento obl\u00edquo:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

A. Falsa\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>ox<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.cos \u03b8=100.0,8=80m\/s.<\/b><\/span><\/span><\/span><\/p>\n

B. Falsa\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.sen\u03b8=100.0,6=60m\/s.<\/b><\/span><\/span><\/span><\/p>\n

C. Falsa\u00a0 —\u00a0 tempo para atingir a altura m\u00e1xima\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>y<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 gt\u00a0 —\u00a0 0=60 \u2013 10t\u00a0 —\u00a0 t=6s\u00a0 —\u00a0 tempo de perman\u00eancia no ar\u00a0 —\u00a0 t=2.6=12s\u00a0 —\u00a0 alcance horizontal\u00a0 —\u00a0 x=V<\/b><\/span><\/span><\/span>ox<\/b><\/span><\/span><\/sub><\/span>t=80.12=960m.<\/b><\/span><\/span><\/span><\/p>\n

D. Falsa\u00a0 —\u00a0 na altura m\u00e1xima t=6s\u00a0 —\u00a0 y=V<\/b><\/span><\/span><\/span>oy<\/b><\/span><\/span><\/sub><\/span>t \u2013 gt<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 = 60.6 \u2013 10.6<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=240m.<\/b><\/span><\/span><\/span><\/p>\n

E Correta\u00a0 —\u00a0 Veja (C).<\/b><\/span><\/span><\/span><\/p>\n

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

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

01-\u00a0<\/b><\/span><\/span><\/span>Velocidade escalar (V) de um MCU\u00a0\"\"<\/b><\/span><\/span><\/span>\u00a0Para qualquer m\u00f3vel em MCU, que percorre espa\u00e7os iguais em intervalos de tempo iguais), o espa\u00e7o percorrido (\u0394S) durante um per\u00edodo (\u0394t=T que \u00e9 o intervalo de tempo que ele demora para efetuar uma volta completa) ser\u00e1\u00a0<\/b><\/span><\/span><\/span>\u0394S=2\"\"\u00a0<\/b><\/span><\/span><\/span>, onde\u00a0<\/b><\/span><\/span><\/span>R \u00e9 o raio da circunfer\u00eancia<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u0394S=2\"\"R e \u0394t=T\u00a0 —\u00a0 V= \u0394S\/\u0394t\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>V=2\"\"R\/T<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Lembrando que a frequ\u00eancia\u00a0f\u00a0de um MCU corresponde ao numero de voltas completas que o m\u00f3vel efetua na unidade de tempo e que a frequ\u00eancia \u00e9 o inverso do per\u00edodo, que\u00a0f=1\/T e que T=1\/f\u00a0voc\u00ea ter\u00e1<\/b><\/span><\/span><\/span><\/p>\n

V=2\"\"R\/(1\/f) —\u00a0V=2\"\"Rf\u00a0— o exerc\u00edcio fornece —\u00a0R=60cm=0,6m\u00a0— f=300rpm=300\/60 rps —\u00a0f=5 Hz\u00a0—\u00a0\"\"\u00a0—\u00a0V=2\"\"Rf=2x3x0,6×5 —\u00a0V=18m\/s\u00a0—<\/b><\/span><\/span><\/span>–\u00a0<\/b><\/span><\/span><\/span>R- B<\/b><\/span><\/span><\/span><\/p>\n

02-\u00a0Se voc\u00ea n\u00e3o domina a teoria, ela est\u00e1 a seguir:<\/b><\/span><\/span><\/span><\/p>\n

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

\"\"\u00a0<\/span>Pode-se 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><\/p>\n

\"\"<\/p>\n

Todos os pontos da correia (admitidos inextens\u00edveis) t\u00eam a mesma velocidade escalar V que todos os pontos da periferia de cada polia, desde que n\u00e3o ocorra deslizamento.<\/b><\/span><\/span><\/span><\/p>\n

O mesmo ocorre com todos os dentes da polia engrenada, que tem a mesma velocidade escalar V.<\/b><\/span><\/span><\/span><\/p>\n

Assim, V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.R<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0= W<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>.R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 2\u03c0\/T<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.R<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>= 2\u03c0\/T<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>.R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 \u00a02\u03c0f<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.R<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0=\u00a0 2\u03c0f<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>.R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0<\/span>Veja na figura abaixo que que, se\u00a0o raio da engrenagem A \u00e9 R<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>, o da B ser\u00e1 R<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=(11 \u2013 R<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>)<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

f<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=375 voltas — f<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=1000 voltas<\/b><\/span><\/span><\/span><\/p>\n

f<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>.R<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=f<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>.R<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0<\/b><\/span><\/span><\/span>— 375.R<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0= 1000.(11 \u2013 R<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>) — 375R<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=11000 \u2013 1000R<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0— 1375R<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=11000 — R<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=<\/b><\/span><\/span><\/span><\/p>\n

