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

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james carsley
james carsley 2019-7-25
编辑： Torsten 2019-7-26
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 个评论
Torsten
Torsten 2019-7-26
编辑：Torsten 2019-7-26
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.

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