## Engineering Mechanics: Statics & Dynamics (14th Edition)

$v_A=22m/s$ $a_{At}=12m/s^2$ $a_{An}=968m/s^2$
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: $a_{An}=968m/s^2$