Step 1

We are given the angular velocity of disc as $\omega = \left( {5{t^2} + 2} \right)\,{\rm{rad/s}}$, the radius of the disc as $r = 0.8\,{\rm{m}}$, and the time of observation as $t = 0.5\,{\rm{s}}$.

We are asked to determine the magnitudes of the velocity and acceleration of point $A$ on the disk when $t = 0.5\,{\rm{s}}$.

 
Step 2

First find the angular velocity of the disk at $t = 0.5\,{\rm{s}}$ using the following relation.

\[\omega = \left( {5{t^2} + 2} \right)\,{\rm{rad/s}}\]

On substituting the known value in the above equation we get,

\[\begin{array}{c} \omega = \left( {5{{\left( {0.5} \right)}^2} + 2} \right)\,{\rm{rad/s}}\\ = \left( {1.25 + 2} \right)\,{\rm{rad/s}}\\ = 3.25\,{\rm{rad/s}} \end{array}\]
 
Step 3

To find the velocity of the disk at $t = 0.5\,{\rm{s}}$ use the following relation.

\[v = \omega r\]

On substituting the known values in the above equation we get,

\[\begin{array}{c} v = \left( {3.25\,{\rm{rad/s}}} \right)\left( {0.8\,{\rm{m}}} \right)\\ = 2.6\,{\rm{m/s}} \end{array}\]
 
Step 4

To find the angular acceleration of the disk at $t = 0.5\,{\rm{s}}$ use the following relation.

\[\alpha = \frac{d}{{dt}}\omega \]

On substituting $5{t^2} + 2$ for $\omega $ and differentiating the above equation we get,

\[\begin{array}{c} \alpha = \frac{d}{{dt}}\left( {5{t^2} + 2} \right)\\ = 10t + 0\\ = \left( {10t} \right)\,{\rm{rad/}}{{\rm{s}}^{\rm{2}}} \end{array}\]

On substituting the known value in the above equation we get,

\[\begin{array}{c} \alpha = \left( {10\left( {0.5} \right)} \right)\,{\rm{rad/}}{{\rm{s}}^{\rm{2}}}\\ = 5\,{\rm{rad/}}{{\rm{s}}^{\rm{2}}} \end{array}\]
 
Step 5

To find the tangential acceleration of point A use the following relation.

\[{a_t} = \alpha r\]

On substituting the values in the above expression we get,

\[\begin{array}{l} {a_t} = \left( {5\;{\rm{rad/}}{{\rm{s}}^{\rm{2}}}} \right)\left( {0.8\;{\rm{m}}} \right)\\ {a_t} = 4\;{\rm{m/}}{{\rm{s}}^{\rm{2}}} \end{array}\]

To find the normal acceleration of point A use the following relation.

\[{a_n} = {\omega ^2}r\]

On substituting the values in the above expression we get,

\[\begin{array}{l} {a_n} = {\left( {3.25\;{\rm{rad/s}}} \right)^2}\left( {0.8\;{\rm{m}}} \right)\\ {a_n} = 8.45\;{\rm{m/}}{{\rm{s}}^{\rm{2}}} \end{array}\]

To find the magnitude of acceleration of point A use the following relation.

\[a = \sqrt {a_t^2 + a_n^2} \]

On substituting the values in the above expression we get,

\[\begin{array}{l} a = \sqrt {{{\left( {4\;{\rm{m/s}}} \right)}^2} + \left( {8.45\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right)} \\ a = 9.35\;{\rm{m/}}{{\rm{s}}^{\rm{2}}} \end{array}\]