How to get the integral of distance(mph) v time(s) graph in meters?

3 views (last 30 days)
I need to find the integral of the equation 'v' below in meters between the time 1 second and 10 seconds but 'v' is in mph.
v=(0.0011*t^6)-(0.02928*t^5)+(0.2807*t^4)-(1.1837*t^3)-(0.8283*t^2)+(41.234*t)-3.3549 % v is in mph
I have converted 'v' into m/s to get the graph. But to find the integral I have used a function of 'v' that is still in mph so the integral doesnt come out in meters. I have tried to convert 'v' within the function to m/s but doesnt work.
clc
clear
%intial conditions
a=1; %lower limit
b=10; %upper limit
n=9; %number of sub-intervals
r_error=0.1;%intial realtive error
i=0; %loop counter
I=0;%integral set to zero
n_separations=zeros(size(n));
t_distance=zeros(size(I));
%while loop to find new integral with a realtive error less than 0.0002%
while r_error>0.00002
h=(b-a)/n; %distnace of each separation
new_i=0; %intially set to zero
i=i+1; %increase counter by 1 every loop
for m=1:n
x_left=a+(i-1)*h;
x_right=a+(i*h);
f_left=function_Q1(x_left);
f_right=function_Q1(x_right);
new_i=((h/2)*(f_left+f_right)+I); %new integral
New_x=new_i; %new distance is equal to new integral
midpoint=(a+b)/2;
end
r_error=((new_i-I)/new_i)*100; %realtive error
n_separations(i)=n;
t_distance(i)=I;
n=n*2; %number of separtions doubled every loop
I=new_i; %new integral is stores as old integral for new loop
fprintf('Integral Value[Distance(m)]=%6.3f\n',I) %new integral value displayed to 3 decimal places
fprintf('Number of Separations=%6.0f\n',n) % displaying number of separations
fprintf('Relative Error=%6.5f\n',r_error) %displaying relative error to 5 decimal places
end
t=linspace(1,10,10);
v=zeros(size(t));
m=1609.344/3600;%conversion from mph to m/s
for i=1:numel(t)
v(i)=((0.0011*(t(i)^6))-(0.02928*(t(i)^05))+(0.2807*(t(i)^4))-(1.1837*(t(i)^3))-(0.8283*(t(i)^2))+(41.234*(t(i))-3.3549))*m;
hold on
end
plot(t,v,'r*-')
title('9 Seconds of Car Driving-Speed(m/s) Vs Time(s)')
xlabel('Time(s)')
ylabel('Speed(m/S)')
xlim([1 10])
Function used
function v=function_Q1(t)
v=((0.0011*t^6)-(0.02928*t^5)+(0.2807*t^4)-(1.1837*t^3)-(0.8283*t^2)+(41.243*t)-3.3549);
end
The desired integral is approx. 483m. Any help much appreciated.
  1 Comment
Torsten
Torsten on 26 Jul 2019
Edited: Torsten on 26 Jul 2019
Which unit has t in your equation for v ?
If t is in seconds,
fun = @(t) (0.0011*t^6-0.02928*t^5+0.2807*t^4-1.1837*t^3-0.8283*t^2+41.243*t-3.3549)*1609.344/3600;
distance = integral(fun,0,10)
should give you the distance in metres.

Sign in to comment.

Answers (0)

Categories

Find more on Polynomials in Help Center and File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!