A = (1 point) A particle is moving with acceleration a(t) = 6t + 8. its position at time t = O is s(0) = 6 and its velocity at time t = 0 is v(O) = 2. What is its position at time t = 7? =

Respuesta :

Answer:

[tex]559[/tex].

Explanation:

Integrate [tex]a(t)[/tex] with respect to time [tex]t[/tex] to find an expression for velocity:

[tex]\begin{aligned} v(t) &= \int a(t)\, d t \\ &= \int (6\, t + 8)\, d t && (\text{power rule}) \\ &= 3\, t^{2} + 8\, t + C_{v} \end{aligned}[/tex].

Note that since this integral is indefinite, the expression for [tex]v(t)[/tex] includes a constant [tex]C_{v}[/tex].

Find the value of [tex]C_{v}[/tex] using the fact that [tex]v(0) = 2[/tex]. Specifically, substitute [tex]t = 0[/tex] into the expression [tex]v(t) = 3\, t^{2} + 8\, t + C_{v}[/tex] and solve for [tex]C_{v}\![/tex]:

[tex]v(0) = 3\, (0)^{2} + 8\, (0) + C_{v} = C_{v}[/tex].

[tex]v(0) = 2[/tex].

[tex]C_{v} = 2[/tex].

In other words, [tex]v(t) = 3\, t^{2} + 8\, t + 2[/tex].

Similarly, integrate [tex]v(t)[/tex] with respect to [tex]t[/tex] to find an expression for position:

[tex]\begin{aligned} s(t) &= \int v(t)\, d t \\ &= \int (3\, t^{2} + 8\, t + 2)\, d t\\ &= t^{3} + 4\, t^{2} + 2\, t + C_{s} \end{aligned}[/tex].

Similarly, find the value of constant [tex]C_{s}[/tex] using the fact that [tex]s(0) = 6[/tex]:

[tex]s(0) = (0)^{3} + 4\, (0)^{2} + 2\, (0) + C_{s} = C_{s}[/tex].

[tex]s(0) = 6[/tex].

[tex]C_{s} = 6[/tex].

In other words, [tex]s(t) = t^{3} + 4\, t^{2} + 2\, t + 6[/tex]. Substitute in [tex]t = 7[/tex] and evaluate to find the position of the particle at that moment:

[tex]s(7) = 7^{3} + 4\, (7)^{2} + 2\, (7) + 6 = 559[/tex].