{"id":1021,"date":"2015-06-29T02:24:27","date_gmt":"2015-06-29T02:24:27","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=1021"},"modified":"2024-08-22T17:41:01","modified_gmt":"2024-08-22T17:41:01","slug":"resolucoes-movimento-acelerado-e-movimento-retardado-aceleracao-media","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/mecanica\/cinematica\/aceleracao-escalar-media\/resolucoes-movimento-acelerado-e-movimento-retardado-aceleracao-media\/","title":{"rendered":"Movimento acelerado e movimento retardado \u2013 acelera\u00e7\u00e3o m\u00e9dia"},"content":{"rendered":"

Resolu\u00e7\u00f5es – <\/b><\/span><\/span><\/span><\/span>Movimento acelerado e movimento retardado \u2013 acelera\u00e7\u00e3o m\u00e9dia<\/b><\/span><\/span><\/span><\/p>\n

 <\/p>\n

01-<\/b><\/span><\/span><\/span>\u00a0a<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/span><\/sub>=(V – V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub>)\/(t \u2013 t<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub>)\u00a0 —\u00a0 a<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/span><\/sub>=(72\/3,6 \u2013 0)\/(2 \u2013 0)\u00a0 —\u00a0<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/span><\/sub>=10m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup><\/p>\n

02-<\/b><\/span><\/span>\u00a0I- a<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=(2v \u2013 v)\/t\u00a0 —\u00a0 a<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=v\/t\u00a0 —\u00a0 II \u2013 a<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=(3v \u2013 v)\/2t\u00a0 —\u00a0 a<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=v\/t\u00a0 —\u00a0 III \u2013 a<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>=(5v \u2013 v)\/5t\u00a0 —\u00a0 a<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>=4v\/5t\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- D<\/b><\/span><\/span><\/p>\n

03-<\/b><\/span><\/span><\/span>\u00a0a<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/span><\/sub>=(0 \u2013 20)\/(4 \u2013 0)\u00a0 —\u00a0 a<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/span><\/sub>=-5m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- C<\/b><\/span><\/span><\/span><\/p>\n

04<\/b><\/span><\/span><\/span>-a<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/span><\/sub>=(V \u2013 V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub>)\/(t \u2013 t<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub>)\u00a0 —\u00a0 2=(108\/3,6 \u2013 0)\/(t )\u00a0 —\u00a0 2=30\/(t )\u00a0 —<\/b><\/span><\/span><\/span>\u00a0t=15s<\/b><\/span><\/span><\/span><\/p>\n

05-<\/b><\/span><\/span>\u00a0Orientando a trajet\u00f3ria, antes de rebater velocidade positiva e depois de rebater, velocidade negativa:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

A<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/span><\/sub>=(-30 \u2013 20)\/0,1\u00a0 —\u00a0 a<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/span><\/sub>=-50\/0,1\u00a0 —\u00a0 a<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/span><\/sub>=-500m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sup>\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- D<\/b><\/span><\/span><\/span><\/p>\n

06-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- B<\/b><\/span><\/span><\/span><\/p>\n

07-<\/b><\/span><\/span>\u00a0Menor tempo \u2013 maior acelera\u00e7\u00e3o \u2013 Corvette\u00a0 —\u00a0\u00a0 a<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=100\/4=25(km\/h)\/s\u00a0 —\u00a0 maior tempo \u2013 menor acelera\u00e7\u00e3o \u2013 Parati\u00a0 —-\u00a0 a<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=100\/33,35=3(km\/h)\/s\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- C<\/b><\/span><\/span><\/p>\n

08-<\/b><\/span><\/span>\u00a0Entre 3s e 9s\u00a0 —\u00a0 a<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=(0 \u2013 12)\/(9 -3)\u00a0 —\u00a0 a<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=-12\/6\u00a0 —<\/b><\/span><\/span>\u00a0a<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=-2m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 entre 0 e 12s\u00a0 —\u00a0 a<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=(-12 – 6)\/(12 \u2013 0) \u00a0—\u00a0 a<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=-18\/12\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>a<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=- 1,5m\/s<\/b><\/span><\/span><\/p>\n

09-\u00a0<\/b><\/span><\/span>a=(2 \u2013 0)\/(4 – 0)\u00a0 —\u00a0 a=1\/2m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- B<\/b><\/span><\/span><\/p>\n

