{"id":203,"date":"2020-07-13T13:28:11","date_gmt":"2020-07-13T17:28:11","guid":{"rendered":"http:\/\/fiziko.net\/?page_id=203"},"modified":"2020-08-14T23:17:51","modified_gmt":"2020-08-15T03:17:51","slug":"movimento-1d","status":"publish","type":"page","link":"https:\/\/fiziko.net\/?page_id=203","title":{"rendered":"Movimento 1D"},"content":{"rendered":"\n<p class=\"has-text-align-left\">A parte da <strong><em>mec\u00e2nica<\/em><\/strong> que estuda a descri\u00e7\u00e3o dos movimentos \u00e9 chamada de <strong><em>cinem\u00e1tica<\/em><\/strong> e aquela que estuda as causas dos movimentos, de <em><strong>din\u00e2mica<\/strong><\/em>.<\/p>\n\n\n\n<p class=\"has-text-align-left\"><strong><em>Part\u00edcula<\/em><\/strong> ou <em><strong>Ponto Material<\/strong><\/em> \u00e9 o tipo de corpo mais simples que pode-se imaginar. Possui dimens\u00f5es desprez\u00edveis, \u00e9 um corpo que em situa\u00e7\u00f5es espec\u00edficas pode ser considerado como um ponto no espa\u00e7o.<\/p>\n\n\n\n<p class=\"has-text-align-left\">Um <strong><em>Sistema F\u00edsico<\/em><\/strong> \u00e9 qualquer parte do universo bem definida.<\/p>\n\n\n\n<p class=\"has-text-align-left\">Um <strong><em>Sistema de Part\u00edculas<\/em><\/strong> \u00e9 um conjunto de part\u00edculas que forma um sistema f\u00edsico.<\/p>\n\n\n\n<p class=\"has-text-align-left\">um <strong><em>corpo r\u00edgido<\/em><\/strong> \u00e9 um conjunto de part\u00edculas cuja  dist\u00e2ncia entre qualquer par de part\u00edculas do conjunto permanece sempre a mesma.<\/p>\n\n\n\n<p class=\"has-text-align-left\">Um <strong><em>Sistema R\u00edgido<\/em><\/strong> \u00e9 um sistema qualquer de part\u00edculas cuja  dist\u00e2ncia entre qualquer par de suas part\u00edculas permanece sempre a mesma.<\/p>\n\n\n\n<p class=\"has-text-align-left\">A reta na qual foram espec\u00edficadas a origem, os semi-eixos positivo e negativo e a correspond\u00eancia entre pontos e n\u00fameros \u00e9 chamada de um <em><strong>Eixo Coordenado<\/strong><\/em>.<\/p>\n\n\n\n<p class=\"has-text-align-left\">Um <strong><em>Sistema de Coordenadas<\/em><\/strong> \u00e9 caracterizado por tr\u00eas eixos coordenados <em>OX<\/em>, <em>OY<\/em> e <em>OZ<\/em>, ortogonais entre si e com origem comum <em>O<\/em>.<\/p>\n\n\n\n<i class=\"large material-icons\">play_circle_filled<\/i>\n\n\n\n<figure class=\"wp-block-embed-youtube aligncenter 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<div class='embed-container'><iframe title=\"Aula 01 - Movimento 1D\" width=\"1920\" height=\"1080\" src=\"https:\/\/www.youtube.com\/embed\/MnnautJtE5I?feature=oembed\" frameborder=\"0\" allow=\"accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen><\/iframe><\/div>\n<\/div><\/figure>\n\n\n\n<p class=\"has-text-align-left\">Dizemos que a trinca de coordenadas d\u00e1 a <strong><em>Posi\u00e7\u00e3o<\/em><\/strong> da part\u00edcula em rela\u00e7\u00e3o a <em>OXYZ<\/em>.<\/p>\n\n\n\n<p class=\"has-text-align-left\">Um <em><strong>Referencial<\/strong><\/em> \u00e9 uma estrutura para medir posi\u00e7\u00f5es e instantes de tempo e \u00e9 formado por um sistema de coordenadas, junto com as r\u00e9guas e os rel\u00f3gios.<\/p>\n\n\n\n<p class=\"has-text-align-left\">Um <strong><em>Observador<\/em><\/strong> \u00e9 um agente fixo em um referencial e \u00e9 capaz de realizar medi\u00e7\u00f5es. O observador pode ser uma pessoa ou um aparelho programado para medir. \u00c9 conveniente supor que os rel\u00f3gios est\u00e3o sincronizados em um dado referencial para que haja um \u00fanico instante do tempo atribuido a um dado evento.<\/p>\n\n\n\n<p class=\"has-text-align-left\">Dizemos que uma part\u00edcula est\u00e1 em <em><strong>Movimento <\/strong><\/em>em rela\u00e7\u00e3o a um referencial quando sua posi\u00e7\u00e3o em rela\u00e7\u00e3o ao referencial muda com o passar do tempo.