A particle moving along the x-axis has its position described by the function x = 12t 3 + 2t + 12 m, where t is in s. At t = 2 s what are the particle’s (a) position, (b) velocity, and (c) acceleration?

The position of the particle is given by the function,$x=\left(2t^3+2t+1\right)m$.

To find the position of the particle at $t=2s$, substitute 2 for $t$into the given function.

$$\begin{gathered} x=2{\left(2\right)}^3+2\left(2\right)+1 \\ x=2\times 8+4+1 \\ x=21m \end{gathered}$$

Therefore, the position of the particle at$t=2s$is $21m$.

The velocity is given by $v_x=\frac{dx}{dt}$.

Differentiate the given function with respect to $t$.

$$\begin{gathered} v_x=\frac{d\left(2t^3+2t+1\right)}{dt} \\ =\frac{d\left(2t^3\right)}{dt}+\frac{d\left(2t\right)}{dt}+\frac{d\left(1\right)}{dt} \\ =2\left(3t^2\right)+2 \\ =6t^2+2 \end{gathered}$$

To find the velocity of the particle at $t=2s$, substitute 2 for $t$into the obtained expression.

$$\begin{gathered} v_x\left(att=2s\right)=6{\left(2\right)}^2+2 \\ =6\times 4+2 \\ =26\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right. \end{gathered}$$

Therefore, the velocity of the particle at$t=2s$is$26\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$.

The acceleration is given by $a_x=\frac{d{v}_x}{dt}$.

Differentiate $v_x=6t^2+2$with respect to $t$.

$$\begin{gathered} a_x=\frac{d\left(6t^2+2\right)}{dt} \\ =\frac{d\left(6t^2\right)}{dt}+\frac{d\left(2\right)}{dt} \\ =6\left(2t\right) \\ =12t \end{gathered}$$

To find the acceleration of the particle at $t=2s$, substitute 2 for $t$into obtained expression.

$$\begin{gathered} a_x\left(att=2s\right)=12\left(2\right) \\ =24\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right. \end{gathered}$$

Therefore, the acceleration of the particle is$24\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.$.