a particle moves along a horizontal line. its position function is $s(t)$ for $tgeq0$. for each problem…

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 $v(t)=s^\prime(t)=-3t^{2}+26t$.

Step2: Set the velocity function equal to zero

Set $v(t)=-3t^{2}+26t = 0$. Factor out $t$: $t(-3t + 26)=0$. So $t = 0$ or $-3t+26 = 0$. Solving $-3t+26 = 0$ gives $t=\frac{26}{3}$.

Step3: Test intervals

We have three intervals to test: $[0,\frac{26}{3})$ and $(\frac{26}{3},\infty)$. For the interval $[0,\frac{26}{3})$, let's choose a test - point, say $t = 1$. Then $v(1)=-3(1)^{2}+26(1)=-3 + 26=23>0$. When $v(t)>0$, the particle is moving to the right. For the interval $(\frac{26}{3},\infty)$, let's choose a test - point, say $t = 9$. Then $v(9)=-3(9)^{2}+26(9)=-243 + 234=-9<0$. When $v(t)<0$, the particle is moving to the left.

Answer:

The particle is moving to the right on the interval $[0,\frac{26}{3})$ and moving to the left on the interval $(\frac{26}{3},\infty)$.