<\/p>\n\n\n\n<h2 class=\"has-text-align-left wp-block-heading\">Part\u00edcula em Movimento 1D<\/h2>\n\n\n\n<p class=\"has-text-align-left\">Podemos considerar qualquer eixo coordenado para representar um movimento em uma dimens\u00e3o (Movimento 1D). Para facilitar nosso estudo, iremos representar a posi\u00e7\u00e3o da uma part\u00edcula que se move na dire\u00e7\u00e3o <em>x<\/em> pela nota\u00e7\u00e3o <\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\">\\begin{cases} { x }_{ 1 },\\quad em \\quad { t }_{ 1 } \\\\ { x }_{ 2 },\\quad em \\quad { t }_{ 2 } \\end{cases}<\/div>\n\n\n\n<p class=\"has-text-align-left\"><strong><em>D<\/em><\/strong><em style=\"font-weight: bold;\">eslocamento<\/em><strong> <\/strong>\u00e9 a diferen\u00e7a entre a posi\u00e7\u00e3o final da part\u00edcula e a posi\u00e7\u00e3o inicial da part\u00edcula.<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\">\\Delta x={ x }_{ 2 }-{ x }_{ 1 } <\/div>\n\n\n\n<p class=\"has-text-align-left\">Como sabemos, na F\u00edsica existem dois tipos de grandezas, as <strong><em>Grandezas Escalares<\/em><\/strong> que s\u00e3o representadas apenas por um n\u00famero, ou valor num\u00e9rico e as<strong><em> Grandezas Vetoriais<\/em><\/strong> que s\u00e3o representadas por um m\u00f3dulo (valor num\u00e9rico), uma dire\u00e7\u00e3o e um sentido.<\/p>\n\n\n\n<p class=\"has-text-align-left\">O deslocamento \u00e9 uma grandeza vetorial, \u0394<em>x<\/em> indica o m\u00f3dulo e a dire\u00e7\u00e3o, enquanto que o sinal ir\u00e1 indicar o sentido. O mesmo vale para \u0394<em>y<\/em> e \u0394<em>z<\/em>.<\/p>\n\n\n\n<p class=\"has-text-align-left\">  A <strong><em>Velocidade M\u00e9dia<\/em><\/strong> \u00e9 a raz\u00e3o entre o deslocamento e o intervalo de tempo decorrido.<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\">{ v }_{ m }=\\frac { \\Delta x }{ \\Delta t } =\\frac { { x }_{ 2 }-{ x }_{ 1 } }{ { t }_{ 2 }-{ t }_{ 1 } } <\/div>\n\n\n\n<p class=\"has-text-align-left\">No Sistema Internacional de Unidades e Medidas (SI) o espa\u00e7o \u00e9 medido em metros (m) e o tempo \u00e9 medido em segundos (s) , logo a velocidade ser\u00e1 dada em metros por segundo (m\/s).<\/p>\n\n\n\n<p class=\"has-text-align-left\">A <strong><em>Velocidade Instant\u00e2nea<\/em><\/strong> \u00e9 definida como o limite da raz\u00e3o \u0394<em>x<\/em>\/\u0394<em>t<\/em> da part\u00edcula quando o tempo tende a zero, ou ainda, \u00e9 a derivada da posi\u00e7\u00e3o da part\u00edcula em rela\u00e7\u00e3o ao tempo. Determina a mudan\u00e7a do espa\u00e7o conforme passa o tempo.<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\">{ v }=\\lim _{ \\Delta t\\rightarrow 0 }{ \\frac { \\Delta x }{ \\Delta t }  } =\\frac { dx }{ dt } <\/div>\n\n\n\n<p class=\"has-text-align-left\">ou<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\">{ v }=\\lim _{ \\Delta t\\rightarrow 0 }{ \\frac { x(t+\\Delta t)-x(t) }{ \\Delta t }  } =\\frac { dx }{ dt } <\/div>\n\n\n\n<p class=\"has-text-align-left\">A <strong><em>Acelera\u00e7\u00e3o m\u00e9dia<\/em><\/strong> \u00e9 a raz\u00e3o entre a varia\u00e7\u00e3o da velocidade e o tempo decorrido.<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\">{ a }_{ m }=\\frac { \\Delta v }{ \\Delta t } =\\frac { { v }_{ 2 }-{ v }_{ 1 } }{ { t }_{ 2 }-{ t }_{ 1 } } <\/div>\n\n\n\n<p class=\"has-text-align-left\">No Sistema Internacional de Unidades e Medidas (SI) o espa\u00e7o \u00e9 medido em metros (m) e o tempo \u00e9 medido em segundos (s) , a velocidade dada em metros por segundo (m\/s) e a acelera\u00e7\u00e3o ser\u00e1 dada em metros por segundo ao quadrado (m\/s\u00b2).<\/p>\n\n\n\n<p class=\"has-text-align-left\">A <strong><em>Acelera\u00e7\u00e3o Instant\u00e2nea<\/em><\/strong> \u00e9 definida como o limite da raz\u00e3o \u0394<em>v<\/em>\/\u0394<em>t<\/em> da part\u00edcula quando o tempo tende a zero, ou ainda, \u00e9 a derivada da velocidade da part\u00edcula em rela\u00e7\u00e3o ao tempo. Determina a mudan\u00e7a da velocidade conforme passa o tempo.