Minimizing the time needed to reach at a point with constraints

Question

Consider a situation of particle located at origin and velocity of $V_°$ along y-axis . . The maximum possible acceleration magnitude which can be attained by that particle is $a_°$ for any moment of time . Now if the particle wants to reach a point $(d,0)$ in shortest possible time , what shape of the trajectory needs to be so as to make that possible ?

My progress : i let the acceleration along x -axis to be $a_x(t)$ and along y-axis to be $a_y(t)$ now $V_°- \int_{0}^{t}a_y(t)dt = v_y$ ..(i) , $\int_{0}^{t}a_x(t)dt =v_x$ ..(ii) and $\sqrt(a_y(t)^2 + a_x(t)^2)$ = $a_°$..(iii) . Now to minimize time i think we need to maximize the average velocity along x -axis during the whole motion . But how do we so i am stuck now as such i think my problem is variables are more and i have less equations .

• Note : Answer is Parabolic trajectory if someone wanted to know .

Answer 1

Consider the situation in the inertial frame of reference where you're initially at the origin with zero velocity. (Note that the acceleration constraint is the same in this moving frame.) The destination point is initially at $(d,0)$, and after time $t$ it will be at $(d,-V_0 t)$. On the other hand, the farthest you can get from the origin in time $t$ is $\frac{1}{2}a_0 t^2$, achievable through maximum acceleration in any fixed direction. So the minimum time is just the minimum positive solution to $\frac{1}{2}a_0t^2=\sqrt{d^2 + V_0^2 t^2}$. Squaring both sides, this becomes a quadratic in $t^2$: $\frac{1}{4}a_0^2 t^4 - V_0^2 t^2 - d^2 = 0$, with solutions $$ t^2 = \frac{V_0^2 \pm \sqrt{V_0^4 + a_0^2 d^2}}{a_0^2/2}. $$ If $d\neq 0$, then only the $+$ sign leads to a solution for $t$: $$ t^{*}=\sqrt{2\left(\frac{V_0}{a_0}\right)^2 + 2\sqrt{\left(\frac{V_0}{a_0}\right)^4 + \left(\frac{d}{a_0}\right)^2}} = \sqrt{2\alpha^2+2\sqrt{\alpha^4 + \beta^4}}=2\alpha\sqrt{\frac{1}{2} + \frac{1}{2}\sqrt{1 + (\beta/\alpha)^4}} $$ where $\alpha=V_0/a_0$ and $\beta=\sqrt{d/a_0}$ are the two parameters in the problem with dimensions of time. And, of course, constant acceleration leads you along a parabolic trajectory in the original frame.

Answer 2

I can't prove it optimal, but it won't be far wrong to use a constant acceleration. Compute the time from your starting conditions to $y=0$ as a function of $a_y$ and the time to $x=d$ as a function of $a_x$. Set those equal and solve the pair simultaneously with $\sqrt{a_x^2+a_y^2}=a$. You may have to do it numerically. Let $\theta$ be the angle of the acceleration vector and use a root finder.

Answer 3

Assumption: Accelaration vector is constant.

Under the assumption above, $\vec{a}=<a_x,-a_y> $ and so we take $a_x^2+a_y^2=a_0^2$ for minumum time. Then, the following equations must be satisfied:

1. $\frac{1}{2}a_xt_{min}^2=d$
2. $-\frac{1}{2}a_yt_{min}^2+v_0t_{min}=0$

If we solve the equations, since $x=x(t)=\frac{1}{2}a_xt^2$ and $y=y(t)=-\frac{1}{2}a_yt^2+v_0t$, we find the parabola $$y=-\frac{\sqrt{a_0^2-a_x^2}}{a_x}x+\frac{\sqrt{2}v_0}{\sqrt{a_x}}\sqrt{x} \;\;\;\text{where}\;\;\; a_x=\frac{\sqrt{v_0^4+d^2a_0^2}-v_0^2}{d}.$$ But, I can't imagine a solution without the assumption. It must involve high level of calculus of variations.