a particle moves along a horizontal line. its position function is s(t) for t≥0. for each problem, find the…

Explanation:

Step1: Find the velocity function

The velocity function $v(t)$ is the derivative of the position function $s(t)$. Given $s(t)=t^{3}-13t^{2}$, using the power - rule $\frac{d}{dt}(t^{n}) = nt^{n - 1}$, we have $v(t)=s^\prime(t)=3t^{2}-26t=t(3t - 26)$.

Step2: Find the acceleration function

The acceleration function $a(t)$ is the derivative of the velocity function. So $a(t)=v^\prime(t)=6t - 26 = 2(3t - 13)$.

Step3: Find the critical points of $v(t)$ and $a(t)$

For $v(t)=t(3t - 26)=0$, the critical points are $t = 0$ and $t=\frac{26}{3}$. For $a(t)=2(3t - 13)=0$, the critical point is $t=\frac{13}{3}$.

Step4: Determine the signs of $v(t)$ and $a(t)$ on intervals

We consider the intervals $[0,\frac{13}{3})$, $[\frac{13}{3},\frac{26}{3})$, and $[\frac{26}{3},\infty)$.

• On the interval $[0,\frac{13}{3})$, $v(t)=t(3t - 26)\leq0$ (since $t\geq0$ and $3t - 26<0$) and $a(t)=2(3t - 13)<0$. Since $v(t)$ and $a(t)$ have the same sign, the particle is speeding up.
• On the interval $[\frac{13}{3},\frac{26}{3})$, $v(t)\leq0$ and $a(t)\geq0$. Since $v(t)$ and $a(t)$ have opposite signs, the particle is slowing down.
• On the interval $[\frac{26}{3},\infty)$, $v(t)\geq0$ and $a(t)\geq0$. Since $v(t)$ and $a(t)$ have the same sign, the particle is speeding up.

Answer:

The particle is slowing down on the interval $[\frac{13}{3},\frac{26}{3})$ and speeding up on the intervals $[0,\frac{13}{3})$ and $[\frac{26}{3},\infty)$.