Answer
$$
V_B=3 \frac{\mathrm{m}}{\mathrm{s}} \rightarrow \\
V_c=0.5864 \mathrm{~m} / \mathrm{s} \rightarrow \\
$$
Work Step by Step
$$
\begin{array}{l}
\begin{array}{l}
\theta=\tan ^{-1}\left(\frac{3}{4}\right) \rightarrow \theta=36.87^{\circ} \\
\vec{V}_B=V_A+w_{A B D} \hat{k} \hat{r}_{B / A} \\
V_B \hat{\imath}=-4 \hat{\jmath}+w_{A B D} \hat{k}\left(-0.55 \cos 36.87^{\circ} \hat{\imath}+0.55 \operatorname{sen} 36.87^{\circ} \hat{\jmath}\right) \\
V_B \hat{\imath}=-4 \hat{\jmath}-0.44 W_{A B D} \hat{\jmath}-0.33 W_{A B D} \hat{\imath} \\
\hat{\jmath}=\hat{\jmath} \\
0=-4-0.44 \omega_{A B D} \\
0.44 w_{A D D}=-4 \\
W_{A B D}=\frac{-4}{0.44} \\
W_{A B D}=-9.09 \\
\sim w_{A B D}=9.09 \frac{\mathrm{rad}}{\mathrm{s}} \circlearrowright \\
\hat{\imath}=\hat{\imath} \\
V_B=-0.33 w_{A B D} \\
V_B=-0.33(-9.09) \\
V_B=3 \frac{\mathrm{m}}{\mathrm{s}} \rightarrow \\
\vec{V}_D=V_B+w_{A B D} \hat{k} \hat{r}_{D / B} \\
\vec{V}_D=3 \hat{\imath}+(-9.09) \hat{k}\left(0.25 \cos 36.87^{\circ} \hat{\imath}-0.25 \operatorname{sen} 36.87^{\circ} \hat{\jmath}\right) \\
\vec{V}_D=3 \hat{\imath} -1.818\hat{\jmath} -1.3635\hat{\imath}\\
\vec{V}_D=1.6364 \hat{\imath}-1.818 \hat{\jmath}\\
\vec{V}_D=V_C+w_{C D E} \hat{k} \hat{r}_{D / C} \\
\end{array}\\
\begin{array}{l}
(1.6364 \hat{\imath}-1.818 \hat{\jmath})=V_C \hat{\imath}+W_{C D E} \hat{k}\left(0.4 \cos 30^{\circ} \hat{\imath}+0.4 \operatorname{sen} 30^{\circ} \hat{\jmath}\right) \\
(1.6364 \hat{\imath}-1.818 \hat{\jmath})=v_C \hat{\imath}+0.3464 w_{C D E} \hat{\jmath}-0.2 w_{\operatorname{CDE}} \hat{\imath} \\
\hat{\jmath}=\hat{\jmath} \\
-1.818=0.34642 W_{C D E} \\
w_{C D E}=\frac{-1.818}{0.3464} \\
w_{C D E}=-5.25 \\
\sim w_{\text {CDE }}=5.25 \frac{\mathrm{rad}}{\mathrm{s}} \circlearrowright \\
\hat{\imath}=\hat{\imath} \\
1.6364=v_c-0.2 \omega_{C D E} \\
V_c=1.6364+0.2(-5.25) \\
V_c=0.5864 \mathrm{~m} / \mathrm{s} \rightarrow \\
\end{array}
\end{array}
$$
