Answer
$\vec{v_C}=-2.4 (ft/s)\hat i$
Work Step by Step
We can determine the required velocity as follows:
$\vec{v_C}=\vec{v_B}+\vec{\omega}\times \vec{r_{C/B}}~$[eq(1)]
As we know that the gear rack $B$ is fixed, therefore $\vec{v_B}=0$
Similarly, $\vec{r_{C/B}}=0.6$
We plug in the known values in eq(1) to obtain:
$\vec{v_C}=0+(4\hat k)(0.6\hat j)$
$\implies \vec{v_C}=-2.4 (ft/s)\hat i$