11000\/1375 —\u00a0<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=8cm\u00a0<\/b><\/span><\/span><\/span>— \u00e9 pedido o\u00a0<\/b><\/span><\/span><\/span>raio da menor que \u00e9 a B — R<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=11 \u2013 8 — R<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=3cm<\/b><\/span><\/span><\/span><\/p>\n

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

03-\u00a001. Correta<\/b><\/span><\/span><\/span>\u00a0— pelo enunciado, o\u00a0<\/b><\/span><\/span><\/span>n\u00famero de<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>voltas da catraca \u00e9 proporcional ao n\u00famero de voltas da coroa<\/b><\/span><\/span><\/span>, com\u00a0<\/b><\/span><\/span><\/span>raz\u00e3o de propor\u00e7\u00e3o igual \u00e0 raz\u00e3o entre os raios da coroa (<\/b><\/span><\/span><\/span>R<\/b><\/i><\/span><\/span><\/span>) e da catraca (<\/b><\/span><\/span><\/span>r<\/b><\/i><\/span><\/span><\/span>) — f<\/b><\/span><\/span><\/span>catraca<\/b><\/span><\/span><\/sub><\/span>\/f<\/b><\/span><\/span><\/span>coroa<\/b><\/span><\/span><\/sub><\/span>\u00a0= R\/r\u00a0<\/b><\/span><\/span><\/span>—\u00a0pelo enunciado f<\/b><\/span><\/span><\/span>coroa<\/b><\/span><\/span><\/sub><\/span>=1 —<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/span>catraca<\/b><\/span><\/span><\/sub><\/span>\/1 = R\/r —\u00a0<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/span>catraca<\/b><\/span><\/span><\/sub><\/span>=R\/r.<\/b><\/span><\/span><\/span><\/p>\n

02. Correta\u00a0<\/b><\/span><\/span><\/span>—\u00a0<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/span>catraca<\/b><\/span><\/span><\/sub><\/span>\/f<\/b><\/span><\/span><\/span>coroa<\/b><\/span><\/span><\/sub><\/span>\u00a0= R\/r\u00a0<\/b><\/span><\/span><\/span>— observe em R\/r que, como r \u00e9 constante, se voc\u00ea diminuir R , o quociente R\/r=k ir\u00e1 diminuir — ent\u00e3o\u00a0<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/span>catraca<\/b><\/span><\/span><\/sub><\/span>\/f<\/b><\/span><\/span><\/span>coroa<\/b><\/span><\/span><\/sub><\/span>=R\/r — f<\/b><\/span><\/span><\/span>catraca\u00a0<\/b><\/span><\/span><\/sub><\/span>=\/f<\/b><\/span><\/span><\/span>coroa<\/b><\/span><\/span><\/sub><\/span>.R\/r\u00a0<\/b><\/span><\/span><\/span>observe que\u00a0<\/b><\/span><\/span><\/span>r<\/b><\/span><\/span><\/span>\u00a0\u00e9\u00a0<\/b><\/span><\/span><\/span>inversamente proporcional \u00e0 f<\/b><\/span><\/span><\/span>catraca<\/b><\/span><\/span><\/sub><\/span>\u00a0— se\u00a0<\/b><\/span><\/span><\/span>r diminui, f<\/b><\/span><\/span><\/span>catraca\u00a0<\/b><\/span><\/span><\/sub><\/span>aumenta<\/b><\/span><\/span><\/span>\u00a0fazendo com que a roda gire mais, percorrendo uma dist\u00e2ncia maior.<\/b><\/span><\/span><\/span><\/p>\n

O4. Falsa<\/b><\/span><\/span><\/span>\u00a0— a de\u00a0<\/b><\/span><\/span><\/span>menor raio<\/b><\/span><\/span><\/span>\u00a0\u201cvarre\u201d\u00a0<\/b><\/span><\/span><\/span>maior \u00e2ngulo no mesmo tempo<\/b><\/span><\/span><\/span>, tendo, portanto\u00a0<\/b><\/span><\/span><\/span>maior velocidade angular.<\/b><\/span><\/span><\/span><\/p>\n