10-<\/b><\/span><\/span>\u00a0Observe que at\u00e9 3s a acelera\u00e7\u00e3o \u00e9 constante e vale 3m\/s<\/b><\/span><\/span>2\u00a0<\/b><\/span><\/span><\/sup>(terreno mais inclinado) e que depois ela cai para 2m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0(terreno menos inclinado)\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- C<\/b><\/span><\/span><\/p>\n

11-<\/b><\/span><\/span>\u00a0Trecho AB\u00a0 —\u00a0 progressivo (V>0) e retardado (o m\u00f3dulo de V est\u00e1 diminuindo, ele vai parar em B \u201cV<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=0\u201d)<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Trecho BC\u00a0 —\u00a0 retr\u00f3grado (V<0) e acelerado (m\u00f3dulo de V est\u00e1 aumentando \u201cem B estava parado\u201d)<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

 <\/p>\n

Trecho CD\u00a0 —\u00a0 Movimento uniforme com velocidade constante e negativa, portanto retr\u00f3grado mas sem acelera\u00e7\u00e3o.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Trecho DE\u00a0 —\u00a0 retr\u00f3grado (V<0) e retardado (m\u00f3dulo de V est\u00e1 diminuindo \u201cp\u00e1ra em E\u201d)<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Trecho EF\u00a0 —\u00a0 progressivo (V>0) e acelerado (m\u00f3dulo de V est\u00e1 aumentando \u201csaiu do repouso em E\u201d)<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

12-<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>R- C\u00a0<\/b><\/span><\/span>\u00a0—\u00a0 entre t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0e t<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>\u00a00 movimento \u00e9 retr\u00f3grado (velocidade negativa) e acelerado (m\u00f3dulo da velocidade est\u00e1 aumentando \u201cestava parado em t<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u201d)<\/b><\/span><\/span><\/p>\n

13-<\/b><\/span><\/span>\u00a0a) entre\u00a0<\/b><\/span><\/span>6s e 16s<\/b><\/span><\/span>, a velocidade \u00e9 constante e a acelera\u00e7\u00e3o \u00e9 nula.<\/b><\/span><\/span><\/p>\n

b) entre<\/b><\/span><\/span>\u00a00 e 6s<\/b><\/span><\/span>\u00a0(movimento progressivo acelerado)<\/b><\/span><\/span><\/p>\n

c) o deslocamento (dist\u00e2ncia percorrida) em todo gr\u00e1fico V X t \u00e9 fornecido pela \u00e1rea entre 0 e 20s\u00a0 —\u00a0 \u00e1rea do tri\u00e2ngulo=b.h\/2=6.12\/2=36m + \u00e1rea do ret\u00e2ngulo=b.h=10.12=120m + \u00e1rea do trap\u00e9zio=(B + b).h\/2=(12 + 10).4\/2=44m\u00a0 —\u00a0 deslocamento = 36 + 120 + 44\u00a0 —<\/b><\/span><\/span>\u00a0d=200m<\/b><\/span><\/span><\/p>\n

d) V<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=d\/t\u00a0 —\u00a0 V<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=200\/20\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>V<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=10m\/s<\/b><\/span><\/span><\/p>\n

14-<\/b><\/span><\/span>\u00a0a)\u00a0 V=90\/3,6=25m\/s<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

b) A dist\u00e2ncia \u00e9 fornecida pela \u00e1rea do trap\u00e9zio\u00a0 —\u00a0 \u0394S=(B + b).h\/2=(50 + 30).25\/2\u00a0 —\u00a0<\/b><\/span><\/span>\u0394S=1.000m<\/b><\/span><\/span><\/p>\n

15-<\/b><\/span><\/span>\u00a0Se o movimento for progressivo o sinal ser\u00e1 positivo, mas se o movimento for retr\u00f3grado o sinal ser\u00e1 negativo.<\/b><\/span><\/span><\/p>\n

16-<\/b><\/span><\/span>\u00a0(01)\u00a0<\/b><\/span><\/span>Correta\u00a0<\/b><\/span><\/span>\u00a0—\u00a0 V \u00e9 constante e vale 12m\/s X 3.6=43,2km\/h e o movimento \u00e9 progressivo (V>0 e a=0)<\/b><\/span><\/span><\/p>\n

(02)\u00a0<\/b><\/span><\/span>Falsa\u00a0<\/b><\/span><\/span>\u00a0—\u00a0 o movimento \u00e9 uniformemente acelerado mas a acelera\u00e7\u00e3o vale a=(20 \u2013 12)\/(25 \u2013 10)\u00a0 —\u00a0 a=0,53m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/p>\n

