There are two objects on a plane: a stationary object P1 and a moving object P2 moving at a known costant velocity vector v2. I also know the relative position vector d.
I want to calculate the direction object P1 should start accelerating at a given constant acceleration a to collide with object P2.
A triangle will be formed where vertex C is the collision point and t being the time when the objects will collide. I can't figure out how to get the angle alpha knowing d, a, v2 and beta.
I could only solve this if P1 should start moving immediately at a constant velocity v1 by dropping a vertical v from point C straight down to side d and using the ratio of the two remaining sides. v2 * t / (v1 * t) = v2 / v1 = sin(beta) / sin(alpha)
In the case of linear acceleration of object P1, I cannot use this approach because t doesn't cancel out.
I would appreciate any help.
The main problem is that there are a lot of terms that don't interact nicely, so we want to try to eliminate as many of those as possible. To do so, I'm going to simplify the problem by assuming that $P_2$ starts at the origin and moves to the right with velocity $v$. Then its equation of motion is given by $P_2(t) = v t\hat\imath$, where $\hat\imath = (1, 0)$ is the unit vector in the positive $x$ direction.
The equation of motion for $P_1$ is then $P_1(t) = \mathbf{d} + \frac{1}{2} \mathbf{a} t^2$, where $\mathbf{d}$ is the initial position vector of $P_1$, and $\mathbf{a}$ is the acceleration vector whose direction we are trying to determine.
At some time $t$, we have the two particles colliding, which we can describe by equating the two equations of motion:
$$\begin{eqnarray} P_1(t) & = & P_2(t) \\ \mathbf{d} + \frac{1}{2} \mathbf{a} t^2 & = & v t \hat\imath \\ \frac{1}{2} \mathbf{a} t^2 - v t \hat\imath + \mathbf{d} & = & 0 \end{eqnarray}$$
This is a quadratic equation in $t$ - in fact, it's two quadratics, one in the $x$ direction and one in the $y$. Because of our transformations, the $y$ direction is the easier one:
$$\frac{1}{2} a_y t^2 + d_y = 0 \implies t = \sqrt{\frac{-2 d_y}{a_y}}$$
We only need to take the positive square root since we don't want our collision to happen in the past. Also, notice that for this to make any sense we need $d_y$ and $a_y$ to have opposite signs, which makes sense - if $P_1$ starts higher than $P_2$ it needs to accelerate downwards, and vice versa.
We can then substitute this into the equation for the $x$ direction, giving us an equation relating our other two unknowns $a_x$ and $a_y$:
$$\begin{eqnarray} \frac{1}{2} a_x t^2 - v t + d_x & = & 0 \\ \frac{1}{2} a_x \left( \frac{-2 d_y}{a_y} \right) - v \sqrt{\frac{-2 d_y}{a_y}} + d_x & = & 0 \\ - \frac{a_x}{a_y} d_y - v \sqrt{\frac{-2 d_y}{a_y}} + d_x & = & 0 \end{eqnarray}$$
This is all still pretty gnarly, but we have one last piece of information: we know that magnitude of $\mathbf{a}$ is the fixed value $a$, so we can substitute $a_x = a \cos \theta$ and $a_y = a \sin \theta$, where $\theta$ is the angle representing the direction that the particle is accelerating in. Using that, we can rewrite and rearrange:
$$\begin{eqnarray} - \frac{a_x}{a_y} d_y - v \sqrt{\frac{-2 d_y}{a_y}} + d_x & = & 0 \\ - \frac{a \cos \theta}{a \sin \theta} d_y - v \sqrt{\frac{-2 d_y}{a \sin \theta}} + d_x & = & 0 \\ v \sqrt{\frac{-2 d_y}{a \sin \theta}} & = & d_x - d_y \cot \theta \\ \frac{-2 v^2 d_y}{a} \csc \theta & = & d_x^2 - 2 d_x d_y \cot \theta + d_y^2 \cot^2 \theta \end{eqnarray}$$
Now to solve this, we need to substitute $\csc \theta = \pm \sqrt{1 + \cot^2 \theta}$, which means that this is actually a quartic polynomial in $\cot \theta$. It's not even a particularly nice one, so it almost certainly doesn't have a general solution that can be written out easily. But it exists, and you can find a numerical approximation for its solutions given values of the various quantities.