# What is the optimal direction of constant acceleration for a moving particle to reach a target point in 2D?

A particle in the plane has initial position $$P$$ and initial velocity $$V$$. It can accelerate at exactly rate $$A$$ in any direction $$\theta$$. What path describes the time-optimal trajectory to a target point, $$G$$?

My intuition leads me to believe there is a single value of $$\theta$$ that if the particle accelerates in that direction, will lead it on the minimal-time trajectory to $$G$$. If that intuition is true, the optimal value for $$\theta$$ can be formulated by solving the system of equations:

$$G_x = P_x + V_x t + \frac{cos(\theta) A t^2}{2}$$

$$G_y = P_y + V_y t + \frac{sin(\theta) A t^2}{2}$$

When I try to solve these on a whiteboard, I quickly wind up in a mess of quartic craziness, and I suspect some vector notation could help clarify the situation.

Is my intuition true regarding the time-optimal path? And, if so, what is the solution for $$\theta$$ in terms of the initial conditions?

• Here's a tutorial and reference for typesetting math on this site. Mar 31, 2020 at 4:17
• The fact that this is in two dimensions isn't mentioned in the body of the question. Mar 31, 2020 at 4:18
• Shouldn't the rate of acceleration $A$ appear in the quadratic terms in the equations? Mar 31, 2020 at 4:24

I’ll write the equations as

$$\vec s+t\vec v+\frac12at^2\vec n_\theta=0\;,$$

where $$\vec s=\vec p-\vec g$$ and $$\vec n_\theta=(\cos\theta,\sin\theta)$$ is the unit vector at angle $$\theta$$ with the positive $$x$$ axis.

This isn’t really an optimization problem. There are only discrete values of $$\theta$$ for which the target is reached at all. To find them, multiply the equation with $$\vec u_\theta=\vec n_{\theta+\frac\pi2}$$, a direction vector orthogonal to $$\vec n_\theta$$:

$$\vec s\cdot\vec u_\theta+t\vec v\cdot\vec u_\theta=0$$

and thus

$$t=-\frac{\vec s\cdot\vec u_\theta}{\vec v\cdot\vec u_\theta}\;.$$

Multiplying the original equation by $$\vec n_\theta$$, substituting $$t$$ from above and multiplying by $$\vec v\cdot\vec u_\theta$$ yields

$$\left(\vec v\cdot\vec u_\theta\right)\left(\vec s\cdot\vec n_\theta\right)-\left(\vec v\cdot\vec n_\theta\right)\left(\vec s\cdot\vec u_\theta\right)+\frac12a\frac{\left(\vec s\cdot\vec u_\theta\right)^2}{\vec v\cdot\vec u_\theta}=0\;.$$

If the first two terms are written out in terms of the trigonometric functions, half the terms cancel and the other half can be combined using $$\cos^2+\sin^2=1$$. Then multiplying by $$\vec v\cdot\vec u_\theta$$ again yields

$$(v_ys_x-v_xs_y)\left(\vec v\cdot\vec u_\theta\right)+\frac12a\left(\vec s\cdot\vec u_\theta\right)^2=0\;.$$

This is a trigonometric equation in $$\theta$$. You can turn it into a quartic algebraic equation either by writing it as an equation in $$\mathrm e^{\mathrm i\theta}$$ or by using $$\sin^2=1-\cos^2$$, bringing the terms proportional to $$\sin\theta$$ to one side, squaring and using $$\sin^2=1-\cos^2$$ again to obtain a quartic equation in $$\cos\theta$$. (The squaring introduces spurious solutions that you need to discard.) The solutions for which $$t$$ (as found above) is positive are the only values of $$\theta$$ for which the target is reached at all. If there is more than one such solution, the optimization consists merely in comparing the corresponding $$t$$ values. I suspect there will always be one solution for positive $$t$$ and one for negative $$t$$, but I’m not sure about that.

• Thank you for simplifying the math with the vector notation. The optimization part of the problem, I think, involves proving that the path followed by picking the single value of $\theta$ that hits the target is the time-optimal path. In other words, is there a shorter path that involves changing the direction of acceleration while traveling? If not, why? Intuitively, I believe the single-acceleration-vector solution is time-optimal, but I don't know how to prove it. Mar 31, 2020 at 16:21