(04) O deslocamento \u00e9 fornecido pela \u00e1rea do trap\u00e9zio\u00a0 —\u00a0 \u0394S=(20 + 12).15\/2\u00a0 —\u00a0 \u0394S=240m\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>Correta<\/b><\/span><\/span><\/p>\n

(08) \u0394S=10t=10.10=100m\u00a0 —\u00a0 entre 0 e 10s \u0394S \u00e9 fornecido pela \u00e1rea do ret\u00e2ngulo=10.12=120m\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0Falsa<\/b><\/span><\/span><\/p>\n

(16)\u00a0<\/b><\/span><\/span>Falsa<\/b><\/span><\/span>\u00a0 —\u00a0 nada se pode afirmar a respeito da trajet\u00f3ria<\/b><\/span><\/span><\/p>\n

Soma=(01 + 04) = 05<\/b><\/span><\/span><\/p>\n

17-<\/b><\/span><\/span>\u00a0Quanto mais inclinada reta representativa da velocidade maior ser\u00e1 a acelera\u00e7\u00e3o\u00a0 —\u00a0 R- A<\/b><\/span><\/span><\/p>\n

18-<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>R- C<\/b><\/span><\/span>\u00a0 —\u00a0 \u00e9 o \u00fanico movimento que \u00e9 retardado com a velocidade diminuindo at\u00e9 que ele pare.<\/b><\/span><\/span><\/p>\n

19-<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>R- D<\/b><\/span><\/span><\/p>\n

20-<\/b><\/span><\/span>\u00a0Movimento progressivo (indo), \u0394S=\u00e1rea do trap\u00e9zio=(2 + 1).2\/2=3m que \u00e9 a mesma que do movimento retr\u00f3grado (voltando)\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- C<\/b><\/span><\/span><\/p>\n

21-\u00a0<\/b><\/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 diminuendo\u00a0 —<\/b><\/span><\/span>\u00a0 R- C<\/b><\/span><\/span><\/p>\n

22-\u00a0<\/b><\/span><\/span>Entre 0 e 6s\u00a0 —\u00a0 V<\/b><\/span><\/span>I<\/b><\/span><\/span><\/sub>=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0+ a<\/b><\/span><\/span>I<\/b><\/span><\/span><\/sub>t<\/b><\/span><\/span>I<\/b><\/span><\/span><\/sub>=2 + 4.6\u00a0 —\u00a0 V<\/b><\/span><\/span>i<\/b><\/span><\/span><\/sub>=26 cm\/s\u00a0 —\u00a0 entre 6s e 10s\u00a0 —\u00a0 V<\/b><\/span><\/span>I<\/b><\/span><\/span><\/sub>=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=26cm\/s\u00a0 —\u00a0 V<\/b><\/span><\/span>II<\/b><\/span><\/span><\/sub>=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0+ a<\/b><\/span><\/span>II<\/b><\/span><\/span><\/sub>t<\/b><\/span><\/span>II<\/b><\/span><\/span><\/sub>=26 + (-3).4\u00a0 —\u00a0 V<\/b><\/span><\/span>II<\/b><\/span><\/span><\/sub>=14 cm\/s\u00a0 —\u00a0 V<\/b><\/span><\/span>II<\/b><\/span><\/span><\/sub>=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=14 cm\/s\u00a0 —\u00a0 V<\/b><\/span><\/span>III<\/b><\/span><\/span><\/sub>=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0+ a<\/b><\/span><\/span>III<\/b><\/span><\/span><\/sub>.t<\/b><\/span><\/span>III<\/b><\/span><\/span><\/sub>=14 + 4.6\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>V<\/b><\/span><\/span>III<\/b><\/span><\/span><\/sub>=38 cm\/s<\/b><\/span><\/span><\/p>\n

