{"id":8179,"date":"2022-02-28T09:00:00","date_gmt":"2022-02-28T12:00:00","guid":{"rendered":"https:\/\/professorpinguim.com.br\/blog\/?p=8179"},"modified":"2023-08-06T21:01:13","modified_gmt":"2023-08-07T00:01:13","slug":"para-treinar-exercicios-de-cinematica-vetorial","status":"publish","type":"post","link":"https:\/\/professorpinguim.com.br\/blog\/para-treinar-exercicios-de-cinematica-vetorial\/","title":{"rendered":"Para treinar &#8211; Exerc\u00edcios de Cinem\u00e1tica Vetorial"},"content":{"rendered":"\n<p class=\"wp-block-paragraph\">Fala, pessoal! Nesta aula, vamos resolver alguns exerc\u00edcios sobre cinem\u00e1tica vetorial, incluindo deslocamento vetorial e velocidade vetorial m\u00e9dia. Mas vamos dar uma revisada no conte\u00fado antes. Acompanhe!<\/p>\n\n\n\n<!--more-->\n\n\n\n<h2 class=\"wp-block-heading\">Deslocamento vetorial<\/h2>\n\n\n\n<p class=\"wp-block-paragraph\">Quando um corpo sai de uma posi\u00e7\u00e3o para outra, temos dois tipos de deslocamento: o deslocamento escalar, que segue a trajet\u00f3ria percorrida (\u0394s), e o deslocamento vetorial, que liga o in\u00edcio do movimento ao final (uma reta), desprezando a trajet\u00f3ria. Veja:<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh3.googleusercontent.com\/tD5Cotcf63jXvhAEY3iHSjVcUd9hb4KsQ6B-XlIiTGixTJmliKUh5hPBdd-WVrTCe6gZbZdMtCfVEr25mjyVftSJ92dsVQtbQF54pTfuXE3glMEGNXo7vsDX3OTSWbDpcFIegTg9\" alt=\"\"\/><\/figure>\n\n\n\n<h2 class=\"wp-block-heading\">Velocidade vetorial m\u00e9dia<\/h2>\n\n\n\n<p class=\"wp-block-paragraph\">Vamos supor que o corpo fa\u00e7a a trajet\u00f3ria descrita na figura abaixo (da esquerda para a direita):<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh5.googleusercontent.com\/jZAXCdO6tN9IqfD_eAJTPkGmDifkwuJPOVc6j4-X2M-R6pQNxZb4uAWl08WGJIkkggWlbNJ5HFI8VI_5HPy4Tc5Ujzl_juZ8l-iTLfkU-_PHbI_C1oa2IJE59F7gTZMXwd4IFrnJ\" alt=\"\"\/><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\">Se quisermos medir a varia\u00e7\u00e3o de espa\u00e7o do corpo pela trajet\u00f3ria, basta contar quantos quarteir\u00f5es ele andou. Por outro lado, como vimos, para o vetor do deslocamento, temos que desenhar uma linha reta entre o in\u00edcio e o final da trajet\u00f3ria.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Nesse sentido, vamos lembrar que a velocidade escalar m\u00e9dia \u00e9 aquela em que utilizamos a trajet\u00f3ria que o corpo descreveu:<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh6.googleusercontent.com\/-AMRgyp_L8k7NUH3FpCbBZwkuVCE7Optg8Rw2br-HFuRdwA4yGnTKH7BhSJbsmYVccT_Z3wtBUT6lBYltPQBYx7U4tUAzQkK61CSABlvaZdykgJjFJVQA6wZvvl4wH39vvO3SGRp\" alt=\"\"\/><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\">J\u00e1 para calcular a velocidade vetorial m\u00e9dia, isto \u00e9, a velocidade m\u00e9dia na reta do deslocamento vetorial, a f\u00f3rmula muda um pouco:<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh4.googleusercontent.com\/3q8WUnPe3OkOVlKjZUtLYGTrRHpHgM5jKwuq2NNgeGCcYNacwLtMjSYoaiK0H-jaS_h4gMFa8iEVag64TQORZ9ToRakZbhLiByZTiv7m13UDNKZLW73BRmn9BE4gNjWNbnP96Cqp\" alt=\"\"\/><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\">O c\u00e1lculo do m\u00f3dulo do vetor deslocamento vai depender da figura que formamos. No nosso exemplo acima, repare que podemos considerar <em>d<\/em> como a hipotenusa de um tri\u00e2ngulo ret\u00e2ngulo:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Assim, bastar\u00edamos saber quanto vale cada quarteir\u00e3o para calcularmos os catetos e utilizarmos o Teorema de Pit\u00e1goras.