Answer
$\dfrac{4(t-1)}{\sqrt{4t^2-8t+5}},\dfrac{2}{\sqrt{4t^2-8t+5}}$
Work Step by Step
Formula to calculate the tangential acceleration component is:
$a_T=\dfrac{r'(t) \cdot r''(t)}{|r'(t)|}$ ....(1)
and
Formula to calculate the normal acceleration component is:
$a_N=\dfrac{r'(t) \times r''(t)}{|r'(t)|}$ ....(2)
Thus, $r'(t)=1i+(2t-2)j$ and $r''(t)=2j$
$|r'(t)|=\sqrt{1+4t^2-8t+4}=\sqrt{4t^2-8t+5}$
$r'(t) \cdot r''(t)=[1i+(2t-2)j] \cdot [2j]=4(t-1)$
and
$r'(t) \times r''(t)=[1i+(2t-2)j] \times [2j]=2k$
From equation (1), we have
$a_T=\dfrac{r'(t) \cdot r''(t)}{|r'(t)|}=\dfrac{4(t-1)}{\sqrt{4t^2-8t+5}}=\dfrac{4(t-1)}{\sqrt{4t^2-8t+5}}$
From equation (2), we have
$a_N=\dfrac{r'(t) \times r''(t)}{|r'(t)|}=\dfrac{2}{\sqrt{4t^2-8t+5}}$