08. Falsa\u00a0<\/b><\/span><\/span><\/span>—\u00a0<\/b><\/span><\/span><\/span>coroa — R=15cm<\/b><\/span><\/span><\/span>\u00a0—\u00a0<\/b><\/span><\/span><\/span>catraca\u00a0<\/b><\/span><\/span><\/span>—\u00a0<\/b><\/span><\/span><\/span>r=<\/b><\/span><\/span><\/span>15\/4=<\/b><\/span><\/span><\/span>3,75cm<\/b><\/span><\/span><\/span>\u00a0—\u00a0<\/b><\/span><\/span><\/span>roda\u00a0<\/b><\/span><\/span><\/span>—\u00a0<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/span>r<\/b><\/span><\/span><\/sub><\/span>=5.r=5.3,75 —<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

R<\/b><\/span><\/span><\/span>r<\/b><\/span><\/span><\/sub><\/span>=18,75cm<\/b><\/span><\/span><\/span>\u00a0— f<\/b><\/span><\/span><\/span>coroa<\/b><\/span><\/span><\/sub><\/span>.R = f<\/b><\/span><\/span><\/span>catraca.<\/b><\/span><\/span><\/sub><\/span>r —\u00a0<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/span>.15 = f<\/b><\/span><\/span><\/span>catraca<\/b><\/span><\/span><\/sub><\/span>.3,75 —\u00a0<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/span>catraca<\/b><\/span><\/span><\/sub><\/span>=4 voltas<\/b><\/span><\/span><\/span>\u00a0—\u00a0<\/b><\/span><\/span><\/span>enquanto a coroa efetua<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>1 volta completa, a catraca e a roda efetua efetuam 4 voltas completas<\/b><\/span><\/span><\/span>\u00a0—\u00a0<\/b><\/span><\/span><\/span>deslocamento da roda ap\u00f3s as 4 voltas<\/b><\/span><\/span><\/span>\u00a0—<\/b><\/span><\/span><\/span>S=<\/b><\/span><\/span><\/span>2\u03c0R<\/b><\/span><\/span><\/span>r<\/b><\/span><\/span><\/sub><\/span>=2.3.18,75×4=450cm=<\/b><\/span><\/span><\/span>4,5m.<\/b><\/span><\/span><\/span><\/p>\n

16. Correta\u00a0<\/b><\/span><\/span><\/span>—\u00a0<\/b><\/span><\/span><\/span>roda — R<\/b><\/span><\/span><\/span>r<\/b><\/span><\/span><\/sub><\/span>=50cm<\/b><\/span><\/span><\/span>\u00a0—\u00a0<\/b><\/span><\/span><\/span>R=2r<\/b><\/span><\/span><\/span>\u00a0— f<\/b><\/span><\/span><\/span>coroa<\/b><\/span><\/span><\/sub><\/span>.R = f<\/b><\/span><\/span><\/span>catraca<\/b><\/span><\/span><\/sub><\/span>.r — 2.2r = f<\/b><\/span><\/span><\/span>catraca<\/b><\/span><\/span><\/sub><\/span>.r —\u00a0<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/span>catraca<\/b><\/span><\/span><\/sub><\/span>=4 voltas<\/b><\/span><\/span><\/span>\u00a0—\u00a0<\/b><\/span><\/span><\/span>dist\u00e2ncia percorrida pela roda ap\u00f3s essas 4 voltas<\/b><\/span><\/span><\/span>\u00a0—\u00a0<\/b><\/span><\/span><\/span>S=<\/b><\/span><\/span><\/span>4×2\u03c0R<\/b><\/span><\/span><\/span>r<\/b><\/span><\/span><\/sub><\/span>=8\u03c050=400\u03c0cm=<\/b><\/span><\/span><\/span>4\u03c0m<\/b><\/span><\/span><\/span>\u00a0—\u00a0<\/b><\/span><\/span><\/span>velocidade (pedida) da roda e consequentemente da bicicleta em t=1s — V=S\/t=4\u03c0m\/1s<\/b><\/span><\/span><\/span>\u00a0—<\/b><\/span><\/span><\/span><\/p>\n

