Answer
$12\dfrac {rad}{s^{2}}$
Work Step by Step
To find the average angular acceleration at a given time interval, we need to find the change in angular velocity for this interval. Therefore, the angular acceleration is:
$\overline {\alpha }=\dfrac {\Delta w}{\Delta t}...........\left( 1\right) $
Lets now calculate the angular velocity at $t=2s$ and $t=4s$:
Angular velocity of the point is :
$w\left( t\right) =\dfrac {\partial \theta }{\partial t}=\dfrac {\partial }{\partial t}\left( 4t-3t^{2}+t^{3}\right) =4-6t+3t^{2}$
So, at time $t=4s$, we get:
$w\left( 4\right) =4-6t+3t^{2}=4-6\times 4+3\times 4^{2}=28\dfrac {rad}{s}.............(2)$
And at time $t=2s$, we get:
$w\left( 2\right) =4-6t+3t^{2}=4-6\times 2+3\times 2^{2}=4\dfrac {rad}{s}..........(3)$
So,
$\Delta t=4s-2s=2s(4)$
Using (1),(2),(3),(4); we get $\overline {\alpha }=\dfrac {\Delta w}{\Delta t}=\dfrac {w\left( 4\right) -w\left( 2\right) }{4s-2s}=\dfrac {28-4}{2}=12\dfrac {rad}{s^{2}}$
