At least one of the answers above is NOT correct.
The position of a particle moving on a horizontal line (where $s$ is in feet to the right of a fixed reference point and $t$ is in seconds after the start of the observation) is
\[
s(t)=t^{4}-10 t+17, \quad t \geq 0 .
\]
(A) Find the velocity, in feet per second, at time $t: v(t)=4 t^{3}-10$
(B) Find the velocity (in $\mathrm{ft} / \mathrm{sec}$ ) of the particle at time $t=3$.
(C) Find all values of $t$ for which the particle is at rest. (If there are no such values, enter none. If there are more than one value, list them separated by commas.)
\[
t=
\]
(D) Use interval notation to indicate when the particle is moving in the positive direction. (If needed, enter inf for $\infty$. If the particle is never moving in the positive direction, enter none .)
(E) Find the total distance (in feet) traveled during the first 8 seconds.

Step 1 ：First, we find the velocity function by taking the derivative of the position function with respect to time: \(v(t) = 4t^3 - 10\).

Step 2 ：Next, we find the velocity of the particle at time \(t=3\). Substituting \(t=3\) into the velocity function, we get \(v(3) = 98\) feet per second.

Step 3 ：Then, we find the values of \(t\) for which the particle is at rest. This is when \(v(t) = 0\). Solving the equation \(4t^3 - 10 = 0\), we get \(t = \sqrt[3]{\frac{5}{2}}\).

Step 4 ：To indicate when the particle is moving in the positive direction, we need to find when \(v(t) > 0\). Solving the inequality \(4t^3 - 10 > 0\), we get \(t > \sqrt[3]{\frac{5}{2}}\). So the particle is moving in the positive direction for \(t \in (\sqrt[3]{\frac{5}{2}}, \infty)\).

Step 5 ：Finally, to find the total distance travelled during the first 8 seconds, we need to find the absolute difference between the position at \(t=0\) and \(t=8\). We get \(s(0) = 17\) and \(s(8) = 4081\). So the total distance travelled is \(\boxed{4081 - 17 = 4064}\) feet.