Calculate maximum possible velocity, given D, V0, A, J

Question

I'm writing a program that controls space ship. I needed to determine total travel time for the following scenario:

• ship has some initial velocity V0
• ship needed to travel distance D with maximum possible velocity, but have zero V at the end of the travel
• so, ship accelerates up to Vmax and then decelerates to zero V
• ship accelerates with acceleration Aa = 0, with constant jerk Ja up to Aamax, up to Vmax
• then it immediately starts to decelerate using other group of thrusters
• thrusters used for acceleration cut off instantly
• ship starts to decelerate with Ad = 0, with constant jerk Jd up to Admax, down to zero velocity
• as for accel thrusters, decel thrusters cut off instantly too

Control program needed to calculate total travel time or Vmax.

Thanks in advance

Edit: clarification

• Ship needed to travel distance D with minimum time possible - it accelerates up to some speed Vmax (which is unknown) then it decelerates. At the end of the travel speed must be zero
• Given: distance D, initial velocity V0 (can be zero), max acceleration Aamax, acceleration jerk Ja, max deceleration Admax, deceleration jerk Jd
• There is no connection between Vmax and maximum acceleration
• Both acceleration and deceleration work in the same manner: firstly, it accelerates under jerk for Tj = Amax / J with A = Jt, then it accelerates for Ta with constant A = Amax
• Total time traveled: T = T1j + T1a + T2j + T2a, T1j = Aamax / Ja, T1a > 0, T2j = Admax / Jd, T2a > 0

Sorry for inconvenience

Answer 1

We have the following 2 equations $$ V_{max}-V_0=\int_0^{T_1} a_1(t)dt \\ V_{max}=\int_0^{T_2} a_2(t)dt $$ where $a_1(t)=\min(J_at,a_{a,max})$, $a_2(t)=\min(J_d t,a_{d,max})$ So the time of achieving maximum acceleration is $a_{max}/J_a$ and we have $$ V_{max}-V_0=\int_0^{a_{a,max}/J_a}J_a tdt +\int_{a_{a,max}/J_a}^{T_1} a_{a,max}dt $$ Thus $$ V_{max}-V_0=1/2 a_{a,max}^2/J_a +a_{a,max}(T_1-a_{a,max}/J_a)=a_{a,max} T_1-1/2 a_{a,max}^2/J_a $$ and finally $$ T_1=(V_{max}-V_0)/a_{a,max}+1/2 a_{a,max}/J_a $$ Similarly $$ V_{max}=\int_0^{a_{d,max}/J_d}J_d tdt +\int_{a_{d,max}/J_d}^{T_2} a_{d,max}dt $$ and $$ T_2=V_{max}/a_{d,max}+1/2 a_{d,max}/J_d $$ The final $T=T_1+T_2$

However, some condition must hold for the above integrals to be true: $$ V_{max}-V_0\ge 1/2 a_{a,max}^2/J_a \\ V_{max}\ge 1/2 a_{d,max}^2/J_d $$

Otherwsie the ship will achieve maximum velocity (or stopped) before it will get maximum acceleration.

EDIT. After clarification I have to proceed the solution. $V_{max}$ is unknown but The full distance $D$ is known so $$ D=D_1+D_2\\ D_1=\int_0^{T_1}V_1(t)dt \\ D_2=\int_0^{T_2} V_2(t)dt $$ where $$ V_1(t)=V_0+\int_0^t a_1(t) dt, ~~ V_1(T_1)=V_{max} \\ V_2(t)=V_{max}-\int_0^t a_2(t) dt, ~~V_2(T_2)=0; $$ So $$ V_1(t)=V_0 +J_a t^2/2 [t< a_{a,max}/J_a]+a_{a,max}(t-a_{a,max}/J_a)(t \ge a_{a,max}/J_a)\\ V_2(t)=V_{max}-J_d t^2/2[t<a_{d,max}/J_d]-a_{d,max}(t-a_{d,max}/J_d)(t \ge a_{d,max}/J_d)\\ $$ So $$ D_1=V_0T_1+a_{a,max}^3/6/J_a^2-a_{a,max}^3/2/J_a^2+a_{a_max}T_1^2/2\\ D_2=V_{max}T_2-a_{d,max}^3/6/J_d^2+a_{d,max}^3/2/J_d^2-a_{d_max}T_2^2/2\\ $$ Since $T_1=(V_{max}-V_0)/a_{a,max}+1/2 a_{a,max}/J_a$ and $T_2=V_{max}/a_{d,max}+1/2 a_{d,max}/J_d$ we can plug this in the above expressions. Finally you will get equation for $V_{max}$ and from there you can get $T_1$ and $T_2$.I hope you can finish this unless you can find an easier solution.