V= 4\u03c0m\/s.<\/b><\/span><\/span><\/span><\/p>\n

R- (01, 02, 16)<\/b><\/span><\/span><\/span><\/p>\n

04-\u00a0Todos os pontos da correia (admitidos inextens\u00edveis) t\u00eam\u00a0a mesma velocidade escalar V que todos os pontos da periferia de cada polia, desde que n\u00e3o ocorra deslizamento—\u00a0\u00a0V=W.R\u00a0\u00a0—\u00a0\u00a0W=2\u03c0\/T\u00a0\u00a0—\u00a0\u00a0assim, V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.R<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0= W<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>.R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 2\u03c0\/T<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.R<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>= 2\u03c0\/T<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>.R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 \u00a02\u03c0f<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.R<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0=\u00a0 2\u03c0f<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>.R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Generalizando—\u00a0\u00a0f.R=constante=K\u00a0\u00a0—\u00a0\u00a0observe que o raio \u00e9 inversamente proporcional \u00e0 freq\u00fc\u00eancia (maior raio corresponde \u00e0 menor freq\u00fc\u00eancia)<\/b><\/span><\/span><\/span><\/p>\n

Note na montagem P que as polias\u00a01 e 2 est\u00e3o acopladas e\u00a0V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0= V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>—o mesmo acontece na montagem<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

ondeV<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0= V<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>—\u00a0\u00a0polias que est\u00e3o fixas no mesmo eixo de rota\u00e7\u00e3o possuem a mesma velocidade angular, (\u201cvarrem\u201d o mesmo \u00e2ngulo no mesmo tempo) assim,\u00a0\u00a0nas montagens P e Q voc\u00ea tem que\u00a0W<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0= W<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span><\/p>\n

\u00a0<\/span>Resumindo:as polias 1 e 3 giram com velocidades lineares iguais em pontos perif\u00e9ricos (V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0= V<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>) e, pelo enunciado, por quest\u00e3o de seguran\u00e7a, \u00e9 necess\u00e1rio que a serra possua menor velocidade linear, o que acontece na montagem Q onde a serra de fita est\u00e1\u00a0\u00a0acoplada \u00e0 polia de menor raio<\/b><\/span><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/span>—\u00a0\u00a0R- A.<\/b><\/span><\/span><\/span><\/p>\n

05-\u00a0Todos os pontos da correia e das periferias (extremidades) das polias B e C possuem a mesma velocidade (escalar, linear, tangencial ) V\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 rela\u00e7\u00e3o entre as velocidades angular W e tangencial V\u00a0 —\u00a0 W=V\/R\u00a0 —\u00a0 V=WR\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>.R<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0= W<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>.R<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0(I)\u00a0 —\u00a0 mas, as polias que giram em torno do mesmo eixo possuem a mesma velocidade angular W, ou seja, W<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=W<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>, que substitu\u00eddo em (I) fornece\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>R<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0= W<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>R<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0\u00a0 W<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=(R<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>\/R<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>).W<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>=(10\/2)W<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=5W<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 mas W=\u2206\u03b8(\u00e2ngulo varrido)\/ \u2206t (intervalo de tempo)\u00a0 —\u00a0 assim, enquanto a polia A efetua uma volta completa , no mesmo tempo a polia C efetuar\u00e1 5 voltas completas\u00a0 —\u00a0 observe ent\u00e3o que, se no tempo de desacelera\u00e7\u00e3o a polia a polia C efetuar n voltas at\u00e9 parar, a polia A efetuar\u00e1 5n voltas nesse mesmo tempo\u00a0 —\u00a0 c\u00e1lculo do n\u00famero de voltas completas efetuadas pela polia A enquanto est\u00e1 sendo desenrolada pelo fio\u00a0 —\u00a0 comprimento do fio, dado do exerc\u00edcio L=10\u03c0 m (II) —\u00a0 dist\u00e2ncia percorrida pela polia A ao efetuar uma volta completa\u00a0 —\u00a0 \u2206S<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>= 2\u03c0R<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=2\u03c0,1=2\u03c0 m (III) —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

Dividindo (II) por (III) voc\u00ea obt\u00e9m o n\u00famero de voltas dadas pela polia A at\u00e9 soltar o fio\u00a0 —\u00a0 10\u03c0\/2\u03c0=5voltas\u00a0 —\u00a0 mas,p\u00f3s essas 5 voltas ela continua girando at\u00e9 parar efetuando mais 2n voltas\u00a0 —\u00a0 n\u00famero total de voltas efetuadas pela polia A\u00a0 —\u00a0\u00a0n<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=5n + 5=5(n + 1)\u00a0 —\u00a0 calculando a acelera\u00e7\u00e3o a pela polia A onde o ponto P da corda percorre L=10\u03c0 m at\u00e9 parar (V=0) com velocidade inicial V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>, quando o fio come\u00e7a a ser puxado\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ 2.(-a),L\u00a0 —<\/b><\/span><\/span><\/span><\/p>\n

