A particle moves according to a law of motion $s=f(t), t \geq 0$, where $t$ is measured in seconds and $s$ in feet. (If an answer does not exist, enter DNE.)
\[
f(t)=t^{3}-9 t^{2}+24 t
\]
(a) Find the velocity (in $\mathrm{ft} / \mathrm{s}$ ) at time $t$.
\[
v(t)=
\]
$\mathrm{ft} / \mathrm{s}$
(b) What is the velocity (in $\mathrm{ft} / \mathrm{s}$ ) after 1 second?
\[
v(1)=
\]
$\mathrm{ft} / \mathrm{s}$
(c) When is the particle at rest? (Enter your answers as a comma-separated list.)
\[
t=
\]
(d) When is the particle moving in the positive direction? (Enter your answer using interval notation.)
(e) Draw a diagram to illustrate the motion of the particle and use it to find the total distance (in ft) traveled during the first 6 seconds.
$\mathrm{ft}$
(f) Find the acceleration (in $\mathrm{ft} / \mathrm{s}^{2}$ ) at time $t$.
\[
a(t)=
\]
\[
\mathrm{ft} / \mathrm{s}^{2}
\]
Find the acceleration (in $\mathrm{ft} / \mathrm{s}^{2}$ ) after 1 second.
\[
a(1)=
\]
\[
\mathrm{ft} / \mathrm{s}^{2}
\]
Answer
\(\boxed{v(t) = 3t^2 - 18t + 24\) ft/s is the velocity of the particle at time t}
Step 1 ：Given the position function of the particle is \(f(t)=t^{3}-9 t^{2}+24 t\)
Step 2 ：To find the velocity of the particle at time t, we need to find the derivative of the position function with respect to time
Step 3 ：The derivative of \(f(t)=t^{3}-9 t^{2}+24 t\) is \(v(t) = 3t^2 - 18t + 24\)
Step 4 ：\(\boxed{v(t) = 3t^2 - 18t + 24\) ft/s is the velocity of the particle at time t}
