Answer
$v_A=8.10m/s$
$a_{At}=4.95m/s^2$
$a_{An}=437.4m/s^2$
Work Step by Step
We can determine the required velocity and acceleration components as follows:
As $\omega_A=5(3)^2+3(3)=54rad/s$
and $\alpha=\frac{d\omega}{dt}=10t+3$
At $t=3s$
$\alpha= 10(3)+3=33 rad/s^2$
The velocity at point A is given as
$v_A=\omega_A r_A$
$\implies v_A=54(0.15)=8.1m/s$
Now the tangential and normal components of acceleration can be calculated as
$a_{At}= \alpha r_A$
We plug in the known values to obtain:
$a_{At}=(33)(0.15)=4.95m/s^2$
and $a_{A_n}=\omega^2 r_A$
We plug in the known values to obtain:
$a_{An}=(54)^2(0.15)=437.4m/s^2$
