HOw to find displacement integrating velocity?

Hi, i need to find the displacement of a moving object x(t) knowing its velocity as a function of the displacement itself v(x). The function i have used is the following

I imposed that dx / dt = v(x) and then I tryed to use ODE 45 to find x(t) but I always get a vector full of zeros but i can't understand where the mistake is. This is the code that I wrote:

%% data
Po=1.5; % [bar]  absolute feeding pressure
P1=1.01315; % [bar]  absolute atmosferic pressure

rho=2700; %[kg/m^3] projectile density
D=0.048; % [m] projectile diameter
Length=0.10; % [m] projectile length
A=(pi/4)*D.^2;  % [m^2] projectile face area
m=rho*A*Length; % [kg] projectile mass

L=0.90; %[m] useful barrell length
L_d=0.4; %[m] decelleration barrel length
Vol=0.0095;  %[m^3] chamber volume
gamma=1.4; % air constant

% friction forces with 0.47 friction coefficient between aluminium and% steel
f=0.47*m*9.81; 

%% model
x_dot=@(t,x) sqrt((2./m).*((Po.*1e5.*Vol)./(gamma-1).*(1-(Vol./(A.*x+Vol)).^(gamma-1))-A.*x.*P1.*1e5-x.*f));
x0=0;
tspan=0:0.1:1;

[t,x]=ode45(x_dot,tspan,x0);
plot(t,x)

[Torsten]

Hint: See what you get for v if you insert x = 0 (your initial condition for x).

Interpretation: For x = 0, your projectile already hit the ground - and it will stay there for all coming centuries.

[Shivam Gothi]

hello @Riccardo,

can you please explain in brief what does this equation do? which system dynamics does it represents ?

I am getting some non-zero imaginery solution for velocity if I put "x0=1".

This is because, I found that the value of expression inside the sqrt() function is coming out to be negative (for positive "x", take x=1 as an example).

So sqrt() applied on negative number gives imaginary value. What does the imaginery "x_dot" represent here ?

[Riccardo]

hi,@Shivam Gothi

This equation should reperesent the velocity of a projectile being accelerated by an adiabatic expansion of gas.

Vol represent the volume of the chamber containing the gas, L the length of the barrel, m the mass of the projectiel and A the cross section of the barrel

Answer 1

[Torsten]

The equation from above describes the stationary velocity profile over the length L. Thus replacing v by dx/dt does not make sense in this case because there is no movement of x: x is the static position over the length L.

Or do you try to compute the position of a massless particle moving in x-direction over time ? Then use

%% data
Po=1.5; % [bar]  absolute feeding pressure
P1=1.01315; % [bar]  absolute atmosferic pressure
rho=2700; %[kg/m^3] projectile density
D=0.048; % [m] projectile diameter
Length=0.10; % [m] projectile length
A=(pi/4)*D.^2;  % [m^2] projectile face area
m=rho*A*Length; % [kg] projectile mass
L=0.90; %[m] useful barrell length
L_d=0.4; %[m] decelleration barrel length
Vol=0.0095;  %[m^3] chamber volume
gamma=1.4; % air constant
% friction forces with 0.47 friction coefficient between aluminium and% steel
f=0.47*m*9.81; 
%% model
x_dot=@(t,x) sqrt((2./m).*((Po.*1e5.*Vol)./(gamma-1).*(1-(Vol./(A.*x+Vol)).^(gamma-1))-A.*x.*P1.*1e5-x.*f));
x0=1e-8;
tspan=0:0.01:1;
options = odeset('Events',@(t,x)event(t,x,L));
[t,x]=ode45(x_dot,tspan,x0,options);
plot(x,t)
xlabel('Length')
ylabel('Time')
grid on

function [position,isterminal,direction] = event(t,x,L)
  position = x(1)-L; % The value that we want to be zero
  isterminal = 1;  % Halt integration 
  direction = 0;   % The zero can be approached from either direction
end

[Riccardo]

@Torsten Thank you very much this is exactly what i needed