Suppose we have two points $A, B \in \mathbb{R}^2$ where $A = (0 | y)$ and $B = (0|-y)$. Also we have a point $P$ which lies on the line between $A$ and $B$. If we can move the two points on the x-axis at velocity $v_x$ and the point $P$ on the line at velocity $v_{\lambda}$ then what is the kinematics equation?
I used $f(\lambda) = \lambda A + (1 - \lambda) B, \lambda \in [0,1]$ to model the line on which the point $P$ can be. I struggled at combining the three movements into one equation because $f(\lambda)$ depends on $A$ and $B$.
We have
$$\vec{v}_{P / \mathcal{I}} = \vec{v}_{A / \mathcal{I}} + \vec{v}_{P / \mathcal{B}} + \vec{\omega}_{\mathcal{B} / \mathcal{I}} \times \vec{r}_{A \to P}$$
where $\mathcal{I}$ is the inertial frame, and $\mathcal{B}$ is a frame attached to line segment $AB$. Note that $\mathcal{B}$ is not rotating in $\mathcal{I}$, so that the last term in the equation above can be ignored. The velocity of $A$ in the inertial frame is given; it is $v_x \hat{i}$ (where $\hat{i}$ is aligned with the positive $x$-axis). The velocity of $P$ in the frame $\mathcal{B}$ is also given; it is $-v_\lambda \hat{i}_\mathcal{B}$, where $\hat{i}_\mathcal{B}$ points in the direction from $B$ to $A$. Note that by coordinate transformation, $\hat{i}_\mathcal{B} = \cos \theta \hat{i} + \sin \theta \hat{j}$, where $\theta$ is the angle that $BA$ makes with the horizontal. Thus, in the inertial basis,
$$\vec{v}_{P / \mathcal{I}} = (v_x - v_\lambda \cos \theta) \hat{i} - v_\lambda \sin \theta \hat{j}$$
Since here, $\theta = \frac{\pi}{4}$, we obtain
$$\vec{v}_{P / \mathcal{I}} = \left( v_x - \frac{\sqrt{2}}{2} v_\lambda \right) \hat{i} - \frac{\sqrt{2}}{2} v_\lambda \hat{j}$$