The Problem
I've been working on a (very basic) game project where a square attempts to evade being touched by the users mouse. I have gotten stumped by the following problem:
Assume the users mouse $M$ is attempting to click on the square $B$.
Given the speed of $M$, an angle along which $B$ should move, and the amount of time $B$ should move for, find the point $P = (x, y)$ which $B$ should move to.
What I've Tried
One possible solution to the speed problem is simple: The speed that $B$ should move at is given by the speed of $M$ multiplied by some difficulty factor $D$ in the range $(0, 1)$. Therefore, $S(B) = MD$.
What has me so stumped is the latter part: Where should $B$ move to?
After searching Google, I found this, with no explanation provided: $$x(t) = x + tv\cos a$$ $$y(t) = y + tv\sin a$$ $where$ $$[x] = [y] = [t, v]$$
I believe that multiplying by $t$ scales the amplitude so that we don't overshoot (the speed portion of the equation), $\cos a$ and $\sin a$ give us how far we should move, and of course $x$ and $y$ give us where we are at now. I believe that $v$ is equivalent to $D$ in my speed formula; it scales the amplitude such that $B$ doesn't move at the same speed as $M$. However, What does $[x] = [y] = [t, v]$ mean?
I'll assume $\alpha$ is the angle between the $Oy$ axis and the position (vector) of B, relative to M, setting the origin of the Cartesian system of axes in M.
We can also write that, initially, $tg$ $ \alpha = \frac{x}{y} $
If we decompose the vector in its horizontal and vertical component ($\vec{v_x}$, respectively $\vec{v_y}$), we can write $\lvert\vec{v_x}\rvert=v*cos(\alpha)$ and $\lvert \vec{v_y}\rvert=v*sin(\alpha)$ and, therefore, the equations of motion for the point, which are what you just presented, are: $$x(t) = x+ t*v_x = x + tv\cos a\text{, or }\cos \alpha$$
$$y(t) = y+ t*v_y = y + tv\sin a\text{, or }\sin \alpha$$