{"id":3593,"date":"2018-02-13T12:12:49","date_gmt":"2018-02-13T12:12:49","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=3593"},"modified":"2024-08-20T23:34:25","modified_gmt":"2024-08-20T23:34:25","slug":"resolucao-comentada-movimento-uniforme-mu","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/enem\/movimento-uniforme-mu\/resolucao-comentada-movimento-uniforme-mu\/","title":{"rendered":"Movimento Uniforme (MU) – Resolu\u00e7\u00e3o"},"content":{"rendered":"

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

01-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Se a velocidade \u00e9 constante, a reta representativa \u00e9 paralela ao eixo dos tempos\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- C<\/b><\/span><\/span><\/span><\/p>\n

02-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>A velocidade m\u00e9dia de uma pessoa caminhando \u00e9 da ordem de grandeza variando entre 4km\/h e 5km\/h\u00a0 —\u00a0 se o tempo estiver em horas a velocidade m\u00e9dia fornecida pelo gr\u00e1fico seria aproximadamente V=10\/2,2=4,5km\/h\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- C<\/b><\/span><\/span><\/span><\/p>\n

03-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>O comprimento de uma circunfer\u00eancia completa sobre a linha do equador vale \u2206S\u2019=2\u03c0R=2.3,14.6.370=40.003.6\u00a0 —\u00a0 \u2206S\u2019\u224840.000km\u00a0 — o avi\u00e3o voando entre essas duas cidades percorre metade dessa circunfer\u00eancia\u00a0 —\u00a0 \u2206S=20.000km\u00a0 —\u00a0 V=\u2206S\/\u2206t\u00a0 —\u00a0 800=20.000\/\u2206t\u00a0 —\u00a0 \u2206t=20.000\/800\u00a0 —\u00a0 \u2206t=25 horas\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- C<\/b><\/span><\/span><\/span><\/p>\n

04-<\/b><\/span><\/span><\/span>A onda mais r\u00e1pida \u00e9 a P (reta mais inclinada, maior velocidade) e a diferen\u00e7a de tempo \u00e9 aproximadamente (24 \u2013 16)=8s\u00a0 — <\/b><\/span><\/span><\/span>R- B<\/b><\/span><\/span><\/span><\/p>\n

05-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>I) azul \u2013(16-4)=12h\u00a0 —\u00a0 vermelho\u00a0 —\u00a0 (18-6)=12h\u00a0 —\u00a0\u00a0Correta\u00a0 —\u00a0 II) Vmazul=(720-0)12=60km\/h\u00a0 —\u00a0 Vmvermelho=(0-720)\/12=-60km\/h\u00a0 —\u00a0\u00a0Correta\u00a0 —\u00a0 III)\u00a0Correta \u2013\u00a0veja o gr\u00e1fico\u00a0 —\u00a0 IV)\u00a0Correta\u00a0\u2013 veja afirma\u00e7\u00f5es anteriores\u00a0 —\u00a0 V)\u00a0Correta\u00a0\u2013 \u00e9 o ponto onde as retas azul e vermelha o se interceptam<\/b><\/span><\/span><\/span><\/p>\n

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

06-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>O gr\u00e1fico sugere\u00a0 — \u00a0movimento progressivo acelerado (corrida para pegar o \u00f4nibus); repouso (espera no ponto); movimento uniforme regressivo (volta para casa); novo repouso (espera pelo t\u00e1xi) e, finalmente, movimento progressivo uniforme (movimento do t\u00e1xi).\u00a0<\/b><\/span><\/span><\/span><\/p>\n

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

07-<\/b><\/span><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/span>Observe que as fotos assinaladas s\u00e3o iguais.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Entre a primeira e a \u00faltima foram tiradas 10 fotos (cuidado: a primeira n\u00e3o conta. Ela \u00e1 o referencial) \u00a0— f=0,5Hz\u00a0 —\u00a0 f=1\/T\u00a0 —\u00a0 0,5=1\/T\u00a0 —\u00a0 T=\u0394t=2,0s\u00a0 —\u00a0 V=\u0394S\/\u0394t=1,5.10\/2\u00a0 —\u00a0 V=7,5m\/s\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- B<\/b><\/span><\/span><\/span><\/p>\n

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

08-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Esquematizando a situa\u00e7\u00e3o inicial<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

SP= 0 + 100t\u00a0 —\u00a0 SM= 10 + 80t\u00a0 —\u00a0 no encontro\u00a0 —\u00a0 SP\u00a0= SM\u00a0 —\u00a0 100t=10 + 80t\u00a0 —\u00a0 t=0,5h\u00a0 —\u00a0 SP=100t=100.0,5=50km\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- D<\/b><\/span><\/span><\/span><\/p>\n

09-\u00a0Fixando um ponto P no final do trem onde coloca-se a origem da trajet\u00f3ria que \u00e9 orientada para a direita\u00a0 —\u00a0 na situa\u00e7\u00e3o inicial<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

deduz-se a equa\u00e7\u00e3o hor\u00e1ria do ponto P e na situa\u00e7\u00e3o final S=320m\u00a0 —\u00a0 inicial \u2013 S=So\u00a0+ Vt\u00a0 —\u00a0 S= 0 + 10t\u00a0 —\u00a0 S= 10t\u00a0 —\u00a0 final \u2013 S=320m\u00a0 —\u00a0 320=10t\u00a0 —\u00a0 t=32s\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- A<\/b><\/span><\/span><\/span><\/p>\n

 <\/p>\n

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

Resolu\u00e7\u00e3o Comentada 01-\u00a0Se a velocidade \u00e9 constante, a reta representativa \u00e9 paralela ao eixo dos tempos\u00a0 —\u00a0\u00a0R- C 02-\u00a0A velocidade m\u00e9dia de uma pessoa caminhando \u00e9 da ordem de grandeza variando entre 4km\/h e 5km\/h\u00a0 —\u00a0 se o tempo estiver em horas a velocidade m\u00e9dia fornecida pelo gr\u00e1fico seria aproximadamente V=10\/2,2=4,5km\/h\u00a0 —\u00a0\u00a0R- C 03-\u00a0O comprimento de uma circunfer\u00eancia completa sobre<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":3591,"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-3593","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3593","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=3593"}],"version-history":[{"count":4,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3593\/revisions"}],"predecessor-version":[{"id":10680,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3593\/revisions\/10680"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3591"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=3593"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}