0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0– 2.(a).10\u03c0\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>=20\u03c0a\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>.R<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>=20\u03c0a\u00a0 —\u00a0 (5W<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>.1<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=20\u03c0a\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>=W<\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/sub><\/span>=2\u03c0 rad\/s (fornecido pelo gr\u00e1fico da figura 3\u00a0 — \u00a0(5.2\u03c0)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>.1<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=20\u03c0a\u00a0 —\u00a0 100\u03c0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=20\u03c0a\u00a0 —\u00a0\u00a0a=5\u03c0 m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>.\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\u00a0<\/span>R- A.<\/b><\/span><\/span><\/span><\/p>\n

06-\u00a0Como as rodas dentadas est\u00e3o perfeitamente ajustadas, o n\u00famero de dentes \u00e9 proporcional ao<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

raio de cada roda\u00a0 —f<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>.R<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=f<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>.R<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>.144=f<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>.126\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=0,875f<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=2\u03c0f<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 0,21=2.3.,875f<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=0,04Hz\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=2\u03c0f<\/b><\/span><\/span><\/span>B\u00a0<\/b><\/span><\/span><\/sub><\/span>\u00a0—W<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=2.3.0,04\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=0,24rad\/s\u00a0 —\u00a0\u00a0R- C.<\/b><\/span><\/span><\/span><\/p>\n

07-\u00a0O n\u00famero de dentes \u00e9 proporcional ao raio de cada roda\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>.R<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>\u00a0= f<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>.R<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span><\/sub><\/span>=6 voltas\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

—\u00a0 f<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>.21=6\/t.49\u00a0 —\u00a0f<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=294\/21\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=14 voltas completas\u00a0 —\u00a0 enquanto a roda do pedal gira 6 voltas completas, no mesmo tempo a roda traseira gira 14 voltas completas\u00a0 —\u00a0 se, em cada volta completa a roda traseira percorre 1,8m, em 14 voltas percorrer\u00e1\u00a0 —\u00a0 \u2206S=14.1,8=25,2 voltas\u00a0 —\u00a0\u00a0R- D<\/b><\/span><\/span><\/span><\/p>\n

08-\u00a0a) Para o port\u00e3o abrir totalmente ele deve girar um \u00e2ngulo \u2206\u03b8=90<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>=\u03c0\/2 rad\u00a0 —\u00a0 W=\u2206\u03b8\/\u2206t=(\u03c0\/2)\/\u2206t\u00a0 —\u00a0\u00a0\u2206t=\u03c0\/2W.<\/b><\/span><\/span><\/span><\/p>\n

b) Fun\u00e7\u00e3o hor\u00e1ria do MUV\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<\/b><\/span><\/span><\/span>\u00a0+ at<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 D= 0 + 0 + at<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 D=a.(\u03c0\/2W)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 D=a.(\u03c0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/4W<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>)\/2\u00a0 — D= \u03c0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>a\/8W<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —<\/b><\/span><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/span>a= 8W<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>D\/\u03c0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

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

RESOLU\u00c7\u00d5ES MOVIMENTO UNIFORME 01-\u00a001-\u00a0I.\u00a0Hor\u00e1rio do rush — carro — Vmc=12km\/h — \u0394Sc=12km — Vmc= \u0394Sc\/\u0394tc\u00a0— 12=12\/\u0394tc\u00a0— \u0394tc=1h — esse tempo \u00e9 o mesmo que o do metr\u00f4 fornecido pelo enunciado (1h) —\u00a0Falsa II.\u00a0Hor\u00e1rio do rush — carro — Vmc=12km\/h — \u0394Sc=12km — Vmc= \u0394Sc\/\u0394tc\u00a0— 12=12\/\u0394tc\u00a0— \u0394tc=1h — metr\u00f4 — Vmm=60km\/h — \u0394Sm=20km — Vmm= \u0394Sm\/\u0394tm\u00a0— 60=20\/\u0394tm\u00a0— \u0394tm=1\/3h —\u00a0Correta III.\u00a0Fora do<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":3242,"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-3244","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3244","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=3244"}],"version-history":[{"count":2,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3244\/revisions"}],"predecessor-version":[{"id":3268,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3244\/revisions\/3268"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3242"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=3244"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}