<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\">{ a }=\\lim _{ \\Delta t\\rightarrow 0 }{ \\frac { \\Delta v }{ \\Delta t }  } =\\frac { dv }{ dt } <\/div>\n\n\n\n<p class=\"has-text-align-left\">ou<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\">{ a }=\\lim _{ \\Delta t\\rightarrow 0 }{ \\frac { v(t+\\Delta t)-v(t) }{ \\Delta t }  } =\\frac { dv }{ dt } <\/div>\n\n\n\n<i class=\"large material-icons\">play_circle_filled<\/i>\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<div class='embed-container'><iframe title=\"Aula 02 - Movimento 1D\" width=\"1920\" height=\"1080\" src=\"https:\/\/www.youtube.com\/embed\/0J7bB2fCCoE?feature=oembed\" frameborder=\"0\" allow=\"accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen><\/iframe><\/div>\n<\/div><\/figure>\n\n\n\n<h2 class=\"has-text-align-left wp-block-heading\">Movimento Retil\u00edneo Uniforme &#8211; MRU<\/h2>\n\n\n\n<p class=\"has-text-align-left\">Temos um MRU quando a velocidade da part\u00edcula \u00e9 constante e sua acelera\u00e7\u00e3o \u00e9 nula. Como o nome informa, a part\u00edcula n\u00e3o executa movimentos curvil\u00edneos. \u00c9 o movimento mais somples de todos.<\/p>\n\n\n\n<p class=\"has-text-align-left\">A <strong><em>Fun\u00e7\u00e3o de Movimento<\/em><\/strong> da part\u00edcula em MRU \u00e9 a regra que descreve a posi\u00e7\u00e3o da part\u00edcula em cada instante de tempo.<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\">x(t)=x_{i}+vt<\/div>\n\n\n\n<p class=\"has-text-align-left\">Onde  <a href=\"https:\/\/www.codecogs.com\/eqnedit.php?latex=x_{i}\" target=\"_blank\" rel=\"noopener noreferrer\"><img decoding=\"async\" title=\"x_{i}\" src=\"https:\/\/latex.codecogs.com\/gif.latex?x_{i}\"><\/a>  \u00e9 a posi\u00e7\u00e3o inicial, <a href=\"https:\/\/www.codecogs.com\/eqnedit.php?latex=v\" target=\"_blank\" rel=\"noopener noreferrer\"><img decoding=\"async\" title=\"v\" src=\"https:\/\/latex.codecogs.com\/gif.latex?v\"><\/a> \u00e9 a velocidade da part\u00edcula e  <a href=\"https:\/\/www.codecogs.com\/eqnedit.php?latex=t\" target=\"_blank\" rel=\"noopener noreferrer\"><img decoding=\"async\" title=\"t\" src=\"https:\/\/latex.codecogs.com\/gif.latex?t\"><\/a> \u00e9 o instante de tempo.<\/p>\n\n\n\n<p class=\"has-text-align-left\"><strong><em>APLICA\u00c7\u00c3O<\/em><\/strong><\/p>\n\n\n\n<p class=\"has-text-align-left\"><strong>Exemplo 1.1<\/strong> &#8211; Vamos considerar uma part\u00edcula cujo movimento \u00e9 descrito pela fun\u00e7\u00e3o de movimento dada por<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\">x(t)=2+5t<\/div>\n\n\n\n<p class=\"has-text-align-left\">onde a posi\u00e7\u00e3o \u00e9 dada em metros (m) e o tempo em segundos (s). Quais s\u00e3o (a) a posi\u00e7\u00e3o da part\u00edcula, (b) a velocidade e (c) a acelera\u00e7\u00e3o em t = 3s?<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\">x(t=3s)=2+5(3)\\\\\nx(t=3s)=2+15\\\\\nx(t=3s)=17m<\/div>\n\n\n\n<p class=\"has-text-align-left\">Como vemos, a part\u00edcula est\u00e1 na posi\u00e7\u00e3o <em>x<\/em> = 17m quando<em> t<\/em> = 3s;<\/p>\n\n\n\n<p class=\"has-text-align-left\">Para calcularmos a velocidade neste instante, tomemos a fun\u00e7\u00e3o <\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\">v= \\lim_{\\Delta t \\rightarrow 0}{\\frac {x(t+\\Delta t)-x(t)}{\\Delta t }}\\\\  \n<\/div>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\">v(t)= \\lim_{\\Delta t \\rightarrow 0}{\\frac{[2+5(t+\\Delta t)]-[2+5t]}{\\Delta t}}\\\\ \n<\/div>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\">v(t)=\\lim_{\\Delta t \\rightarrow 0}{\\frac{5\\Delta t}{\\Delta t}}\\\\\n<\/div>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\">v(t)=\\lim_{\\Delta t \\rightarrow 0}{5}\\\\ <\/div>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\">v(t)=5 m\/s<\/div>\n\n\n\n<p class=\"has-text-align-left\">Como vemos, a velocidade da part\u00edcula \u00e9 constante e igual a <em>v = <\/em>5 m\/s, desta forma a acelera\u00e7\u00e3o \u00e9 nula.