The diameter of the wheel is\(D = 61\;{\rm{cm}}\).

The initial revolutions per minute are\({N_1} = 120\;{\rm{rpm}}\).

The final revolutions per minute are\({N_2} = 280\;{\rm{rpm}}\).

The time taken is\(t = 4\;{\rm{s}}\).

The time after acceleration starts is\({t_0} = 2\;{\rm{s}}\).

In order to calculate the angular acceleration, the kinematic relation will be used, and for the component of the acceleration, first, instantaneous velocity is required.

The relation to find the angular acceleration is given by:

\(\alpha = \frac{{{\omega _2} - {\omega _1}}}{t}\)

Here, \({\omega _1}\) and \({\omega _1}\) are the initial and final angular velocities.

On plugging the values in the above relation, you get:

\(\begin{aligned}{l}\alpha &= \left( {\frac{{\left( {280\;{\rm{rpm}} \times \frac{{2\pi \;{\rm{rad}}}}{{1\;{\rm{rev}}}} \times \frac{{1\;{\rm{min}}}}{{60\;{\rm{s}}}}} \right) - \left( {120\;{\rm{rpm}} \times \frac{{2\pi \;{\rm{rad}}}}{{1\;{\rm{rev}}}} \times \frac{{1\;{\rm{min}}}}{{60\;{\rm{s}}}}} \right)}}{{\left( {4\;{\rm{s}}} \right)}}} \right)\\\alpha &= \left( {\frac{{\left( {29.3\;{\rm{rad/s}}} \right) - \left( {12.5\;{\rm{rad/s}}} \right)}}{{\left( {4\;{\rm{s}}} \right)}}} \right)\\\alpha &= 4.2\;{\rm{rad/}}{{\rm{s}}^{\rm{2}}}\end{aligned}\)

Thus, \(\alpha = 4.2\;{\rm{rad/}}{{\rm{s}}^{\rm{2}}}\) is the angular acceleration.

The relation to find the angular velocity is given by:

\(\begin{aligned}{l}{\omega _0} &= {\omega _1} + \alpha {t_0}\\{\omega _0} &= \left( {12.5\;{\rm{rad/s}}} \right) + \left( {4.2\;{\rm{rad/}}{{\rm{s}}^2}} \right)\left( {2\;{\rm{s}}} \right)\\{\omega _0} &= 20.9\;{\rm{rad/s}}\end{aligned}\)

The relation to find the radial acceleration is given by:

\(\begin{aligned}{l}{a_{\rm{r}}} &= {\omega _0}^2r\\{a_{\rm{r}}} &= {\omega _0}^2\left( {\frac{D}{2}} \right)\end{aligned}\)

On plugging the values in the above relation, you get:

\(\begin{aligned}{l}{a_{\rm{r}}} &= {\left( {20.9\;{\rm{rad/s}}} \right)^2}\left( {\frac{{61\;{\rm{cm}} \times \frac{{1\;{\rm{m}}}}{{100\;{\rm{cm}}}}}}{2}} \right)\\{a_{\rm{r}}} &= 133.2\;{\rm{m/}}{{\rm{s}}^2}\end{aligned}\)

The relation to find the tangential acceleration is given by:

\(\begin{aligned}{l}{a_{\rm{t}}} &= \alpha r\\{a_{\rm{t}}} &= \alpha \left( {\frac{D}{2}} \right)\end{aligned}\)

On plugging the values in the above relation, you get:

\(\begin{aligned}{l}{a_{\rm{t}}} &= \left( {4.2\;{\rm{rad/}}{{\rm{s}}^2}} \right)\left( {\frac{{61\;{\rm{cm}} \times \frac{{1\;{\rm{m}}}}{{100\;{\rm{cm}}}}}}{2}} \right)\\{a_{\rm{t}}} &= 1.28\;{\rm{m/}}{{\rm{s}}^2}\end{aligned}\)

Thus, \({a_{\rm{r}}} = 133.2\;{\rm{m/}}{{\rm{s}}^2}\) and \({a_{\rm{t}}} = 1.28\;{\rm{m/}}{{\rm{s}}^2}\) are the radial and tangential acceleration.