{"id":3600,"date":"2018-02-13T12:21:10","date_gmt":"2018-02-13T12:21:10","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=3600"},"modified":"2024-08-20T15:53:44","modified_gmt":"2024-08-20T15:53:44","slug":"resolucao-comentada-movimento-uniformemente-variado-muv","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/enem\/movimento-uniformemente-variado-muv\/resolucao-comentada-movimento-uniformemente-variado-muv\/","title":{"rendered":"Movimento Uniformemente Variado – Resolu\u00e7\u00e3o"},"content":{"rendered":"

Resolu\u00e7\u00e3o Comentada\u00a0<\/b><\/span><\/span><\/span><\/p>\n

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

01-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Quanto mais inclinada reta representativa da velocidade maior ser\u00e1 a acelera\u00e7\u00e3o, pois a=\u2206V\/\u2206t,ou seja, quanto maior \u2206V e menor \u2206t, maior ser\u00e1 a acelera\u00e7\u00e3o a\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- A<\/b><\/span><\/span><\/span><\/p>\n

02-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Trata-se da equa\u00e7\u00e3o de Torricelli, V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0= V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ 2.a.h\u00a0 — com V=0\u00a0 —\u00a0 cuja dedu\u00e7\u00e3o \u00e9 feita a partir de Vm = \u0394S\/\u0394t\u00a0 —\u00a0 onde\u00a0\u00a0 —\u00a0 Vm = (V + V0)\/2=h\/t\u00a0 —\u00a0 h=(V + Vo)t\/2\u00a0 —\u00a0 h = Vm.t\/2\u00a0 —\u00a0 observe nessa equa\u00e7\u00e3o que h \u00e9 proporcional a Vm\u00a0e at\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- E<\/b><\/span><\/span><\/span><\/p>\n

03-<\/b><\/span><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/span>Observe atentamente que o \u00fanico gr\u00e1fico coerente com o enunciado \u00e9 o da alternativa E\u00a0 —\u00a0\u00a0em cada trecho, ele acelera, mant\u00e9m velocidade constante, freia e para e assim por diante\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- E<\/b><\/span><\/span><\/span><\/p>\n

04-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe no enunciado que ele come\u00e7ou a reduzir a velocidade de 360kmh para 270kmh quando apagou\u00a0 —\u00a0 movimento retardado com velocidade diminuindo\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- C<\/b><\/span><\/span><\/span><\/p>\n

05-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Menor tempo \u2013 maior acelera\u00e7\u00e3o \u2013 Corvette\u00a0 —\u00a0\u00a0 a<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>=100\/4=25(km\/h)\/s\u00a0 —\u00a0 maior tempo \u2013 menor acelera\u00e7\u00e3o \u2013 Parati\u00a0 —-\u00a0 a<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>=100\/33,35=3(km\/h)\/s\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- C<\/b><\/span><\/span><\/span><\/p>\n

06-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>I \u2013 Falsa \u2013 \u00e9 no sentido contr\u00e1rio ao do movimento \u2013 a velocidade est\u00e1 diminuindo<\/b><\/span><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

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

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

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

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

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

07-<\/b><\/span><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/span>Observe a figura abaixo que ilustra a situa\u00e7\u00e3o descrita no instante t = 0\u00a0 —\u00a0 o ciclista ingl\u00eas (I) <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

executa movimento uniforme e o ciclista brasileiro (B) executa movimento uniformemente variado\u00a0 —\u00a0 a partir do instante mostrado (t = 0), as respectivas fun\u00e7\u00f5es hor\u00e1rias dos espa\u00e7os s\u00e3o\u00a0 —\u00a0 SI\u00a0= 15 + 22 t\u00a0 e\u00a0 SB\u00a0= 24 t\u00a0 + 0,4.t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 no encontro, essas equa\u00e7\u00f5es s\u00e3o igualadas\u00a0 —\u00a0 \u00a0<\/b><\/span><\/span><\/span><\/p>\n

