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

$v_A=8.10m/s$ $a_{At}=4.95m/s^2$ $a_{An}=437.4m/s^2$
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: $a_{An}=(54)^2(0.15)=437.4m/s^2$