<\/p>\n\n\n\n<hr class=\"wp-block-separator\"\/>\n\n\n\n<h2 class=\"wp-block-heading\">Exerc\u00edcios sobre Cinem\u00e1tica Vetorial<\/h2>\n\n\n\n<h3 class=\"wp-block-heading\">Quest\u00e3o 1<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>(Mackenzie) Um avi\u00e3o, ap\u00f3s deslocar-se 120 km para nordeste (NE), desloca-se 160 km para sudeste (SE). Sendo um quarto de hora, o tempo total dessa viagem, o m\u00f3dulo da velocidade vetorial m\u00e9dia do avi\u00e3o, nesse tempo, foi de<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">a) 320 km\/h<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">b) 480 km\/h<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">c) 540 km\/h<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">d) 640 km\/h<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">e) 800 km\/h<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>RESOLU\u00c7\u00c3O:<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Em quest\u00f5es com vetores, o ideal \u00e9 sempre desenhar os trajetos que s\u00e3o fornecidos, at\u00e9 porque geralmente formar\u00e3o uma figura. Nesse caso, se nos basearmos na rosa dos ventos, o \u00e2ngulo formado entre o deslocamento para nordeste e para sudeste ser\u00e1 de 90\u00ba.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Ou seja, o deslocamento vetorial que buscamos ser\u00e1 a hipotenusa de um tri\u00e2ngulo ret\u00e2ngulo cujos catetos s\u00e3o os deslocamentos fornecidos no enunciado. Algo como:<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh4.googleusercontent.com\/KYx2-Ib1-PjWK8RmAME_hzahmjdYAyhgL9JWfjKsYa0tg3heZoc8bkdGLKMhzJ_xFBxoANuhA8dZcXwz_nVgpp3V_Ro7-Fdn3ZMgs0Z08bo0fDW4I7bZsHdsYTfpk3jLOiD3HRuj\" alt=\"\"\/><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\">Agora, basta aplicar Pit\u00e1goras:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">d\u00b2 = d<sub>1<\/sub>\u00b2 + d<sub>2<\/sub>\u00b2<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">d\u00b2 = 120 \u00b2 + 160\u00b2<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">d = 200 km<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Agora podemos calcular a velocidade vetorial m\u00e9dia. E repare que a quest\u00e3o nos pede em m\u00f3dulo. Ent\u00e3o:<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh5.googleusercontent.com\/AdCpvzSMAMGwtj9j8ilSEAlnetgl4_NCf4YzCHvmoUzQhK1UuowFav2xcx3QoaCn9uuhXWqIfVX9Lc9HdUMs3jMvHVzwp-bmZuuT-R1MVQ2qke1kcf0WOVPfEVBUDg1TsehA9ht3\" alt=\"\"\/><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>RESPOSTA: E<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Quest\u00e3o 2<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>(Uesc) Considere um m\u00f3vel que percorre a metade de uma pista circular de raio igual a 10,0 m em 10,0 s. Adotando-se \u221a2 como sendo 1,4 e \u03c0 igual a 3, \u00e9 correto afirmar:<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">a) O espa\u00e7o percorrido pelo m\u00f3vel \u00e9 igual a 60,0 m.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">b) O deslocamento vetorial do m\u00f3vel tem m\u00f3dulo igual a 10,0 m.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">c) A velocidade vetorial m\u00e9dia do m\u00f3vel tem m\u00f3dulo igual a 2,0 m\/s.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">d) O m\u00f3dulo da velocidade escalar m\u00e9dia do m\u00f3vel \u00e9 igual a 1,5 m\/s.