<\/p>\n\n\n\n<p class=\"has-text-align-left\"><strong>Resolu\u00e7\u00e3o de Exerc\u00edcio &#8211; 01<\/strong><\/p>\n\n\n\n<i class=\"large material-icons\">play_circle_filled<\/i>\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<div class='embed-container'><iframe title=\"Exerc\u00edcio 01 - F\u00edsica 1\" width=\"1920\" height=\"1080\" src=\"https:\/\/www.youtube.com\/embed\/JcidE4ICDDM?feature=oembed\" frameborder=\"0\" allow=\"accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen><\/iframe><\/div>\n<\/div><\/figure>\n\n\n\n<h2 class=\"has-text-align-left wp-block-heading\">Movimento Retil\u00edneo Uniformemente Variado &#8211; MRUV<\/h2>\n\n\n\n<p class=\"has-text-align-left\">Considere agora o movimento retil\u00edneo no qual a velocidade varia uniformemente com o tempo, em outras palavras, a acelera\u00e7\u00e3o da part\u00edcula \u00e9 constante.<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\">a=\\frac{v_{f}-v_{i}}{t_{f}-t_{i}}\\\\\nv_f-v_i=at\\\\\nv_f=v_i+at<\/div>\n\n\n\n<p class=\"has-text-align-left\">A equa\u00e7\u00e3o acima \u00e9 obtida tomando-se o intante de tempo inicial igual a zero e o instante de tempo final igual a im instante qualquer. Esta equa\u00e7\u00e3o \u00e9 conhecida como a <strong><em>fun\u00e7\u00e3o hor\u00e1ria da velocidade<\/em><\/strong> para o MRUV.<\/p>\n\n\n\n<p class=\"has-text-align-left\">A <strong><em>fun\u00e7\u00e3o de movimento<\/em><\/strong> da part\u00edcula em MRUV \u00e9 dada por<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\">x_f=x_i+v_it+ \\frac{1}{2}at^2<\/div>\n\n\n\n<p class=\"has-text-align-left\">A equa\u00e7\u00e3o de <strong><em>Torricelli<\/em><\/strong> \u00e9 obtida combinando-se as fun\u00e7\u00f5es hor\u00e1rias da posi\u00e7\u00e3o e da velocidade e n\u00e3o \u00e9 dependente do tempo.<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\">v^2_{f}=v^2_{i}+2a\\Delta x<\/div>\n\n\n\n<p class=\"has-text-align-left\"><em><strong>APLICA\u00c7\u00c3O<\/strong><\/em><\/p>\n\n\n\n<p class=\"has-text-align-left\"><strong>Exemplo 1.2<\/strong> &#8211; Vamos considerar uma part\u00edcula cujo movimento \u00e9 descrito pela fun\u00e7\u00e3o de movimento dada por<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\">x(t)=2+5t^2<\/div>\n\n\n\n<p class=\"has-text-align-left\">onde a posi\u00e7\u00e3o \u00e9 dada em metros (m) e o tempo em segundos (s). Quais s\u00e3o (a) a posi\u00e7\u00e3o da part\u00edcula, (b) a velocidade e (c) a acelera\u00e7\u00e3o em t = 3s?<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\">x(t=3s)=2+5(3)^2 \\\\\nx(t=3s)=2+5 \\times 9 \\\\\nx(t=3s)=2+45 \\\\\nx(t=3s)=47 m<\/div>\n\n\n\n<p class=\"has-text-align-left\">Como vemos, a part\u00edcula est\u00e1 na posi\u00e7\u00e3o <em>x<\/em> = 47m quando<em> t<\/em> = 3s;<\/p>\n\n\n\n<p class=\"has-text-align-left\">Para calcularmos a velocidade neste instante, tomemos a fun\u00e7\u00e3o <\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\">v=\\lim_{\\Delta t \\rightarrow 0}{ \\frac {x(t+\\Delta t)-x(t)}{\\Delta t }}<\/div>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\">v(t=3s)= \\lim_{\\Delta t \\rightarrow 0}{\\frac{[2+5(3+\\Delta t)^2]-[2+5(3)^2]}{\\Delta t}}<\/div>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\">v(t=3s)=\\lim_{\\Delta t \\rightarrow 0}{\\frac{[2+5(9+6\\Delta t+\\Delta t^2)]-[2+45]}{\\Delta t}}\\\\\n<\/div>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\">v(t=3s)=\\lim_{\\Delta t \\rightarrow 0}{\\frac{5(6\\Delta t+\\Delta t^2)}{\\Delta t}}\\\\\n<\/div>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\">v(t=3s)=\\lim_{\\Delta t \\rightarrow 0}{(30+5\\Delta t)}\\\\\n<\/div>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\">v(t=3s)=30 m\/s<\/div>\n\n\n\n<p>Temos que a velocidade da part\u00edcula no instante <em>t<\/em> = 3s \u00e9 de <em>v<\/em> = 30 m\/s.