a particle moves along the x - axis so that at time t ≥ 0 its position is given by x(t)=t³ - 12t² - 99t…

Explanation:

Step1: Find the velocity function

The velocity $v(t)$ is the derivative of the position function $x(t)$. Given $x(t)=t^{3}-12t^{2}-99t$, using the power - rule $\frac{d}{dt}(t^{n}) = nt^{n - 1}$, we have $v(t)=x^\prime(t)=3t^{2}-24t - 99 = 3(t^{2}-8t - 33)=3(t - 11)(t+3)$.

Step2: Find the acceleration function

The acceleration $a(t)$ is the derivative of the velocity function. So, $a(t)=v^\prime(t)=6t-24 = 6(t - 4)$.

Step3: Determine when speed is decreasing

Speed is decreasing when $v(t)$ and $a(t)$ have opposite signs. First, find the critical points of $v(t)$ and $a(t)$. For $v(t)=3(t - 11)(t + 3)=0$, the critical points are $t=-3$ and $t = 11$. Since $t\geq0$, we consider $t = 11$. For $a(t)=6(t - 4)=0$, the critical point is $t = 4$. We consider the intervals $[0,4)$ and $(4,11)$.

• On the interval $[0,4)$:
  • $v(0)=3(0 - 11)(0 + 3)=-99<0$
  • $a(0)=6(0 - 4)=-24<0$. Since $v(t)$ and $a(t)$ have the same sign, speed is not decreasing on $[0,4)$.
• On the interval $(4,11)$:
  • $v(5)=3(5 - 11)(5 + 3)=3\times(-6)\times8=-144<0$
  • $a(5)=6(5 - 4)=6>0$. Since $v(t)$ and $a(t)$ have opposite signs, speed is decreasing on $(4,11)$.

Answer:

$(4,11)$