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">e) A velocidade vetorial m\u00e9dia e a velocidade escalar m\u00e9dia do m\u00f3vel t\u00eam a mesma intensidade<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>RESOLU\u00c7\u00c3O:<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">a) Veja que a quest\u00e3o afirma que o m\u00f3vel percorre metade de uma pista circular. Esse deslocamento ser\u00e1 do tipo escalar (\u0394s), que, como o enunciado informou, vale metade do per\u00edmetro da circunfer\u00eancia:<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh6.googleusercontent.com\/TY0bumLWiprUBuuTBueNzqwsdAWhiMMsRb2hFcJj8-cS-UP3zj9uXnXpXH8ybpMXj-ApDAN9aEf-J0gYTpA0OVxNRmFvLsYxNX-Yz0b7WlRkKokIfO__dBkuKOmtc5zilRT80c2Z\" alt=\"\"\/><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\">\u0394s = 30 m<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">b) Para encontrarmos o deslocamento vetorial, temos que desenhar. Isso vai nos ajudar a entender a situa\u00e7\u00e3o da quest\u00e3o. Rabiscando temos:<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh6.googleusercontent.com\/qLoHs82Ohs4d4Yx4xlQyVkexQf_3pKFoS8LV4y8xebd5CEiNypCVKom2gLnfxws7AAShYn9iSkta2kQlpftneluUN6JO8gSRUQio3vGftJEyECbjLfuWwJ1C4kdYXmbPdJnTCXvQ\" alt=\"\"\/><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\">Repare que <em>d<\/em> nada mais \u00e9 do que o di\u00e2metro, ou seja:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">d = 2 . R<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">d = 20 m<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">c) Agora, vamos calcular a velocidade vetorial m\u00e9dia. Vamos utilizar a f\u00f3rmula:<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh4.googleusercontent.com\/SxMLv8pHcZISkQJJiY9VxYlzJZbz87pUnO6czheXOZOvQmM9MNTCogirtRbK_UEFnjZaWrK8IEwBwaumFj2ZiR_7teEkUZd_GKXg02NZ01VIEuzqI2iQoWdayjxS7JG3Q6fXiOlo\" alt=\"\"\/><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\">d) Para encontrarmos a velocidade escalar m\u00e9dia, basta utilizar a f\u00f3rmula:<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh4.googleusercontent.com\/efA6QRcMnpjYVgknZDkcXru6W4bZ0sKP_QC0NtCl-L64GvNifnkm8j05KCdmHceu5MCJAoI_CskKYybgrw9_aoLjeMgaqpbag5aXr1je5ndwEvMsGe4ChIib17xXMDoeUg2TCaEi\" alt=\"\"\/><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>RESPOSTA: C<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Quest\u00e3o 3<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>(Mack) A figura em escala mostra os vetores deslocamento de uma formiga, que, saindo do ponto A, chegou ao ponto B, ap\u00f3s 3 minutos e 20 segundos. O m\u00f3dulo do vetor velocidade m\u00e9dia do movimento da formiga, nesse trajeto, foi de:<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh6.googleusercontent.com\/F2c75ZfUWK5gkEGq62iZlqEonYDItfumCCDrVv0iO7o454psJw2o96ow0XFi-BbtKByIp0dgxaREuHRLD3JzHwCnbuBl-cA09j6CQ2sG17FZhFS6TOfgo1ZEpelybjToJcYGjJ48\" alt=\"\"\/><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\">a) 0,15 cm\/s<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">b) 0,20 cm\/s<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">c) 0,25 cm\/s<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">d) 0,30 cm\/s<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">e) 0,40 cm\/s<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>RESOLU\u00c7\u00c3O:<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">A primeira coisa: 3 minutos e 20 segundos equivalem a 200 s. Esse \u00e9 o nosso \u0394t. Em seguida, temos que desenhar o vetor do deslocamento. Repare, no entanto, no que vai acontecer:<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh4.googleusercontent.com\/5T5FKZ2raF-PAXFuTpWJJ6cs8FYrrO1Yq_c7DBZlHCXEw8JEVQ1IcpoY4qAWr7dPeCRFXGOnqSssBaHX8-lvx5eP32V26jlOm5Fv_CmqjdrS9gHds28tlloyjgjTHtRnbCih-cGS\" alt=\"\"\/><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\">Temos o vetor do deslocamento como a hipotenusa de um tri\u00e2ngulo