<\/p>\n\n\n\n<p>A velocidade para um instante qualquer \u00e9 dada por<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\">v= \\lim_{\\Delta t \\rightarrow 0}{\\frac {x(t+\\Delta t)-x(t)}{\\Delta t }}<\/div>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\">v(t)= \\lim_{\\Delta t \\rightarrow 0}{\\frac{[2+5(t+\\Delta t)^2]-[2+5t^2]}{\\Delta t}}<\/div>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\">v(t)=\\lim_{\\Delta t \\rightarrow 0}{\\frac{[2+5(t^2+2t\\Delta t+\\Delta t^2)]-[2+5t^2]}{\\Delta t}}<\/div>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\">v(t)=\\lim_{\\Delta t \\rightarrow 0}{\\frac{5(2t\\Delta t+\\Delta t^2)}{\\Delta t}}<\/div>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\">v(t)=\\lim_{\\Delta t \\rightarrow 0}{(10t+5\\Delta t)}<\/div>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\">v(t)=10t<\/div>\n\n\n\n<p class=\"has-text-align-left\">Podemos usar o mesmo procedimento para calcularmos a acelera\u00e7\u00e3o instant\u00e2nea.<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\">a=\\lim_{\\Delta t \\rightarrow 0}{ \\frac {v(t+\\Delta t)-v(t)}{\\Delta t }}<\/div>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\">a= \\lim_{\\Delta t \\rightarrow 0}{\\frac{[10(t+\\Delta t)]-[10t]}{\\Delta t}}<\/div>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\">a(t=3s)=\\lim_{\\Delta t \\rightarrow 0}{\\frac{[10t+10\\Delta t]-[10t]}{\\Delta t}}<\/div>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\">a(t=3s)=\\lim_{\\Delta t \\rightarrow 0}{10}<\/div>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\">a(t=3s)=10 m\/s^2<\/div>\n\n\n\n<p class=\"has-text-align-left\">Neste caso a acelera\u00e7\u00e3o \u00e9 constante e igual a 10 m\/s\u00b2.<\/p>\n\n\n\n<p class=\"has-text-align-left\"><strong>Resolu\u00e7\u00e3o de Exerc\u00edcio &#8211; 02<\/strong><\/p>\n\n\n\n<i class=\"large material-icons\">play_circle_filled<\/i>\n\n\n\n<figure class=\"wp-block-embed-youtube wp-block-embed is-type-video is-provider-youtube wp-embed-aspect-4-3 wp-has-aspect-ratio wp-embed-aspect-16-9\"><div class=\"wp-block-embed__wrapper\">\n<div class='embed-container'><iframe title=\"Exerc\u00edcio 02  - F\u00edsica 1\" width=\"1920\" height=\"1080\" src=\"https:\/\/www.youtube.com\/embed\/LO4kVTG5A5Q?feature=oembed\" frameborder=\"0\" allow=\"accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen><\/iframe><\/div>\n<\/div><\/figure>\n\n\n\n<hr class=\"wp-block-separator\"\/>\n\n\n\n<p class=\"has-text-align-left\"><strong>Resolu\u00e7\u00e3o de Exerc\u00edcio &#8211; 03<\/strong><\/p>\n\n\n\n<i class=\"large material-icons\">play_circle_filled<\/i>\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<div class='embed-container'><iframe title=\"Exerc\u00edcio 01 - F\u00edsica 1\" width=\"1920\" height=\"1080\" src=\"https:\/\/www.youtube.com\/embed\/N3TC3A8Wld8?feature=oembed\" frameborder=\"0\" allow=\"accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen><\/iframe><\/div>\n<\/div><\/figure>\n\n\n\n<hr class=\"wp-block-separator\"\/>\n\n\n\n<p class=\"has-text-align-left\"><strong>Resolu\u00e7\u00e3o de Exerc\u00edcio &#8211; 0<\/strong>4<\/p>\n\n\n\n<i class=\"large material-icons\">play_circle_filled<\/i>\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<div class='embed-container'><iframe title=\"Exerc\u00edcio 17   F\u00edsica 1\" width=\"1920\" height=\"1080\" src=\"https:\/\/www.youtube.com\/embed\/QrCB4S-deME?feature=oembed\" frameborder=\"0\" allow=\"accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen><\/iframe><\/div>\n<\/div><\/figure>\n\n\n\n<h2 class=\"has-text-align-left wp-block-heading\">Movimento 1D &#8211; Formula\u00e7\u00e3o Geral<\/h2>\n\n\n\n<p class=\"has-text-align-left\">\u00c9 importante escrevermos as denifi\u00e7\u00f5es das grandezas da cinem\u00e1tica na formula\u00e7\u00e3o que utiliza os conceitos de deriva\u00e7\u00e3o e integra\u00e7\u00e3o.