Cumulative distance vs time graph using a velocity vs time graph

Hi, i have been given a formula for a velocity and a time interval and i have been tasked with producing a velocity vs time graph , a cumulative distance vs time graph as well as total distance travelled. I have currently got the velocity vs time graph working and im trying to plot a cumulative distance vs time graph by means of intergration, but i can only manage to get the total distance. I am unsure on how to go about this problem. any help would be appreciated.

Thank you

clc;
clear;
close;
hold on;
grid on;
t=0:60;
a=0;
b=60;
n=200;
area=zeros(1,60);

speed=-0.073*(t.^2)+6.1802*(t);

plot(t,speed);




dx=(b-a)/n;
x=a:dx:b;
H=zeros(1,n);
for i=1:n
    fxmiddle=-0.073*((x(i)^2+x(i+1)^2))/2+6.1803*((x(i)+x(i+1))/2);
    fxLeft=-0.073*(x(i)^2)+6.1802*(x(i));
    fxright=-0.073*(x(i+1)^2)+6.1802*(x(i+1));
    H(i)=(fxLeft+4*fxmiddle+fxright)/6;
end
%simpsons rule
integ=0;
for i=1:n
    A=H(i)*dx;
    integ=integ+A;
    

end

plot(t,integ);

syms x f(x)

f(x)=-0.073*(x^2)+6.1802*(x);
truearea=int(f(x),x,0,60);

relative_err_frac=((integ-truearea)/truearea)*100;

relative_err=double(relative_err_frac);

if relative_err>0.06

    print('ERROR, RELATIVE ERROR TOO HIGH, PLEASE RE INPUT VALUES');

end

Answer 1

[Les Beckham]

You need to save the value of integ on each step of the integration.

See changes below marked with % <<<

clc;
clear;
close;
t=0:60;
a=0;
b=60;
n=200;
area=zeros(1,60);

speed=-0.073*(t.^2)+6.1802*(t);

plot(t,speed);
grid on;

dx=(b-a)/n;
x=linspace(a, b, n); % <<< make x the same size and H (and thus integ)
H=zeros(1,n);

for i=1:n-1 % <<<
    fxmiddle=-0.073*((x(i)^2+x(i+1)^2))/2+6.1803*((x(i)+x(i+1))/2);
    fxLeft=-0.073*(x(i)^2)+6.1802*(x(i));
    fxright=-0.073*(x(i+1)^2)+6.1802*(x(i+1));
    H(i)=(fxLeft+4*fxmiddle+fxright)/6;
end

%simpsons rule
integ=zeros(size(H)); % <<< Preallocate integ
for i=2:n
    A=H(i)*dx;
    integ(i)=integ(i-1) + A; % <<< Save each sample of the integration
end
plot(x,integ); % <<< use x here instead of t (I'm not sure why you just didn't make b and n the same)
grid on

syms x f(x)

f(x)=-0.073*(x^2)+6.1802*(x);
truearea=int(f(x),x,0,60);

relative_err_frac=((integ-truearea)/truearea)*100;

relative_err=double(relative_err_frac);

if relative_err>0.06
    print('ERROR, RELATIVE ERROR TOO HIGH, PLEASE RE INPUT VALUES');
end

[Takura Nyatsuro]

That makes sense , thanks for the help :)

[Les Beckham]

You are quite welcome.