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Finding the Velocity at Different intervals

Asked by Kudzanai Sekerere on 24 Aug 2019 at 16:17
Latest activity Commented on by Star Strider
on 26 Aug 2019 at 12:38
tryng to write a code to display the velocity and displacement given a time interval and acceleration at those intervals using integration. getting an error in my for loop what might be the reason
clear all
close all
% Acceleration Vector
a = [0 2 4 7 11 17 24 32 41 48 51];
% define time interval
k =1;
%define time vector
t = 0:k:10;
%initialize velocity
for ii = 1:11
v(ii) = trapz(a(1:ii),k);
end
% calculate the displacement d at each time interval
for jj = 1:11
d(jj) = trapz(v(1:jj),k);
end
% display a table of the velocity values
v = table(t,v,'VariableNames',{'Time(Sec)' 'Velocity(m/s)'})
% display a table of the displacement values
d = table(t',d','VariableNames',{'Time(Sec)' 'Displacement(m)'})

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2 Answers

Answer by Star Strider
on 24 Aug 2019 at 16:42
 Accepted Answer

You can solve the problem of varying numbers of elements in each iteration by using cell arrays for ‘v’ and ‘d’:
for ii = 1:11
v{ii} = trapz(a(1:ii),k);
end
% calculate the displacement d at each time interval
for jj = 1:11
d{jj} = trapz(v{jj},k);
end
The best solution would likely be to use the cumtrapz function instead, and avoid the loops entirely.

  10 Comments

The code works yes but it is producing the wrong results after the second iteration. look at the two tables I sent above. on time = 2secs the value I am getting for displacement is 5 and not 3. I already have the answer to the question but this code is not producing the correct response. so not sure what needs to be editted inorder for it to work correctly.
@Star Strider thank you for your assistance finally noticed that when using the function trapz or cumtrapz interchanging what starts within the bracket affects the results it is (t,v) and not (v,t) and (t,a) not (a,t) that was my main error
Star Strider
on 25 Aug 2019 at 12:02
As always, my pleasure.
I posted the correct code, and didn’t take a close look at yours. I just ran it to see if there were problems, and it ran without error.

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Answer by Kudzanai Sekerere on 25 Aug 2019 at 7:47

After abit of research have come up with these 3 codes that produce the desired results
%Matlab Code (Third Variation)
clear all
close all
% Acceleration Vector
a = [0 2 4 7 11 17 24 32 41 48 51];
%define time vector
t = 0:10;
%initialize velocity
v(1)=0;
for ii = 1:10
v(ii+1) = trapz(t(ii:ii+1),a(ii:ii+1))+v(ii);
end
% calculate the displacement d at each time interval
d(1)= 0;
for jj = 1:10
d(jj+1) = trapz(t(jj:jj+1),v(jj:jj+1))+d(jj);
end
% display a table of the velocity values
tv = table(t(:),v(:),'VariableNames',{'Time_sec' 'Velocity_mps'})
% display a table of the displacement values
td = table(t(:),d(:),'VariableNames',{'Time' 'Displacement'})
%plot graph of the change of the acceleration with time
plot(t,a)
title('Acceleration against time Graph')
xlabel('Time(Sec)')
ylabel('Acceleration(m/s^2)')
%plot graph of the change of the velocity with time
figure(2)
plot(t,v)
title('Velocity against time Graph')
xlabel('Time(Sec)')
ylabel('Velocity(m/s)')
%plot graph of the change of the displacement with time
figure(3)
plot(t,d)
title('Displacement against time Graph')
xlabel('Time(Sec)')
ylabel('Displacement(m)')
%Matlab Code (Second Variation)
clear all
close all
% Acceleration Vector
a = [0 2 4 7 11 17 24 32 41 48 51];
%define time vector
t = 0:10;
%initialize velocity
for ii = 2:11
v(ii) = trapz(t(1:ii),a(1:ii));
end
% calculate the displacement d at each time interval
for jj = 2:11
d(jj) = trapz(t(1:jj),v(1:jj));
end
% display a table of the velocity values
tv = table(t(:),v(:),'VariableNames',{'Time_sec' 'Velocity_mps'})
% display a table of the displacement values
td = table(t(:),d(:),'VariableNames',{'Time' 'Displacement'})
%plot graph of the change of the acceleration with time
plot(t,a)
title('Acceleration against time Graph')
xlabel('Time(Sec)')
ylabel('Acceleration(m/s^2)')
%plot graph of the change of the velocity with time
figure(2)
plot(t,v)
title('Velocity against time Graph')
xlabel('Time(Sec)')
ylabel('Velocity(m/s)')
%plot graph of the change of the displacement with time
figure(3)
plot(t,d)
title('Displacement against time Graph')
xlabel('Time(Sec)')
ylabel('Displacement(m)')
%Matlab Code (Third Variation)
clear all
close all
clc
% initialize time interval
k = 1;
% define time vector
t = 0:k:10;
% define acceleration vector
a = [0,2,4,7,11,17,24,32,41,48,51];
% calculate the velocity
v = cumtrapz(t,a);
d = cumtrapz(t,v);
% display a table of the velocity values
tv = table(t(:),v(:),'VariableNames',{'Time_Sec' 'Velocity_mps'})
% display a table of the displacement values
td = table(t(:),d(:),'VariableNames',{'Time_Sec' 'Displacement_m'})
%plot graph of the change of the acceleration with time
plot(t,a)
title('Acceleration against time Graph')
xlabel('Time(Sec)')
ylabel('Acceleration(m/s^2)')
%plot graph of the change of the velocity with time
figure(2)
plot(t,v)
title('Velocity against time Graph')
xlabel('Time(Sec)')
ylabel('Velocity(m/s)')
%plot graph of the change of the displacement with time
figure(3)
plot(t,d)
title('Displacement against time Graph')
xlabel('Time(Sec)')
ylabel('Displacement(m)')

  3 Comments

Star Strider
on 25 Aug 2019 at 12:06
I don’t understand the need for a loop with trapz, since (if I understand what you are doing), cumtrapz should do what you want, without any loops.
Yes cumtrapz works just fine thanx. and even a loop works. just tried both out and ultimately it is the same result
Star Strider
on 26 Aug 2019 at 12:38
The cumtrapz function is more efficient. I would use it and eliminate the loops.

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