Answer
$4t,4$
Work Step by Step
From the question, we solve for $r'(t)= 4ti+(2t^2-2) j$ and $r''(t)=4i+4tj$
This yields,$|r'(t)|=\sqrt{(4t)^2+(2t^2-2)}=2(t^2+1)$
Also, $r'(t) \cdot r''(t)=[4ti+(2t^2-2) j] \cdot [4i+4tj]=8t^3+8t$
Also,$r'(t) \times r''(t)=[4ti+(2t^2-2) j] \times [4i+4tj]$
or, $=(8t^2+8)k$
Tangential acceleration component: $a_T=\dfrac{r'(t) \cdot r''(t)}{|r'(t)|}=\dfrac{8t^3+8t}{2(t^2+1)}=4t$
Normal acceleration component:$a_N=\dfrac{|r'(t) \times r''(t)|}{|r'(t)|}=\dfrac{(8t^2+8)}{2(t^2+1)}=4$
