Answer
$v_A=26m/s$
$a_{At}=10m/s^2$
$a_{An}=1352m/s^2$
Work Step by Step
We know that
$\omega=\omega_{\circ}+\alpha t$
We plug in the known values to obtain:
$\omega=12+20(2)$
$\omega=52rad/s$
Now, $v_A=\omega r_A$
We plug in the known values to obtain:
$v_A=(52)(0.5)$
$\implies v_A=26m/s$
The tangential and normal components of acceleration are given as
$a_{At}=\alpha r_A$
$\implies a_{At}=\alpha r_A$
We plug in the known values to obtain:
$a_{At}=(20)(0.5)=10m/s^2$
Similarly, $a_{An}=\omega^2 r_A$
We plug in the known values to obtain:
$a_{An}=(52)^2(0.5)$
$\implies a_{An}=1352m/s^2$