Answer
The decomposition of ${\bf{a}}$ at $t=3$:
${\bf{a}}\left( 3 \right) = 3{\bf{T}}\left( 3 \right) + \frac{{27}}{{400}}{\bf{N}}\left( 3 \right)$,
where ${\bf{T}}\left( 3 \right) \simeq \left( { - 0.015,0.999} \right)$
and ${\bf{N}}\left( 3 \right) \simeq \left( { - 0.999, - 0.015} \right)$
The acceleration at $t=3$:
${\bf{a}}\left( 3 \right) \simeq \left( { - 0.112,2.999} \right)$
Work Step by Step
We are given the radius of the circle $R=300$ and the tangential component of the acceleration ${a_{\bf{T}}} = v'\left( t \right) = t$.
The speed of the car is obtained by integrating $v'\left( t \right)$:
$v\left( t \right) = \frac{1}{2}{t^2} + {v_0}$
Since it starts at rest, so $v\left( 0 \right) = {v_0} = 0$. Therefore, $v\left( t \right) = \frac{1}{2}{t^2}$.
By Eq. (1), the normal component of the acceleration is
${a_{\bf{N}}} = \frac{{v{{\left( t \right)}^2}}}{R} = \frac{1}{{1200}}{t^4}$
Thus, the acceleration vector is decomposed into
${\bf{a}}\left( t \right) = t{\bf{T}} + \frac{1}{{1200}}{t^4}{\bf{N}}$
At $t=3$, we get the acceleration vector:
(1) ${\ \ \ }$ ${\bf{a}}\left( 3 \right) = 3{\bf{T}}\left( 3 \right) + \frac{{27}}{{400}}{\bf{N}}\left( 3 \right)$
Next we find the unit tangent vector and the normal vector:
We can parametrize the circle of radius $R=300$ by ${\bf{r}}\left( t \right) = \left( {300\cos \theta \left( t \right),300\sin \theta \left( t \right)} \right)$, where $\theta \left( t \right)$ is the angle of the particle at time $t$. Thus, the arc length traced by the particle from $0$ to $t$ is
$length = \mathop \smallint \limits_0^t v\left( u \right){\rm{d}}u = \mathop \smallint \limits_0^t \frac{1}{2}{u^2}{\rm{d}}u = \frac{1}{6}{t^3}$
It follows that $R\theta \left( t \right) = \frac{1}{6}{t^3}$. So,
$\theta \left( t \right) = \frac{1}{{1800}}{t^3}$
Thus,
${\bf{r}}\left( t \right) = \left( {300\cos \frac{{{t^3}}}{{1800}},300\sin \frac{{{t^3}}}{{1800}}} \right)$
So, the velocity and acceleration vectors are
${\bf{v}}\left( t \right) = {\bf{r}}'\left( t \right) = \left( { - \frac{1}{2}{t^2}\sin \frac{{{t^3}}}{{1800}},\frac{1}{2}{t^2}\cos \frac{{{t^3}}}{{1800}}} \right)$
${\bf{a}}\left( t \right) = {\bf{r}}{\rm{''}}\left( t \right) = ( - t\sin \frac{{{t^3}}}{{1800}} - \frac{1}{{1200}}{t^4}\cos \frac{{{t^3}}}{{1800}},$
$t\cos \frac{{{t^3}}}{{1800}} - \frac{1}{{1200}}{t^4}\sin \frac{{{t^3}}}{{1800}})$
At $t=3$, we get
${\bf{v}}\left( 3 \right) = - \frac{9}{2}\sin \frac{3}{{200}},\frac{9}{2}\cos \frac{3}{{200}} \simeq \left( { - 0.067,4.499} \right)$
${\bf{a}}\left( 3 \right) = \left( { - \frac{{27}}{{400}}\cos \frac{3}{{200}} - 3\sin \frac{3}{{200}},3\cos \frac{3}{{200}} - \frac{{27}}{{400}}\sin \frac{3}{{200}}} \right)$
$ \simeq \left( { - 0.112,2.999} \right)$
Thus, the unit tangent vector is
${\bf{T}} = \frac{{\bf{v}}}{{||{\bf{v}}||}} = \frac{{\left( { - \frac{9}{2}\sin \frac{3}{{200}},\frac{9}{2}\cos \frac{3}{{200}}} \right)}}{{\sqrt {\left( { - \frac{9}{2}\sin \frac{3}{{200}},\frac{9}{2}\cos \frac{3}{{200}}} \right)\cdot\left( { - \frac{9}{2}\sin \frac{3}{{200}},\frac{9}{2}\cos \frac{3}{{200}}} \right)} }}$
${\bf{T}} = \left( { - \sin \frac{3}{{200}},\cos \frac{3}{{200}}} \right) \simeq \left( { - 0.015,0.999} \right)$
Next, we use Eq. (3) of Theorem 1 to find
${a_{\bf{N}}}{\bf{N}} = {\bf{a}} - {a_{\bf{T}}}{\bf{T}}$
${a_{\bf{N}}}{\bf{N}} = \left( { - \frac{{27}}{{400}}\cos \frac{3}{{200}} - 3\sin \frac{3}{{200}},3\cos \frac{3}{{200}} - \frac{{27}}{{400}}\sin \frac{3}{{200}}} \right)$
$ - 3\left( { - \sin \frac{3}{{200}},\cos \frac{3}{{200}}} \right)$
${a_{\bf{N}}}{\bf{N}} = \left( { - \frac{{27}}{{400}}\cos \frac{3}{{200}}, - \frac{{27}}{{400}}\sin \frac{3}{{200}}} \right) \simeq \left( { - 0.067,0.001} \right)$
Since ${\bf{N}}$ is an unit vector, so
${a_{\bf{N}}} = ||{a_{\bf{N}}}{\bf{N}}||$
${a_{\bf{N}}} = \sqrt {{{\left( { - \frac{{27}}{{400}}\cos \frac{3}{{200}}} \right)}^2} + {{\left( { - \frac{{27}}{{400}}\sin \frac{3}{{200}}} \right)}^2}} $
${a_{\bf{N}}} = \frac{{27}}{{400}}$
We find ${\bf{N}}$ by using the equation
${\bf{N}} = \frac{{{a_{\bf{N}}}{\bf{N}}}}{{{a_{\bf{N}}}}} = \frac{{400}}{{27}}\left( { - \frac{{27}}{{400}}\cos \frac{3}{{200}}, - \frac{{27}}{{400}}\sin \frac{3}{{200}}} \right)$
${\bf{N}} = \left( { - \cos \frac{3}{{200}}, - \sin \frac{3}{{200}}} \right) \simeq \left( { - 0.999, - 0.015} \right)$
Substituting ${\bf{T}}\left( 3 \right)$ and ${\bf{N}}\left( 3 \right)$ in equation (1), we obtain the decomposition of ${\bf{a}}$ at $t=3$:
${\bf{a}}\left( 3 \right) = 3{\bf{T}}\left( 3 \right) + \frac{{27}}{{400}}{\bf{N}}\left( 3 \right)$,
where ${\bf{T}}\left( 3 \right) = \left( { - \sin \frac{3}{{200}},\cos \frac{3}{{200}}} \right) \simeq \left( { - 0.015,0.999} \right)$
and ${\bf{N}}\left( 3 \right) = \left( { - \cos \frac{3}{{200}}, - \sin \frac{3}{{200}}} \right) \simeq \left( { - 0.999, - 0.015} \right)$
${\bf{a}}\left( 3 \right) \simeq \left( { - 0.112,2.999} \right)$
