So its a classic problem:
Object $A$ starts at the origin $(0,0)$ and moves straight up the $y$ axis with a speed $v$.
Object $B$ starts at point $(1,0)$, always moves towards object $A$ and has a speed of $2v$.
There are answers on Stack Exchange to the question of finding $y_b (x) $ where $y_b$ is the $y$ coordinate of object $b$.
But I want to find $y_b(t)$ and $x_b(t)$ where $t$ is time
These, I think are the two equations you get from the problem have:
Since the velocity of object $b$ is always $2v$ $$ \sqrt{x_b'(t)^2 + y_b'(t)^2} = 2v$$ Since the slope of object $B$ is always pointing toward object $A$, whose coordinates are $(0, vt)$ $$ \frac{y_b'(t)}{x_b'(t)} = \frac {y_b(t) - vt}{x_b(t)}$$
So far, I have had no luck solving this, and I did try mathematica, but that didn't work. I'm not even sure if there is a reasonable solution.
If you must, you can let $v = 1$, and bonus points if you can solve this for more generic coordinates.
Note: I've only just finished calculus $AB$, so if could keep that in mind with your responses, that would be helpful :)