ret\u00e2ngulo. Portanto:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">d\u00b2 = 30\u00b2 + 40\u00b2<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">d = 50 cm<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Ent\u00e3o, para descobrirmos a velocidade vetorial m\u00e9dia:<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh5.googleusercontent.com\/pFwa_PqpV5JMyVcL1Huhu-5X3NzbprcgxQ9MJHJ5L2n7oAFfRUo0TbUGc7k0rUNeiUNKqA3qgQGRZfFWaj7_xnN3mwTrhrda2fbONvfgxES8h1nL_4ommrDnuvaHlLCEnIHLWEGm\" alt=\"\"\/><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>RESPOSTA: C<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Quest\u00e3o 4<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>(Pucrj) Um veleiro deixa o porto navegando 70 km em dire\u00e7\u00e3o leste. Em seguida, para atingir seu destino, navega mais 100 km na dire\u00e7\u00e3o nordeste. Desprezando a curvatura da terra e admitindo que todos os deslocamentos s\u00e3o coplanares, determine o deslocamento total do veleiro em rela\u00e7\u00e3o ao porto de origem.<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>(Considere \u221a2 = 1,40 e \u221a5 = 2,20)<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">a) 106 km<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">b) 34 km<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">c) 154 km<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">d) 284 km<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">e) 217 km<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>RESOLU\u00c7\u00c3O:<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Novamente vamos ter que desenhar os deslocamentos. Lembre-se de que a dire\u00e7\u00e3o nordeste est\u00e1 a exatos 45\u00ba da dire\u00e7\u00e3o leste. Com isso, o \u00e2ngulo suplementar ser\u00e1 135\u00ba:<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh6.googleusercontent.com\/5gdEgpxOO9oJyUnrQYuQU3vciOAaY37LmIrIFjU36thxGymcjNPxuVjNiBna4VsPvD3unE0DHbu7hAdWVW8PAGmQiO4Ur4XdduWSFt67jJgkUxQ-tGH61NqJ63UQduuh5mjCYuCe\" alt=\"\"\/><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\">Portanto, basta aplicarmos a lei dos cossenos, tendo em mente que o sinal da parte final da f\u00f3rmula dever\u00e1 ser negativo, pois estamos calculando o lado oposto ao \u00e2ngulo dado:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">d\u00b2 = d<sub>1<\/sub>\u00b2 + d<sub>2<\/sub>\u00b2 &#8211; 2 . d<sub>1<\/sub> . d<sub>2<\/sub> . cos 135\u00ba<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/lh3.googleusercontent.com\/COywhmBqxH48-o-_Kcwnjv6E1BgcMRnLY3j4w-1-an1DesikwawHTJCYWuqbZiolHWshIIGDEWR_Gp0t2iD1zNDFSh5vuzFxOvMea6Uvx3C0TFaz6w-OTHBz41shPisZP_Fvm6al\" alt=\"\"\/><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\">d\u00b2 = 4900 + 10000 + 7000 . 1,40<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">d\u00b2 = 24700<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">d \u2245 157 km<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>RESPOSTA: C<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator\"\/>\n\n\n\n<p class=\"wp-block-paragraph\">Espero que voc\u00ea tenha entendido um pouco melhor como resolver <strong>exerc\u00edcios de cinem\u00e1tica vetorial<\/strong>. E se quiser praticar mais, confira minha live de resolu\u00e7\u00e3o de mais exerc\u00edcios sobre o tema. Assista:<\/p>\n\n\n\n<figure class=\"wp-block-embed-youtube wp-block-embed is-type-video is-provider-youtube wp-embed-aspect-16-9 wp-has-aspect-ratio\"><div class=\"wp-block-embed__wrapper\">\n<iframe title=\"Como estudar MEC\u00c2NICA para Vestibulares Paulistas | Professor