Particle in a time varying curve

$\begin{aligned} \vec{v}&=\left(3{t}^2-4\right)\hat{i}+\left(2t+4\right)\hat{j}+\left(16t-9{t}^2 \right)\hat{k}\\\vec{a}&=\left(6t\right)\hat{i}+\left(2\right)\hat{j}+\left(16-18t\right)\hat{k}\end{aligned}$

At $t=2$

$\begin{aligned} \vec{v}&=8\hat{i}+8\hat{j}-4\hat{k}\\\vec{a}&=12\hat{i}+2\hat{j}-20\hat{k}\end{aligned}$

Velocity vector is always tangent to the curve of motion. Magnitude of acceleration along tangent is given by

$\begin{aligned}a_t &=\vec{a}\cdot\hat{v}\\&=\dfrac{\vec{a}\cdot\vec{v}}{\left| \vec{v} \right|}\\&=16\end{aligned}$

Magnitude of acceleration along the normal is given by

$\begin{aligned} a_n&=\sqrt{{\left|\vec{a}\right|}^2-{16}^2}\\&=\sqrt{548-256}\\&\huge\color{#3D99F6}{=\boxed{2\sqrt{73}}}\end{aligned}$

$\vec{a}={\vec{a}_{t }}+{\vec{a}_{n }}$

${ a }_{ n }=\sqrt { ||{ a }_{ t }||^{ 2 } -||a||^2}$

$\vec { v } =\frac { d\vec { R } }{ dt } =(3t^{ 2 }-4)\vec { i } +(2t+4)\vec { k } +(16t-9t^ 2)\vec{k}$

$\vec { a } (t)=\frac { d\vec { v } }{ dt } =6t\vec { i } +2\vec { j } -(16-18t)\vec { k }$

$\vec { a } (2)=12\vec{i}+2\vec{j}-20\vec{k}$

$||\vec { a }(2) ||=\sqrt { 12^2+2^2+(-20)^2 } =\boxed{2\sqrt { 137 }}$

$||\vec{v}||=\sqrt { (3t^2-4t)^2+(2t+4)^2+(16t-9t^2)^2 }$

${ { a } }_{ t }(t)=\frac { dv }{ dt } =\frac { d }{ dt } (\sqrt { (3t^{ 2 }-4t)^{ 2 }+(2t+4)^{ 2 }+(16t-9t^{ 2 })^{ 2 } } )=\frac { 2(45{ t }^{ 3 }-108{ t }^{ 2 }+59t+2) }{ \sqrt { \frac { 45{ t }^{ 4 } }{ 2 } -72{ t }^{ 3 }+59{ t }^{ 2 }+4t+8 } }$

${ { a } }_{ t }(2)=\boxed{16}$

$\therefore||{ a }_{ n }||=\sqrt { { (2\sqrt { 137 } ) }^{ 2 }-16^{ 2 } } = 2\sqrt { 73 }$

Let's write our curve as $\displaystyle \overline{R}(t)=(t^3-4t,t^2+4t,8t^2-3t^3)$ , so that

$\displaystyle \overline{R}'(t)=(3t^2-4,2t+4,16t-9t^2) \Longrightarrow \overline{R}'(t=2)=(8,8,-4) \\ \overline{R}''(t)=(6t,2,16-18t) \Longrightarrow \overline{R}'(t=2)=(12,2,-20)$

The centripetal acceleration $\overline{a}_n(t)$ is defined as

$\displaystyle ||\overline{a}_n(t)||=\frac{||\overline{R}'(t)||^2}{r}$

where $r$ is the curvature radius and $\displaystyle \kappa=\frac{1}{r}$ is the curvature. Since

$\displaystyle \kappa=\frac{||\overline{R}'(t) \times \overline{R}''(t)||}{||\overline{R}'(t)||^3}$

then

$\displaystyle ||\overline{a}_n(t)||=||\overline{R}'(t)||^2\kappa=\frac{||\overline{R}'(t) \times \overline{R}''(t)||}{||\overline{R}'(t)||} \Bigg|_{t=2}=\frac{||\det{((8,8,-4),(12,2,-20))}||}{12}=\frac{24 \sqrt{73}}{12}=\boxed{2\sqrt{73}}$