Answer
The decomposition of ${\bf{a}}$:
${\bf{a}}\left( t \right) = {a_{\bf{T}}}{\bf{T}}\left( t \right) + {a_{\bf{N}}}{\bf{N}}\left( t \right)$
${\bf{a}}\left( t \right) = 0\cdot{\bf{T}}\left( t \right) + \frac{1}{4}{\bf{N}}\left( t \right)$,
where ${\bf{T}}\left( t \right) = \left( { - \sin \frac{t}{{20}},\cos \frac{t}{{20}}} \right)$ and ${\bf{N}}\left( t \right) = \left( { - \cos \frac{t}{{20}}, - \sin \frac{t}{{20}}} \right)$.
Work Step by Step
We can parametrize the circle of radius $R=100$ by ${\bf{r}}\left( t \right) = \left( {100\cos \omega t,100\sin \omega t} \right)$, where $\omega$ is the angular speed of the particle. So, the velocity and acceleration vectors are ${\bf{v}}\left( t \right) = {\bf{r}}'\left( t \right) = \left( { - 100\omega \sin \omega t,100\omega \cos \omega t} \right)$ and ${\bf{a}}\left( t \right) = {\bf{r}}{\rm{''}}\left( t \right) = \left( { - 100{\omega ^2}\cos \omega t, - 100{\omega ^2}\sin \omega t} \right)$, respectively.
The speed of the particle is
$v = \sqrt {{\bf{v}}\left( t \right)\cdot{\bf{v}}\left( t \right)} = \sqrt {{{\left( { - 100\omega \sin \omega t} \right)}^2} + {{\left( {100\cos \omega t} \right)}^2}} $
$v = 100\omega $
Since the particle is moving with constant speed ${v_0} = 5$ cm/s, we get
$100\omega = 5$, ${\ \ \ }$ $\omega = \frac{1}{{20}}$
Thus,
${\bf{v}}\left( t \right) = \left( { - 5\sin \frac{t}{{20}},5\cos \frac{t}{{20}}} \right)$
${\bf{a}}\left( t \right) = \left( { - \frac{1}{4}\cos \frac{t}{{20}}, - \frac{1}{4}\sin \frac{t}{{20}}} \right)$
Thus, the unit tangent vector is
${\bf{T}} = \frac{{\bf{v}}}{{||{\bf{v}}||}} = \frac{{\left( { - 5\sin \frac{t}{{20}},5\cos \frac{t}{{20}}} \right)}}{{\sqrt {\left( { - 5\sin \frac{t}{{20}},5\cos \frac{t}{{20}}} \right)\cdot\left( { - 5\sin \frac{t}{{20}},5\cos \frac{t}{{20}}} \right)} }}$
${\bf{T}} = \left( { - \sin \frac{t}{{20}},\cos \frac{t}{{20}}} \right)$
By Eq. (2) of Theorem 1 we have
${a_{\bf{T}}} = {\bf{a}}\cdot{\bf{T}} = \left( { - \frac{1}{4}\cos \frac{t}{{20}}, - \frac{1}{4}\sin \frac{t}{{20}}} \right)\cdot\left( { - \sin \frac{t}{{20}},\cos \frac{t}{{20}}} \right)$
${a_{\bf{T}}} = 0$
This is expected since the speed is constant, so the tangential component of the acceleration is zero.
Next, we use Eq. (3) of Theorem 1 to find
${a_{\bf{N}}}{\bf{N}} = {\bf{a}} - {a_{\bf{T}}}{\bf{T}}$
${a_{\bf{N}}}{\bf{N}} = \left( { - \frac{1}{4}\cos \frac{t}{{20}}, - \frac{1}{4}\sin \frac{t}{{20}}} \right)$
Since ${\bf{N}}$ is an unit vector, so
${a_{\bf{N}}} = ||{a_{\bf{N}}}{\bf{N}}|| = \sqrt {{{\left( { - \frac{1}{4}\cos \frac{t}{{20}}} \right)}^2} + {{\left( { - \frac{1}{4}\sin \frac{t}{{20}}} \right)}^2}} $
${a_{\bf{N}}} = \frac{1}{4}$
We find ${\bf{N}}$ by using the equation
${\bf{N}} = \frac{{{a_{\bf{N}}}{\bf{N}}}}{{{a_{\bf{N}}}}} = 4\left( { - \frac{1}{4}\cos \frac{t}{{20}}, - \frac{1}{4}\sin \frac{t}{{20}}} \right)$
${\bf{N}} = \left( { - \cos \frac{t}{{20}}, - \sin \frac{t}{{20}}} \right)$
Thus, we obtain the decomposition of ${\bf{a}}$:
${\bf{a}}\left( t \right) = {a_{\bf{T}}}{\bf{T}}\left( t \right) + {a_{\bf{N}}}{\bf{N}}\left( t \right)$
${\bf{a}}\left( t \right) = 0\cdot{\bf{T}}\left( t \right) + \frac{1}{4}{\bf{N}}\left( t \right)$,
where ${\bf{T}}\left( t \right) = \left( { - \sin \frac{t}{{20}},\cos \frac{t}{{20}}} \right)$ and ${\bf{N}}\left( t \right) = \left( { - \cos \frac{t}{{20}}, - \sin \frac{t}{{20}}} \right)$.