Engineering Mechanics: Statics & Dynamics (14th Edition)

$\vec{v_C}=-2.4 (ft/s)\hat i$
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: $\vec{v_C}=0+(4\hat k)(0.6\hat j)$ $\implies \vec{v_C}=-2.4 (ft/s)\hat i$