How to find velocity and position of this ODE
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Hello all.
I've been trying to find the velocity and position of an object with an accelration rate of a(x)=-2x+1 m/s^2 with the initial conditions of v(0) = 4 m/s, and x(0) = 5 m, all at a time of 5 seconds.. I did my code on it, and believed to have found position with my current plot, however I can't seem to figure out how to calculate and plot my x versus t graph for velocity. Can someone help me out please?
Code below:
[T,s]=ode45(@eom,linspace(0,5,101),[5,4]);
plot(T,s(:,l),'-k')
xlabel('Time, sec')
ylabel('x,m')
function ds=eom(T,s)
ds(1,1)=s(2)
ds(2,1)=-2*s(1)+1
end
%Above function was for position, which seems to work.
%Below was my attempt at the velocity plot.
function ds=eom(T,s)
ds(1,1)=s(2)
ds(2,1)=-2*s(2)^2+s(1)_
end
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Risposte (1)
David Goodmanson
il 25 Gen 2019
Hi Peter,
since you have defined s(2) = ds(1)/dt, s(2) is the velocity function already. No more work required, exceptto plot it,
[T,s]=ode45(@eom,linspace(0,5,101),[5,4]);
plot(T,s)
xlabel('Time, sec')
legend('x, m','v ,m/s','location','southeast')
function ds=eom(T,s)
ds(1,1)=s(2)
ds(2,1)=-2*s(1)+1
end
Here the plot function plots out the colums of s independently, so you get x and v.
1 Commento
madhan ravi
il 25 Gen 2019
ode45 with no output arguments plots by default thereby avoiding the plot call.
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