<\/p>\n\n\n\n<p class=\"has-text-align-left\">Em um sistema em que o objeto, ou part\u00edcula, est\u00e1 se movendo em uma dimens\u00e3o, podemos definir a acelera\u00e7\u00e3o instant\u00e2nea pela forma descrita abaixo:<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\">a_{x}=\\frac{dv_{x}}{dt}<\/div>\n\n\n\n<p class=\"has-text-align-left\">Podemos reescrever a equa\u00e7\u00e3o acima na forma da integra\u00e7\u00e3o, que \u00e9 a opera\u00e7\u00e3o inversa \u00e0 deriva\u00e7\u00e3o,<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\">\\int{dv_{x}}=\\int{a_{x}dt}<\/div>\n\n\n\n<p class=\"has-text-align-left\">O mesmo vale para a defini\u00e7\u00e3o de velocidade, temos<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\">v_{x}=\\frac{dx}{dt}<\/div>\n\n\n\n<p class=\"has-text-align-left\">Logo, reescrevendo na forma integral<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\">\\int{dx}=\\int{v_{x}dt}<\/div>\n\n\n\n<p class=\"has-text-align-left\">Para resolvermos a derivada de um mon\u00f4mio, temos que, se <em>m<\/em> e <em>n<\/em> s\u00e3o n\u00fameros reais, temos que<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\">\\frac{d(mt^n)}{dt}=nmt^{n-1}<\/div>\n\n\n\n<p class=\"has-text-align-left\">e para resolvermos a integral de um mon\u00f4mio, basta realizar as opera\u00e7\u00e3es inversas, assim<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\">\\int{mt^n}dt=\\frac{mt^{n+1}}{n+1}<\/div>\n\n\n\n<p class=\"has-text-align-left\">A \u00e1rea alg\u00e9brica sob o gr\u00e1fico de uma fun\u00e7\u00e3o qualquer em um certo intervalo do dom\u00ednio e igual \u00e0 integral da fun\u00e7\u00e3o nesse intervalo.<\/p>\n\n\n\n<p class=\"has-text-align-left\"><em><strong>APLICA\u00c7\u00c3O<\/strong><\/em><\/p>\n\n\n\n<p class=\"has-text-align-left\"><strong>Exemplo 1.2<\/strong> &#8211; Considere duas part\u00edcula, A e B, se movimentando em linha reta pelo eixo <em>OX<\/em>, com fun\u00e7\u00f5es hor\u00e1rias dadas por<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\">x_{A}(t)=-12+8t-3t^2 \\\\\ne \\\\\na_{B}(t)=\\frac{t^3}{3}<\/div>\n\n\n\n<p class=\"has-text-align-left\"> Note que, sabemos a fun\u00e7\u00e3o hor\u00e1ria da posi\u00e7\u00e3o para a part\u00edcula A e a fun\u00e7\u00e3o hor\u00e1ria da acelera\u00e7\u00e3o para a part\u00edcula B. Queremos encontrar as fun\u00e7\u00f5es <em>x(t)<\/em>, <em>v(t)<\/em> e <em>a(t)<\/em> para ambas as part\u00edcula. Iniciaremos com a part\u00edcula A, logo<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\">v_{A}(t)=\\frac{dx_{A}}{dt}=\\frac{d}{dt}(-12+8t-3t^2)=0+8.1t^{1-1}-3.2t^{2-1}=8-6t<\/div>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\">v_{A}(t)=8-6t<\/div>\n\n\n\n<p class=\"has-text-align-left\">a derivada de uma constante \u00e9 zero. A velocidade varia linearmente com o tempo.<\/p>\n\n\n\n<p class=\"has-text-align-left\">Calculando-se a acelera\u00e7\u00e3o da part\u00edcula A, tem-se<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\">a_{A}(t)=\\frac{dv_{A}(t)}{dt}=\\frac{d}{dt}(8-6t)=0-6.1t^{1-1}=-6<\/div>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\">a_{A}(t)=-6<\/div>\n\n\n\n<p class=\"has-text-align-left\">a acelera\u00e7\u00e3o \u00e9 constante. Note que n\u00e3o estamos utilizando um sistema de unidades. Assim,<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\">\\begin{cases} { x }_{ A }(t)=-12+8t-3{ t }^{ 2 } \\\\ { v }_{ A }(t)=8-6t \\\\ { a }_{ A }(t)=-6 \\end{cases}<\/div>\n\n\n\n<p class=\"has-text-align-left\">Para a part\u00edcula B, temos<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\">\\int{dv_{B}}=\\int{a_{B}(t)dt}<\/div>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\">\\int_{0}^{v_{B}}{dv_{B}}=\\int_{0}^{t}{\\frac{t^3}{3}dt}<\/div>\n\n\n\n<p class=\"has-text-align-left\">Neste caso, a velocidade e o tempo iniciais s\u00e3o nulos.<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\">v_{B}(t)=\\frac{1}{3} \\left ( \\frac{t^{3+1}}{3+1} \\right)_{0}^{t}=\\frac{t^4}{12}<\/div>\n\n\n\n<p class=\"has-text-align-left\">Para determinarmos a fun\u00e7\u00e3o que descreve a posi\u00e7\u00e3o da part\u00edcula, temos<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\">\\int{dx_{B}}=\\int{v_{B}(t)dt}<\/div>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\">\\int_{0}^{x_{B}}{dx_{B}}=\\int_{0}^{t}{\\left ( \\frac{t^4}{12} \\right) dt}<\/div>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\">x_{B}(t)=\\frac{1}{12}\\left (\\frac{t^{4+1}}{4+1}\\right)_{0}^{t}=\\frac{t^5}{60}<\/div>\n\n\n\n<p