Problem
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. \[ a(1)= \] \[ \mathrm{ft} / \mathrm{s}^{2} \]
Solution
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}
