2D Pathfinding with Momentum
MATLAB
Try it online
function tf=Intercept(p,a)
X=@(t,a,v,x).5*a*t.^2+v.*t+x;
t=0;
while 1
    
    a1=(0:0.01:1)';
    a2=1-a1;
    theta2=-360:360;
    
    x1=X(t,a(1),0,p(1));
    y1=X(t,a(2),0,p(2));
    theta1=atan(y1/x1)*180/pi;
    T=(t-abs(theta1)/180-abs(theta2-theta1)./180);
    T(T&lt;0)=0;
    x2=0.5*a1.^2*cosd(theta1).*T.^2+0.5*a2.^2*cosd(theta2).*T.^2+a1.*a2*cosd(theta1).*T.^2+a1.*cosd(theta1)*T.*abs(theta2-theta1)/180;
    y2=0.5*a1.^2*sind(theta1).*T.^2+0.5*a2.^2*sind(theta2).*T.^2+a1.*a2*sind(theta1).*T.^2+a1.*sind(theta1)*T.*abs(theta2-theta1)/180;
    x=x1-x2;
    y=y1-y2;
    vx=a(1)*t-(a1*cosd(theta1)*T+a2*cosd(theta2).*T);
    vy=a(2)*t-(a1*sind(theta1)*T+a2*sind(theta2).*T);
    V=sqrt(vx.^2+vy.^2);
    D=sqrt(x.^2+y.^2);
    
    [ida1,idtheta2]=find(D&lt;1&amp;V&lt;1,1);
    
    if ~isempty(idtheta2)
        a1=a1(ida1);
        theta2=theta2(idtheta2);
        break
    end
    
    t=t+0.01;
    
end
if length(theta2) &gt; 1
    return
end
tf=t;
t1=abs(theta1/180);
t3=abs(theta2-theta1)/180;
t3=t3-0*(t3&gt;1);
t2=a1.*(tf-t1-t3);
t4=tf-t3-t2-t1;
x1=a(1)/2*tf^2+p(1);
y1=a(2)/2*tf^2+p(2);
v1=a*tf;
x2=cosd(theta1)/2.*t2.^2+cosd(theta1).*t2.*(t3+t4)+cosd(theta2)./2.*t4.^2;
y2=sind(theta1)/2.*t2.^2+sind(theta1).*t2.*(t3+t4)+sind(theta2)./2.*t4.^2;
v2=[cosd(theta1)*t2+cosd(theta2)*t4;sind(theta1)*t2+sind(theta2)*t4];
V=vecnorm(v1'-v2);
P=vecnorm([x1;y1]-[x2;y2]);
cmds=[theta1,t1,t2,theta2-theta1,t3,t4,tf];
fprintf('Turn %.2f degrees in %.2f seconds\nAccelerate for %.2f seconds\nTurn %.2f degrees in %.2f seconds\nAccelerate for %.2f seconds\nMission Time: %.2f seconds\n\n',cmds); 
fprintf('x1=%.2f, x2=%.2f\ny1=%.2f, y2=%.2f\nv1=[%.2f %.2f], v2=[%.2f %.2f]\nV=%.2f\nD=%.2f\n',x1,x2,y1,y2,v1,v2,V,P);

Explanation:
My methodology is to start by assuming the total mission length \$t\$ starting from 0 (in the case where we already fit the parameters at \$t\$=0) and apply a standard mission profile. The mission has 4 stages:

turn towards where the target will be at time \$t\$
accelerate towards the target
turn to an angle that will cancel velocities so that we are within parameters by time \$t\$
accelerate for the remaining time

Using this profile, there are 4 time values to be concerned about. \$t_1\$ is the first turn, \$t_2\$ is the first burn, \$t_3\$ is the second turn, and \$t_4\$ is the last burn. \$t_1\$ is known since the initial angle can be easily calculated. \$t_2\$ and \$t_4\$ are calculated as percentages \$a_1\$ and \$a_2\$ of \$(t_2+t_4)\$, in order to reduce them to \$a_1(t_2+t_4)\$ and \$(1-a_1)(t_2+t_4)\$, thus making both values a function of \$a_1\$. \$t_3\$ is a function of the second turn angle \$\theta_2\$. The next thing is to calculate the x and y positions and velocities at time \$t\$ as a function of \$a_1\$ and the second angle \$\theta_2\$, the only other unknown. I know that these are limited to a specific range of 0:1 and -360:360  respectively, so I calculate all possible positions as a function of \$a_1\$ and \$\theta_2\$ and find the first one that fits the criteria. If none fit, then \$t\$ is increased by 0.01 and the process is repeated. This method takes advantage of the vectorized computing capabilities of MATLAB. And honestly, it's probably the first code golf challenge where MATLAB is more ideal.
I also have some test code that has all of the test cases already, and times it as well according to your scoring criteria. Since the output of the program is the total mission time, and tic/toc calculates time since tic was called at time toc, this calculates the score as well.
% % Test Intercept*
*edited so that toc comes after the function call, so that time is calculated after the function finishes and not before. Oversight on my part.
tic
(Intercept([10,0], [0,0])*toc)^2;

tic
(Intercept([6,-3], [0,0.5])*toc)^2

tic
(Intercept([-2,8], [0.1,-0.2])*toc)^2

tic
(Intercept([5,5], [0.3,0.3])*toc)^2

tic
(Intercept([15,18], [0.1, 0.3])*toc)^2

tic
(Intercept([20,5], [0.2,-0.2])*toc)^2