23-\u00a0<\/b><\/span><\/span>a) Supondo que nos intervalos de 0 e 4 s e de 12 s a 16 s a velocidade permane\u00e7a constante e que, nos intervalos de 4 s a 8 s e de 16 s a 20 s as varia\u00e7\u00f5es de velocidade sejam constantes, o gr\u00e1fico pedido est\u00e1 representado abaixo.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

b) Com base no gr\u00e1fico obtido no item( a) podemos descrever o movimento do carrinho da seguinte maneira:<\/b><\/span><\/span><\/p>\n

de t = 0 a t = 4 s o movimento \u00e9 progressivo e uniforme;<\/b><\/span><\/span><\/p>\n

de t = 4 s a t = 8 s o movimento \u00e9 progressivo e uniformemente retardado;<\/b><\/span><\/span><\/p>\n

de t = 8 s a t = 12 s o movimento \u00e9 retr\u00f3grado e uniformemente acelerado;<\/b><\/span><\/span><\/p>\n

de t = 12 s a t = 16 s o movimento \u00e9 retr\u00f3grado e uniforme,<\/b><\/span><\/span><\/p>\n

de t = 16 s a t = 20 s o movimento \u00e9 retr\u00f3grado e uniformemente retardado.\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

24-\u00a0<\/b><\/span><\/span>Sendo a trajet\u00f3ria \u00e9 retil\u00ednea, a acelera\u00e7\u00e3o restringe-se \u00e0 componente tangencial\u00a0\"\", que, em m\u00f3dulo, \u00e9 igual a acelera\u00e7\u00e3o escalar (a), dada pela taxa de varia\u00e7\u00e3o da velocidade (Dv) em rela\u00e7\u00e3o ao tempo (Dt)\u00a0 —\u00a0 a=\u0394V\/\u0394t\u00a0 —\u00a0 para cada intervalo voc\u00ea tem\u00a0 —\u00a0 I. a<\/b><\/span><\/span>I<\/b><\/span><\/span><\/sub>=(40 \u2013 0)\/(4 \u2013 0)\u00a0 —<\/b><\/span><\/span>\u00a0a<\/b><\/span><\/span>I<\/b><\/span><\/span><\/sub>=10m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 II.\u00a0<\/b><\/span><\/span>A<\/b><\/span><\/span>II<\/b><\/span><\/span><\/sub>\u00a0= 0<\/b><\/span><\/span>\u00a0(n\u00e3o houve varia\u00e7\u00e3o de velocidade)\u00a0 —\u00a0 III. a<\/b><\/span><\/span>III<\/b><\/span><\/span><\/sub>=(0 \u2013 40)\/(14 \u2013 6)\u00a0 —<\/b><\/span><\/span><\/p>\n

a<\/b><\/span><\/span>III<\/b><\/span><\/span><\/sub>= – 5m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- C<\/b><\/span><\/span><\/p>\n

25-\u00a0<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/span><\/sub>=<\/b><\/span><\/span><\/span>\u0394<\/b><\/span><\/span>V\/<\/b><\/span><\/span><\/span>\u0394<\/b><\/span><\/span>t\u00a0 —\u00a0 como\u00a0<\/b><\/span><\/span><\/span>\u0394<\/b><\/span><\/span>V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/span><\/sub>=2\u00a0<\/b><\/span><\/span><\/span>\u0394<\/b><\/span><\/span>V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sub>\u00a0 —\u00a0 a<\/b><\/span><\/span><\/span>m1<\/b><\/span><\/span><\/span><\/sub>=2a<\/b><\/span><\/span><\/span>m2<\/b><\/span><\/span><\/span><\/sub>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- D<\/b><\/span><\/span><\/span><\/p>\n

 <\/p>\n

26-<\/b><\/span><\/span>\u00a0C\u00e1lculo da acelera\u00e7\u00e3o de retardamento do carro\u00a0\u00a0\u00a0 —\u00a0 V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=54km\/h\/3,6=15m\/s\u00a0 —\u00a0 V=0 (p\u00e1ra)\u00a0 —\u00a0 t=5s\u00a0 —\u00a0 V= V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0+ a.t\u00a0 —\u00a0 0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

= 15 + a.5\u00a0 —\u00a0 a= – 3m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0(o sinal negativo de a significa que o carro est\u00e1 freando)\u00a0 —\u00a0 em m\u00f3dulo\u00a0 —\u00a0 a=3m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 depois que ele acionou os freios a intensidade da for\u00e7a resultante sobre o carro \u00e9 a pr\u00f3pria for\u00e7a de atrito\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 m.a = \u03bc.N\u00a0 —\u00a0 m.a = \u03bc.m.g\u00a0 —\u00a0 3= \u03bc.10\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

\u03bc = 0,3\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- A<\/b><\/span><\/span><\/p>\n

Voltar para os Exerc\u00edcios<\/span><\/a><\/h2>\n","protected":false},"excerpt":{"rendered":"

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