At a particular instant, a particle traveling in a circular path of the radius \[0.5m\] has a speed \[V = {\text{ }}1.6{\text{ }}m/s\] and an angular acceleration $\alpha = 16{\text{ }}rad/{s^2}.$
$(1)$ Find the acceleration of the particle.
\[{\text{A}}.\;\;16\;m/{s^2}\]
\[{\text{B}}.\;\;5.1{\text{ }}m/{s^2}\]
\[{\text{C}}.\;\;10\;m/{s^2}\]
\[{\text{D}}.\;\;9.5{\text{ }}m/{s^2}\]

$(2)$ What angle does the acceleration vector make with a tangent to the path?
\[{\text{A}}.\;\;\;ta{n^{ - 1}}\left( {\dfrac{{9.5}}{8}} \right)\]
\[{\text{B}}{\text{. }}ta{n^{ - 1}}\left( {\dfrac{{5.12}}{8}} \right)\]
\[{\text{C}}{\text{. }}ta{n^{ - 1}}\left( {\dfrac{8}{{9.5}}} \right)\]
\[{\text{D}}{\text{. }}ta{n^{ - 1}}\left( {\dfrac{8}{{5.12}}} \right)\]

Hint:
1. Calculate the component of acceleration towards the center from the given velocity of the circular motion of the particle. Also, calculate another component of the acceleration along the tangent to the path from the angular acceleration.
2. Find the resultant acceleration of the two components of the acceleration.
3. To find the required angle use the two components of the acceleration.

Formula used:
the component of acceleration along the tangent, ${a_{tangent}} = \alpha r$
where $\alpha $ is the angular acceleration and $r$ is the radius of the circular path.
The component of acceleration towards the center, ${a_{center}} = \dfrac{{{v^2}}}{r}$
Where $v$ is the velocity of the circular path.
The resultant acceleration, \[a = \sqrt {{{({a_{tangent}})}^2} + {{({a_{center}})}^2}} \]
$\tan \theta = \dfrac{{{a_{tangent}}}}{{{a_{center}}}}$

Complete step by step answer:
The particle is moving in a circular path.
the component of acceleration along the tangent, ${a_{tangent}} = \alpha r$
given, the angular acceleration $\alpha = 16rad/{s^2}$
the radius of the circular path $r = 0.5m$
$\therefore {a_{\tan gent}} = 16 \times 0.5 = 8m/{s^2}$
Also, the component of acceleration towards the center, ${a_{center}} = \dfrac{{{v^2}}}{r}$
given, the velocity of the circular path, $v = 1.6m/s$
${a_{center}} = \dfrac{{1.6 \times 1.6}}{{0.5}} = 5.12m/{s^2}$
The resultant acceleration, \[a = \sqrt {{{({a_{tangent}})}^2} + {{({a_{center}})}^2}} \]
\[ \Rightarrow a = \sqrt {{8^2} + {{5.12}^2}} \]
\[ \Rightarrow a = \sqrt {90.21} \]
\[ \Rightarrow a = 9.49 \simeq 9.5m/{s^2}\]

$(1) \Rightarrow $ Hence the right answer is in option $D$.

If the angle is $\theta $ that the acceleration vector make with a tangent to the path,
$\tan \theta = \dfrac{{{a_{center}}}}{{{a_{tangent}}}}$
$ \Rightarrow \tan \theta = \dfrac{{5.12}}{8}$
$ \Rightarrow \theta = {\tan ^{ - 1}}\left( {\dfrac{{5.12}}{8}} \right)$

$(2)$ Hence the right answer is in the option $B$.

Note: The net acceleration is acting at an angle $\theta $ with the tangent hence the angle between the net acceleration and the component that is acting towards the center will be $(90 - \theta )$.
Since the acceleration is a vector quantity, we use the vector addition formula for two components of a vector in which if $R$ is the resultant vector of two vectors $A$ and $B$ , then
$\overrightarrow R = \overrightarrow A + \overrightarrow B $
$R = \sqrt {{A^2} + {B^2}} $ .