6-Problemas de Velocidade, Tempo e Dist\303\242ncia Vamos estudar um m\303\251todo de montagem para solu\303\247\303\243o de problemas envolvendo a rela\303\247\303\243o entre tempo e dist\303\242ncia.

restart; with(plots):

Warning, the name changecoords has been redefined

RateTableEmpty := proc(name1, name2) local a, b, rt, M, equation; rt := cr ; #(a*ar + b*br)/(a+b); a := x; b := c-a; M := array( [ [``, `Padr\303\243o *`, ` Tempo`,` = Dist\303\242ncia`], [ name1, `?`, `?`, `?` ], [ name2, `?`, `?`, `?` ] ]); end proc: RateTableRT := proc(name1,r1,t1,name2,r2,t2) local a, b, rt, M, equation; M := array( [ [``, `Padr\303\243o *`, ` Tempo`,` = Dist\303\242ncia`], [ name1, r1, t1, r1*t1 ], [ name2, r2, t2, r2*t2 ] ]); end proc: RateTableRD := proc(name1,r1,d1,name2,r2,d2) local a, b, rt, M, equation; M := array( [ [``, `Padr\303\243o *`, ` Tempo`,` = Dist\303\242ncia`], [ name1, r1, r1/d1, d1 ], [ name2, r2, r2/d2, d2 ] ]); end proc: RateTableTD := proc(name1,t1,d1,name2,t2,d2) local a, b, rt, M, equation; M := array( [ [``, `Padr\303\243o *`, ` Tempo`,` = Dist\303\242ncia`], [ name1, d1/t1, t1, d1 ], [ name2, d2/t2, t2, d2 ] ]); end proc:

O procedimento comum para solucionar esses problemas \303\251 criar uma tabela para listar os fatores conhecidos e os desconhecidos. Geralmente, esses problemas envolvem corpos como carros, trens, avi\303\265es, atletas objetos em movimento. RateTableEmpty( trem1, trem2);

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RateTableEmpty( `primeiro carro`, `segundo carro` );

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RateTableEmpty( plano1, plano2);

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RateTableEmpty( `Jo\303\243o`, `Maria` );

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A seguir, vamos preencher as lacunas com informa\303\247\303\265es. Se soub\303\251ssemos as velocidades e os intervalos de tempo, ter\303\255amos algo assim: RateTableRT( carro1, 50, 3, carro2, 65, 2);

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RateTableRT( `Jo\303\243o`, 5, 3/2, `Maria`, 4.5, 2 );

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Problema 2.1 : Dois carros viajam do Rio de Janeiro para S\303\243o Paulo e chegam ao mesmo tempo. O primeiro carro vai a 83 Km/h e o segundo, que parte duas horas depois, vai a105 Km/h. A. Descubra quanto tempo o primeiro carro viaja. B. Encontre a dist\303\242ncia entre as duas cidades. Primeiro fa\303\247a uma tabela vazia (se for resolver o problema \303\240 m\303\243o). RateTableEmpty(carro1, carro2);

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Podemos tentar um palpite - o que - veremos que estar\303\241 errado. RateTableRT(carro1, 83, 3, carro2, 105, 2);

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Para resolver o problema, teremos t como o tempo desconhecido que o carro 1 viaja. Dessa forma, podemos fornecer os tempos e as velocidades de ambos os carros. As dist\303\242ncias ser\303\243o computadas como produto das duas. RateTableRT(carro1, 83, t, carro2, 105, t-2);

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

J\303\241 que as duas express\303\265es de dist\303\242ncia referem-se \303\240 mesma dist\303\242ncia, elas s\303\243o iguais. Isso significa que temos uma equa\303\247\303\243o que podemos resolver - para encontrar t. Ap\303\263s encontrarmos t, podemos achar d. 83*t = 105*t - 130; t = solve( %, t);

