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lesya [120]
2 years ago
10

Distancia (en metros) que recorre un proyectil disparado desde un avión de combate, se puede describir según la función: () = 8,

6t+1,25t2 + 0,6t3, donde t son los segundos transcurridos. A) ¿Cuál es la rapidez instantánea del proyectil a los 3,7 segundos de ser disparado? B) ¿Cuál es su aceleración instantánea a los 4,1segundos de ser disparado?
English
1 answer:
Stella [2.4K]2 years ago
8 0

Answer:

a) La rapidez instantánea del proyectil a los 3,7 segundos de ser disparado es de 42,492 metros por segundo.

b) La aceleración instantánea del proyectil a los 4,1 segundos de ser disparado es de 17,26 metros por segundo al cuadrado.

Explanation:

El enunciado está incompleto puede prestarse a confusión, la forma correcta del mismo se presenta a continuación:

<em>La distancia en metros que recorre un proyectil disparado desde un avión de combate se puede describir según la función </em>s(t) = 8,6\cdot t + 1,25\cdot t^{2} + 0,6\cdot t^{3}<em>, donde </em>t<em> son los segundos transcurridos desde el lanzamiento del proyectil. </em><em>a)</em><em> ¿Cuál es la rapidez instantánea del proyectil a los 3,7 segundos de ser disparado? </em><em>b)</em><em> ¿Cuál es su aceleración instantánea a los 4,1 segundos de ser disparado? </em>

a) La rapidez instantánea del proyectil, en metros por segundo, es la primera derivada de la función distancia con respecto al tiempo, es decir:

v(t) = \frac{ds(t)}{dt} = 8,6 + 2,50\cdot t + 1,8\cdot t^{2} (1)

Si sabemos que t = 3,7\,s, entonces la rapidez instantánea del proyectil a los 3,7 segundos de ser disparado es:

v(3,7\,s) = 8,6 + 2,50\cdot (3,7\,s) + 1,8\cdot (3,7\,s)^{2}

v(3,7\,s) = 42,492\,\frac{m}{s}

La rapidez instantánea del proyectil a los 3,7 segundos de ser disparado es de 42,492 metros por segundo.

b) La aceleración instantánea del proyectil, en metros por segundo al cuadrado, es la segunda derivada de la función distancia con respecto al tiempo, es decir:

a(t) = \frac{dv(t)}{dt} = \frac{d^{2}s(t)}{dt^{2}} = 2,50 + 3,6\cdot t (2)

Si sabemos que t = 4,1\,s, entonces la aceleración instantánea del proyectil a los 4,1 segundos de ser disparado es:

a(4,1\,s) = 2,50 + 3,6\cdot (4,1\,s)

a(4,1\,s) = 17,26\,\frac{m}{s^{2}}

La aceleración instantánea del proyectil a los 4,1 segundos de ser disparado es de 17,26 metros por segundo al cuadrado.

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