Answer
$v_A=22m/s$
$a_{At}=12m/s^2$
$a_{An}=968m/s^2$
Work Step by Step
We can determine the required velocity and acceleration as follows:
As given that
$a=\frac{d\omega}{dt}=3t^2+12$
$\implies \int_{\omega_{\circ}}^{\omega} d\omega=\int_0^t(3t^2+12)dt$
$\omega-\omega_{\circ}=(\frac{3t^3}{3}+12t)|_0^t$
$\implies \omega=t^3+12t+12$
Now, $v_A=\omega r_A$
$\implies v_A=(t^3+12t+12)r_A$
We plug in the known values to obtain:
$v_A=[(2)^3+12(2)+12](0.5)$
$\implies v_A=22m/s$
Similarly
$a_{At}=\alpha r_A$
$\implies a_{At}=(3t^2+12)r_A$
We plug in the known values to obtain:
$a_{At}=[3(2)^2+12](0.5)$
$\implies a_{At}=12m/s^2$
and $a_{An}=\omega^ 2 r_A$
$\implies a_{An}=(t^3+12t+12)r_A$
We plug in the known values to obtain:
$a_{An}=[(2)^3+12(2)+12](0.5)$
This simplifies to:
$a_{An}=968m/s^2$