class=\"has-text-align-left\">a posi\u00e7\u00e3o e o tempo iniciais s\u00e3o nulos. Portanto,<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\">\\begin{cases} { x }_{ B }(t)=\\frac{ t^5 }{ 60 } \\\\ { v }_{ B }(t)=\\frac{t^4}{12} \\\\ { a }_{ B }(t)=\\frac{t^3}{3} \\end{cases}<\/div>\n\n\n\n<p class=\"has-text-align-left\"><strong>Resolu\u00e7\u00e3o de Exerc\u00edcio &#8211; 04<\/strong><\/p>\n\n\n\n<i class=\"large material-icons\">play_circle_filled<\/i>\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<div class='embed-container'><iframe title=\"Exerc\u00edcio 02 - F\u00edsica 1\" width=\"1920\" height=\"1080\" src=\"https:\/\/www.youtube.com\/embed\/h8XOS63QuQ0?feature=oembed\" frameborder=\"0\" allow=\"accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen><\/iframe><\/div>\n<\/div><\/figure>\n\n\n\n<hr class=\"wp-block-separator\"\/>\n\n\n\n<p class=\"has-text-align-left\"><strong>Resolu\u00e7\u00e3o de Exerc\u00edcio &#8211; 05<\/strong><\/p>\n\n\n\n<i class=\"large material-icons\">play_circle_filled<\/i>\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<div class='embed-container'><iframe title=\"Exerc\u00edcio 03 - F\u00edsica 1\" width=\"1920\" height=\"1080\" src=\"https:\/\/www.youtube.com\/embed\/PeyFKqCHs7Q?feature=oembed\" frameborder=\"0\" allow=\"accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen><\/iframe><\/div>\n<\/div><\/figure>\n\n\n\n<h2 class=\"has-text-align-left wp-block-heading\">Movimento de Queda Livre<\/h2>\n\n\n\n<p class=\"has-text-align-left\">De uma forma gen\u00e9rica, dizemos que um corpo est\u00e1 executando um <em><strong>movimento de queda livre<\/strong><\/em> se o seu movimento for vertical, isto \u00e9, se tiver a dire\u00e7\u00e3o que passe pelo centro da Terra, e se n\u00e3o sofrer nenhuma a\u00e7\u00e3o que n\u00e3o seja a da for\u00e7a gravitacional da Terra.<\/p>\n\n\n\n<p class=\"has-text-align-left\">O Movimento de queda livre \u00e9 uma aplica\u00e7\u00e3o do Movimento Retil\u00edneo Uniformemente Variado, pois a acelera\u00e7\u00e3o da part\u00edcula \u00e9 aproximadamente constante e igual a<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\">g=9,8 m\/s^2<\/div>\n\n\n\n<figure class=\"wp-block-gallery columns-1 is-cropped wp-block-gallery-1 is-layout-flex wp-block-gallery-is-layout-flex\"><ul class=\"blocks-gallery-grid\"><li class=\"blocks-gallery-item\"><figure><img fetchpriority=\"high\" decoding=\"async\" width=\"1024\" height=\"576\" src=\"http:\/\/fiziko.net\/wp-content\/uploads\/2020\/07\/quedaLivre-1024x576.jpg\" alt=\"\" data-id=\"236\" data-full-url=\"http:\/\/fiziko.net\/wp-content\/uploads\/2020\/07\/quedaLivre.jpg\" data-link=\"http:\/\/fiziko.net\/?attachment_id=236\" class=\"wp-image-236\" srcset=\"https:\/\/fiziko.net\/wp-content\/uploads\/2020\/07\/quedaLivre-1024x576.jpg 1024w, https:\/\/fiziko.net\/wp-content\/uploads\/2020\/07\/quedaLivre-300x169.jpg 300w, https:\/\/fiziko.net\/wp-content\/uploads\/2020\/07\/quedaLivre-768x432.jpg 768w, https:\/\/fiziko.net\/wp-content\/uploads\/2020\/07\/quedaLivre.jpg 1280w\" sizes=\"(max-width: 1024px) 100vw, 1024px\" \/><\/figure><\/li><\/ul><figcaption class=\"blocks-gallery-caption\">Objeto em queda livre e an\u00e1lise da acelera\u00e7\u00e3o da gravidade terrestre.<\/figcaption><\/figure>\n\n\n\n<p class=\"has-text-align-left\">As <em><strong>equa\u00e7\u00f5es do movimento de queda livre<\/strong><\/em> s\u00e3o dadas pelas equa\u00e7\u00f5es do MRUV, tamb\u00e9m s\u00e3o v\u00e1lidas para o chamado <em><strong>movimento vertical<\/strong><\/em>, que s\u00e3o<\/p>\n\n\n\n<div class=\"wp-block-katex-display-block katex-eq\" data-katex-display=\"true\">\\begin{cases} g=9,8 m\/{ s }^{ 2 } \\\\ { y }_{ f }={ y }_{ i }+{ v }_{ iy }t-\\frac { 1 }{ 2 } g{ t }^{ 2 } \\\\ { v }_{ fy }={ v }_{ iy }-gt \\\\ { v }_{ fy }^{ 2 }={ v }_{ iy }^{ 2 }-2g\\Delta y \\end{cases}<\/div>\n\n\n\n<p class=\"has-text-align-left\"><strong>Resolu\u00e7\u00e3o de Exerc\u00edcio &#8211; 06<\/strong><\/p>\n\n\n\n<i class=\"large