The position of a particle that traverses a straight line is expressed as $$x(t)=t^3+2t^2-t+10$$, where x is in feet and t in seconds. What is the acceleration (ft/$$sec^2$$) of the particle when the velocity is zero?

The acceleration can be found by taking the second derivative of position function. In addition, velocity is found by taking the first derivative of the position function. Let's first find the first and second derivatives of the function in order to get expressions of both our velocity and acceleration:

$$\begin{gathered} x\left(t\right) = t^3+2t^2-t+10 \\ x\text{'}\left(t\right) = 3t^2+4t-1 \\ x\text{'}\text{'}\left(t\right)=6t+4 \end{gathered}$$

Therefore:

$$\begin{gathered} x\text{'}\left(t\right) = v\left(t\right) \\ v\left(t\right) = 3t^2+4t-1 \\ x\text{'}\text{'}\left(t\right) = a\left(t\right) \\ a\left(t\right) = 6t+4 \end{gathered}$$

Now that we have expressions for both velocity and acceleration, we are interested in finding at what value of t the velocity is equal to 0. Let's plug 0 into our expression for velocity and solve for t:

$$\begin{gathered} v\left(t\right) = 3t^2+4t-1 \\ 0 =3t^2+4t-1 \end{gathered}$$

Using the quadratic formula to solve for the roots:

$$\begin{gathered} x=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ t=\frac{-4\pm \sqrt{4^2-4\left(3\right)(-1)}}{2\left(3\right)} \\ t=\frac{-4\pm \sqrt{28}}{6} \\ t=\frac{-4+5.29}{6}, \frac{-4-5.29}{6} \\ t=\frac{-4+5.29}{6}, \frac{-4-5.29}{6} \\ t=.215s, -1.548s \end{gathered}$$

Since time can't be negative, our time value when the velocity is equal to 0 is equal to .215 seconds. Now, we simply plug this into our expression for acceleration to get our answer:

$$\begin{gathered} a\left(t\right) = 6t+4 \\ a(.215) = 6(.215)+4 \\ a = 5.29 \end{gathered}$$

Therefore, our acceleration of the particle is approximately 5.29 ft/s^2.