NiMvLCQqJiIjJCkiIiIlInRHIiIiIiIiLCYqJiIkMCIiIiIlInRHIiIiIiIiIiRJIiEiIg==NiMvJSJ0RyMiI2wiIzY=

d = 83*10;

NiMvJSJkRyIkSSk=

Problema 3.1 : Dois avi\303\265es viajam 3155 Km. O segundo avi\303\243o viaja 157 Km/h mais r\303\241pido e leva uma hora a menos para percorrer a dist\303\242ncia. Encontre a velocidade m\303\251dia e o intervalo de tempo do primeiro avi\303\243o. RateTableEmpty( plano1, plano2);

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

O que resulta em termos: RateTableRT(carro1, r, t, carro2, r+157, t-1);

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Ambas as express\303\265es para a dist\303\242ncia s\303\243o iguais a 3155. Agora, temos duas equa\303\247\303\265es com duas inc\303\263gnitas. EQ1 := r*t = 3155; EQ2 := (r+157)*(t-1) = 3155;

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Resolva as equa\303\247\303\265es para r. Ent\303\243o equacione as solu\303\247\303\265es, pois ambas s\303\243o iguais a r, e resolva. solve(EQ1,r); solve(EQ2, r); EQ3 := % = %%;

NiMsJComIiViSiIiIiUidEchIiIiIiI=NiMsJComLCYqJiIkZCIiIiIlInRHIiIiIiIiIiU3TCEiIiIiIiwmJSJ0RyIiIiIiIiEiIiEiIiEiIg==NiM+JSRFUTNHLywkKiYsJiomIiRkIiIiIiUidEciIiIiIiIiJTdMISIiIiIiLCYlInRHIiIiIiIiISIiISIiISIiLCQqJiIlYkoiIiIlInRHISIiIiIi

solve(%,t);

NiQsJiMiIiIiIiNGJSomIiQ5JCEiIikiKCopZisjRiRGJUYpLCZGJEYlRidGJQ==

t = 5;

NiMvJSJ0RyIiJg==

r = 1960/5;

NiMvJSJyRyIkI1I=

Problema 4.1: Um ciclista e um patinador viajam durante tr\303\252s horas. O ciclista viaja duas vezes mais r\303\241pido que o patinador e p\303\241ra tr\303\252s quil\303\264metros na sua frente. Qual foi a dist\303\242ncia percorrida pelo ciclista? A chave desse problema \303\251 colocar as express\303\265es certas nos lugares certos. A velocidade e a dist\303\242ncia percorrida pelo ciclista est\303\243o ambas expressadas em fun\303\247\303\243o das informa\303\247\303\265es do patinador. Ent\303\243o devemos considerar r = velocidade do patinador, e d = dist\303\242ncia do patinador. Agora, podemos expressar a velocidade e a dist\303\242ncia do ciclista relativas a estas informa\303\247\303\265es. RateTableRD( bicicleta, 2*r, d+2, rollerblader, r, d);

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

ambas as express\303\265es de tempo s\303\243o iguais a 3. 2*r/(d+2) = 3; r/d = 3;

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solve(%%, r); solve(%%,r);

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Desta maneira, podemos encontrar d, a dist\303\242ncia percorrida pelo patinador equacionando essas duas express\303\265es para r e solucionando-as. Ent\303\243o, para responder \303\240 quest\303\243o, precisamos adicionar 3 para, obter a dist\303\242ncia percorrida pelo ciclista. d = solve( % = %%, d);

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d + 3;

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Problema 5.1 : Felipe e Carlos correm \303\240 mesma velocidade. Felipe pode correr 50 quil\303\264metros se correr quatro horas a mais que Carlos quando percorreu 30 quil\303\264metros. Encontre a velocidade m\303\251dia. RateTableTD( Felipe, t, 18, Carlos, t+4, 30);

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Temos aqui outros exemplos. 18/t = 30/(t+4);

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t = solve(%, t);

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r = 18/6;

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