$[3]$ The disk rolls without slipping and has the angular motion shown in the figure below. Determine the acceleration of point $A$ at this instant

$=6ra\frac{d}{s}$

$=4ra\frac{d}{s^2}$

$a_A=4ra\frac{d}{s^2}*0\mathrm{.}5ft-(6ra\frac{d}{s})$

$=2ra\frac{d}{s^2}-(36ra\frac{d}{s})(0\mathrm{.}5Ft)$

$=2-18$

$a_A=16ra\frac{d}{s^2}$