writing second degree ODE

Hello everyone,

I'm pretty new to matlab and writing ODE codes. I just finished my first grader project, and i want to ask a question about ODE writing.

Here is the code that was not work well;

function dYdt = racecarfun(t,Y);
    Y1=Y(1);
    Y2 = Y(2);
    Y3=Y(3);
    dY1dt=Y2;
    dY2dt=Y3;
    F = 4000.0;            % force on car,     [N]
    m = 1300.0;            % mass of car,      [kg]
    A = 3.0;               % frontal area,     [m^2]
    Cd = 0.6;              % drag coefficient, [-]
    rho = 1.23; 
    dY3dt=((F-(Cd*rho*A*0.5*Y2.^2))/m);
    dYdt=[dY1dt;dY2dt;dY3dt];
end

and here is the code i wrote after a quick search;

function dydt = racecarfun(t,y);
    dydt=zeros(2,1);
    dydt(1)=y(2);
    F = 4000.0;            % force on car,     [N]
    m = 1300.0;            % mass of car,      [kg]
    A = 3.0;               % frontal area,     [m^2]
    Cd = 0.6;              % drag coefficient, [-]
    rho = 1.23; 
    dydt(2)=((F-(Cd*rho*A*0.5*y(2).^2))/m);
end

In the first one, i wanted to write all x,v and a in formula however, the example goes with only v and a in the second one. I wonder what is the problem with the first code?

Case Study: Modeling Race Car Motion

Answer 1

[Jan]

In your first code, (F-(Cd*rho*A*0.5*Y2.^2))/m is the 3rd derivative of the position. From a physical point of view, this value is surely the acceleration, so it must be the 2nd derivative of the position. Therefore the 2nd code has a physical meaning.

[Rahan Öztürk]

i am not sure i understand it clearly.

Y1=Y(1); ---> i wanted Y(1) is my x,

Y2 = Y(2);

Y3=Y(3);

dY1dt=Y2; ---> so, dxdt can be v,

dY2dt=Y3 ---> and dvdt is the a

what i am thinking wrong?

[Jan]

Your function dYdt = racecarfun(t,Y) calculate the derivative to the point [t,Y]. The position is not a part of the derivative, but only the velocity and the acceleration. So you are right partially: dY1dt=Y(2) is the velocity, dY2dt=Y(3) is the accleration, but then dY3dt is the "jerk", the 3rd derivative. But you provide a formula according to a = F/m, so this is the acceleration, the 2nd derivative, not the 3rd one.

If you have a equation for the 3rd derivative, your code would work again. But this equation cannot be a force devided by the mass.

[Rahan Öztürk]

So, Y(1) is already first derivative of x. Therefore i need Y(1) for "v" and dY1dt=Y(2) for "a".

Thank your for your time Jan, have a good day.

[Torsten]

No. Y1 is X, dY1/dt = Y2 is v and dY2/dt is a.