material-icons\">play_circle_filled<\/i>\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<div class='embed-container'><iframe title=\"Exerc\u00edcio 03  - F\u00edsica 1\" width=\"1920\" height=\"1080\" src=\"https:\/\/www.youtube.com\/embed\/lB9YWwiVIfY?feature=oembed\" frameborder=\"0\" allow=\"accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen><\/iframe><\/div>\n<\/div><\/figure>\n\n\n\n<hr class=\"wp-block-separator\"\/>\n\n\n\n<p class=\"has-text-align-left\"><strong>Resolu\u00e7\u00e3o de Exerc\u00edcio &#8211; 07<\/strong><\/p>\n\n\n\n<i class=\"large material-icons\">play_circle_filled<\/i>\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<div class='embed-container'><iframe title=\"Exerc\u00edcio 04  - F\u00edsica 1\" width=\"1920\" height=\"1080\" src=\"https:\/\/www.youtube.com\/embed\/oinZGcRxsHg?feature=oembed\" frameborder=\"0\" allow=\"accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen><\/iframe><\/div>\n<\/div><\/figure>\n\n\n\n<hr class=\"wp-block-separator\"\/>\n\n\n\n<p class=\"has-text-align-left\"><strong>Resolu\u00e7\u00e3o de Exerc\u00edcio &#8211; 08<\/strong><\/p>\n\n\n\n<i class=\"large material-icons\">play_circle_filled<\/i>\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<div class='embed-container'><iframe title=\"Exerc\u00edcio 05  - F\u00edsica 1\" width=\"1920\" height=\"1080\" src=\"https:\/\/www.youtube.com\/embed\/wTG9_OwnGFA?feature=oembed\" frameborder=\"0\" allow=\"accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen><\/iframe><\/div>\n<\/div><\/figure>\n\n\n\n<p class=\"has-text-align-left\">Obrigado e bons estudos!<\/p>\n\n\n\n<p><strong>BIBLIOGRAFIA<\/strong><\/p>\n\n\n\n<p>HIBBELER, R. C. <strong>Din\u00e2mica: Mec\u00e2nica para Engenharia<\/strong>. Edi\u00e7\u00e3o: 12 ed. S\u00e3o Paulo (SP): Pearson Universidades, 2010.<\/p>\n\n\n\n<p>NUSSENZVEIG, H. M. <strong>Curso de F\u00edsica B\u00e1sica: Mec\u00e2nica<\/strong>. Edi\u00e7\u00e3o: 5 ed. [s.l.] Blucher, 2013.<\/p>\n\n\n\n<p>RESNICK, R.; WALKER, J.; HALLIDAY, D. <strong>Fundamentos de F\u00edsica &#8211; Volume 1 &#8211; Mec\u00e2nica<\/strong>. Edi\u00e7\u00e3o: 10 ed. [s.l.] LTC, 2016.<\/p>\n\n\n\n<p>SERWAY, R.; JEWETT, J. <strong>Princ\u00edpios de f\u00edsica &#8211; vol. I: Volume 1<\/strong>. Edi\u00e7\u00e3o: 2 ed. [s.l.] Cengage Learning, 2014.<\/p>\n\n\n\n<p>YOUNG, H. D.; FREEDMAN, R. A. <strong>F\u00edsica de Sears &amp; Zemansky: Volume I: Mec\u00e2nica: Volume 1<\/strong>. Edi\u00e7\u00e3o: 14 ed. [s.l.] Pearson Universidades, 2015.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>A parte da mec\u00e2nica que estuda a descri\u00e7\u00e3o dos movimentos \u00e9 chamada de cinem\u00e1tica e aquela que estuda as causas dos movimentos, de din\u00e2mica. Part\u00edcula ou Ponto Material \u00e9 o tipo de corpo mais simples que pode-se imaginar. Possui dimens\u00f5es desprez\u00edveis, \u00e9 um corpo que em situa\u00e7\u00f5es espec\u00edficas pode ser considerado como um ponto no&hellip; <br \/> <a class=\"read-more\" href=\"https:\/\/fiziko.net\/?page_id=203\">Leia mais<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"class_list":["post-203","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/fiziko.net\/index.php?rest_route=\/wp\/v2\/pages\/203","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/fiziko.net\/index.php?rest_route=\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/fiziko.net\/index.php?rest_route=\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/fiziko.net\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/fiziko.net\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=203"}],"version-history":[{"count":35,"href":"https:\/\/fiziko.net\/index.php?rest_route=\/wp\/v2\/pages\/203\/revisions"}],"predecessor-version":[{"id":390,"href":"https:\/\/fiziko.net\/index.php?rest_route=\/wp\/v2\/pages\/203\/revisions\/390"}],"wp:attachment":[{"href":"https:\/\/fiziko.net\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=203"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}