Pinguim\" width=\"800\" height=\"450\" src=\"https:\/\/www.youtube.com\/embed\/zqy1GbT9lOM?feature=oembed\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share\" referrerpolicy=\"strict-origin-when-cross-origin\" allowfullscreen><\/iframe>\n<\/div><\/figure>\n\n\n\n<hr class=\"wp-block-separator\"\/>\n\n\n\n<figure class=\"wp-block-image\"><a href=\"https:\/\/curso.professorpinguim.com.br\/pt\/comprar\/produto\/curso-gratuito-matematica-para-aprender-fisica\" target=\"_blank\" rel=\"noopener noreferrer\"><img decoding=\"async\" src=\"https:\/\/professorpinguim.com.br\/blog\/wp-content\/uploads\/2022\/02\/curso-fisica-para-matematica-professor-pinguim.png\" alt=\"Curso gr\u00e1tis de Matem\u00e1tica para aprender F\u00edsica - Professor Pinguim\"\/><\/a><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>SAIBA MAIS<\/strong><br>\ud83d\udc27 <strong><a rel=\"noreferrer noopener\" href=\"https:\/\/professorpinguim.com.br\/blog\/5-questoes-resolvidas-sobre-movimento-uniformemente-variado-muv\/\" target=\"_blank\">Exerc\u00edcios sobre Movimento Uniformemente Variado (MUV)<\/a><\/strong><br>\ud83d\udc27 <strong><a rel=\"noreferrer noopener\" href=\"https:\/\/professorpinguim.com.br\/blog\/questoes-para-praticar-de-velocidade-relativa\/\" target=\"_blank\">Exerc\u00edcios sobre velocidade relativa<\/a><\/strong><br>\ud83d\udc27 <a rel=\"noreferrer noopener\" href=\"https:\/\/professorpinguim.com.br\/blog\/pinguimrevisa-cinematica-sp\/\" target=\"_blank\"><strong>Tudo sobre Cinem\u00e1tica<\/strong><\/a><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Me acompanhe nas redes sociais: curta a minha&nbsp;<a href=\"https:\/\/www.facebook.com\/professorpinguim\/\"><strong>p\u00e1gina no Facebook<\/strong><\/a><strong>,<\/strong>&nbsp;me&nbsp;siga no&nbsp;<a href=\"https:\/\/www.instagram.com\/professorpinguim\/\"><strong>Instagram<\/strong><\/a><strong>,<\/strong>&nbsp;se inscreva no&nbsp;<a href=\"https:\/\/www.youtube.com\/c\/PinguimV%C3%ADdeoF%C3%ADsica\/?sub_confirmation=1\"><strong>Youtube<\/strong><\/a>&nbsp;e participe do meu canal oficial no&nbsp;<a href=\"https:\/\/professorpinguim.com.br\/telegram\"><strong>Telegram<\/strong><\/a>.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Nesta aula, vamos resolver alguns exerc\u00edcios sobre cinem\u00e1tica vetorial, incluindo deslocamento vetorial e velocidade vetorial m\u00e9dia. Mas vamos dar uma revisada no conte\u00fado antes. Acompanhe!<\/p>\n","protected":false},"author":1,"featured_media":8181,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[420,744,355],"tags":[784,728,217,785],"class_list":["post-8179","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-cinematica","category-exercicios-resolvidos","category-mecanica","tag-cinematica-vetorial","tag-deslocamento-vetorial","tag-exercicios-resolvidos","tag-velocidade-vetorial-media"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v20.12 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Para treinar - Exerc\u00edcios de Cinem\u00e1tica Vetorial - Professor Pinguim | F\u00edsica<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/professorpinguim.com.br\/blog\/para-treinar-exercicios-de-cinematica-vetorial\/\" \/>\n<meta property=\"og:locale\" content=\"pt_BR\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Para treinar - Exerc\u00edcios de Cinem\u00e1tica Vetorial - Professor Pinguim | F\u00edsica\" \/>\n<meta property=\"og:description\" content=\"Nesta aula, vamos resolver alguns exerc\u00edcios sobre cinem\u00e1tica vetorial, incluindo deslocamento vetorial e velocidade vetorial m\u00e9dia. 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