24 t + 0,2 t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 = 15 + 22 t\u00a0 —\u00a0\u00a0 0,2 t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ 2 t \u2013 15 = 0\u00a0 —\u00a0 Resolvendo essa equa\u00e7\u00e3o do 2\u00ba grau\u00a0 —\u00a0 t<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0= -15 s\u00a0 e\u00a0 t<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0= 5 s\u00a0 —\u00a0\u00a0 t = 5 s\u00a0 —\u00a0 o ciclista brasileiro alcan\u00e7a o ciclista ingl\u00eas no instante t = 5 s\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R-\u00a0E<\/b><\/span><\/span><\/span><\/p>\n

<\/a>08-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>C\u00e1lculo da dist\u00e2ncia percorrida pelo\u00a0predador\u00a0 —\u00a0 entre 0 e 4s\u00a0 —\u00a0 acelerado com a=(54\/3,6 \u2013 0)\/(4 \u2013 0)=3,75m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 \u0394S<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>t + at<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=0.4 +3,75.16\/2\u00a0 —\u00a0 \u0394S<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=90m\u00a0 —\u00a0 entre 4s e 10s \u00e9 um MU com V=15ms\u00a0 —\u00a0 V= \u0394S\/\u0394t\u00a0 —\u00a0 15= \u0394S<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\/6\u00a0 —\u00a0 \u0394S<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=90m\u00a0 —\u00a0\u0394S<\/b><\/span><\/span><\/span>predador<\/b><\/span><\/span><\/sub><\/span>=90 + 90\u00a0 —\u00a0\u00a0\u0394S<\/b><\/span><\/span><\/span>predador<\/b><\/span><\/span><\/sub><\/span>=180m<\/b><\/span><\/span><\/span><\/p>\n

C\u00e1lculo da dist\u00e2ncia percorrida pela\u00a0presa\u00a0\u00a0—\u00a0 entre 0 e 5s\u00a0 —\u00a0 acelerado com a=(12 \u2013 0)\/(5 \u2013 0)=2,4m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 \u0394S<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>t + at<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=0.5 +2,4.25\/2\u00a0 —\u00a0 \u0394S<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=30m\u00a0 —\u00a0 entre 5s e 10=14s \u00e9 um MU com V=12ms\u00a0 —\u00a0 V= \u0394S\/\u0394t\u00a0 —\u00a0 12= \u0394S<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\/9\u00a0 —\u00a0 \u0394S<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=108m\u00a0 —\u00a0 \u0394S<\/b><\/span><\/span><\/span>predador<\/b><\/span><\/span><\/sub><\/span>=30 + 108\u00a0 —\u00a0\u00a0\u0394S<\/b><\/span><\/span><\/span>presa<\/b><\/span><\/span><\/sub><\/span>=138m<\/b><\/span><\/span><\/span><\/p>\n

Resposta\u00a0 —\u00a0 \u0394S=180 \u2013 138\u00a0 —\u00a0 \u0394S=42m\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- C<\/b><\/span><\/span><\/span><\/p>\n

 <\/p>\n

Voltar para os exerc\u00edcios<\/span><\/a><\/h3>\n

\u00a0<\/span><\/p>\n","protected":false},"excerpt":{"rendered":"

Resolu\u00e7\u00e3o Comentada\u00a0 Movimento Uniformemente Variado (MUV) 01-\u00a0Quanto mais inclinada reta representativa da velocidade maior ser\u00e1 a acelera\u00e7\u00e3o, pois a=\u2206V\/\u2206t,ou seja, quanto maior \u2206V e menor \u2206t, maior ser\u00e1 a acelera\u00e7\u00e3o a\u00a0 —\u00a0\u00a0R- A 02-\u00a0Trata-se da equa\u00e7\u00e3o de Torricelli, V2\u00a0= Vo2\u00a0+ 2.a.h\u00a0 — com V=0\u00a0 —\u00a0 cuja dedu\u00e7\u00e3o \u00e9 feita a partir de Vm = \u0394S\/\u0394t\u00a0 —\u00a0 onde\u00a0\u00a0 —\u00a0 Vm =<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":3598,"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-3600","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3600","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=3600"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3600\/revisions"}],"predecessor-version":[{"id":10627,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3600\/revisions\/10627"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3598"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=3600"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}