# Given a speed and angle, find a point at time T.

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: $$x(t) = x+ t*v_x = x + tv\cos a\text{, or }\cos \alpha$$
$$y(t) = y+ t*v_y = y + tv\sin a\text{, or }\sin \alpha$$