The relationship between points B and C
$$\vec{v}_C = \vec{v}_B + \vec{W}_{BC} \times \vec{r}_{BC} \tag{eq.1}$$
Additionally you know that
$$\vec{v}_D = \vec{v}_C + \vec{W}_{CD} \times \vec{r}_{CD} \tag{eq.2}$$
However you already know that:
- $v_B = \begin{bmatrix}0 \\-3j\\0 \end{bmatrix}$
- $v_D = \begin{bmatrix}0 \\0\\0 \end{bmatrix}$
If you do the calcs you get from the equations (1, 2):
\begin{equation}
\vec{v}_D = \vec{v}_B + \vec{W}_{BC} \times \vec{r}_{BC} + \vec{W}_{CD} \times \vec{r}_{CD}
\end{equation}
if you do the math you should get:
$$
\begin{bmatrix}0 \\0\\0 \end{bmatrix} = \begin{bmatrix}-w_{BC} - 0.5*w_{CD} \\-0.5*w_{CD} - 3\\0 \end{bmatrix}
$$
From which you can obtain:
- $w_{CD}= -6 [\frac{rad}{s}]$
- $w_{BC}= +3 [\frac{rad}{s}]$
Similarly the (chain) relationships for the acceleration of point B, C, D are:
$$a_B = a_A + α_{AB}\times r_{AB} + w_{AB}\times(w_{AB}\times r_{AB}) \Leftrightarrow a_B = -6j - 9 i$$
$$a_C = a_B + α_{BC}\times r_{BC} + w_{BC}\times(w_{BC}\times r_{BC}) $$
$$a_D = a_C + α_{CD}\times r_{CD} + w_{CD}\times(w_{CD}\times r_{CD}) $$
However as before, $a_D$ should be equal to zero. So if you do the math you should get to:
$$\begin{bmatrix}0 \\0\\0 \end{bmatrix} =
\begin{bmatrix}
-a_{BC} - 0.5*a_{CD} - 4.5 \\
-0.5*a_{CD} - 19.5\\
0 \end{bmatrix}$$
The result I am getting is:
$$a_{CD}= -39[rad/s^2]\qquad
a_{BC}= 15[rad/s^2]
$$