Una partícula se mueve a lo largo del eje xxx. La función x(t)x(t)x, left parenthesis, t, right parenthesis da la posición de la partícula en cualquier tiempo t\geq 0t≥0t, is greater than or equal to, 0

x(t)=t^4-9t^3+2t-1x(t)=t

4

−9t

3

+2t−1x, left parenthesis, t, right parenthesis, equals, t, start superscript, 4, end superscript, minus, 9, t, cubed, plus, 2, t, minus, 1

¿Cuál es la aceleración de la partícula, a(t)a(t)a, left parenthesis, t, right parenthesis, en t=4t=4t, equals, 4?

Answers

Answer 1

Answer:

La partícula tiene una aceleración de -24 cuando t = 4.

Explanation:

Sea [tex]x(t) = t^{4}-9\cdot t^{3}+2\cdot t-1[/tex], donde [tex]x[/tex] y [tex]t[/tex] son la posición y el tiempo, respectivamente.

Físicamente, entendemos que la aceleración es la segunda derivada de la posición y la primera derivada de la velocidad. Entonces tenemos que la funciones de velocidad y posición son, respectivamente.

Velocidad

[tex]v(t) = 4\cdot t^{3}-27\cdot t^{2}+2[/tex]

Aceleración

[tex]a(t) = 12\cdot t^{2}-54\cdot t[/tex]

Si conocemos que [tex]t = 4[/tex], entonces evaluamos la segunda derivada de la posición:

[tex]a(4) = 12\cdot (4)^{2}-54\cdot (4)[/tex]

[tex]a(4) = -24[/tex]

La partícula tiene una aceleración de -24 cuando t = 4.


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