MATLAB: How to create a vector

Hello everyone, I hope you are doing well. I need some help with my matlab project. I''m finding the velocity and time after each 1 meter displacement. It''s good when I''m doing that and I''m actually getting correct results. However, I couldn''t find the velocity and time after for example 1.5 or 1.6 meter. this is my program, so can you please help me with it.

function [v,t]=freefall(h)  
g=9.81;
%Gravity in m^2/s 
for k=1:h
    v(k)=sqrt(2*g*k);
end 
for
k=1:h
    t(k)=sqrt(2*k/g);
end 
plot(t,v)
xlabel(''Time
(s)'')
ylabel(''Velocity (m/s)'')
title(''Free fall: velocity vs time'')

Best Answer

Use the interp1 function.

If I understand your code correctly, this will work. Add these lines to the end of your code:

k = 1:h;                                                    % Displacement
dspl_intrp = [1.5; 1.6];                                    % Displacements To Interpolate
vt_intrp = interp1(k, [v' t'], [1.5; 1.6], 'linear');       % Interpolate Velocity & Time
fprintf(1, 'Displacement %.2f,  Velocity = %.2f,  Time = %.2f\n',  [dspl_intrp vt_intrp]')
Displacement 1.50,  Velocity = 5.35,  Time = 0.55
Displacement 1.60,  Velocity = 5.53,  Time = 0.56

I added the fprintf call to show that the result appears to be correct.

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MATLAB: My code is just about there, but there is something not right…I can’t figure it out. Help?!

I am pretty sure this is the correct code now. As for the sum(x>1000), the answer was 44. Is that the number of time steps it took for x to become greater than 1000? Here's the code:

m=1428.8;%mass of car
  a_i=3;%initial acceleration
  b=1.04;%constant
  MaxSpeed= 65 * .4474;%mph - m/s

  SpeedLimit= 50 * .4474;%mph - m/s
  dt=.5;%time step
  t=[0:dt:50];%time
  N=length(t);%legthg of time
  a= NaN(1,length(t));%empty acceleration vector
  v= NaN(1,length(t));%empty velocity vector
  x= NaN(1,length(t));%empty position vector
  a(1)=a_i;%initial acceleration = 3
  v(1)=0;%initial velocity = 0
  x(1)=0;%initial position =0 
  for k=2:N
      F_drag=.5*b*((v(k-1))^2);%drag force 
      a(k)=a(k-1)-(F_drag/m);%net acceleration 
      v(k)= v(k-1)+ ((a(k)+a(k-1))/2)*dt;%velocity 
      x(k)=x(k-1)+ dt*((v(k)+v(k-1)/2));%position 
      %a)
      if x(k)>150%when acceleration should equal zero 
          a(k)=0;%acceleration = 0
          v(k)=MaxSpeed;%velocity equals max speed
          x(k)=x(k-1)+ dt*((v(k)+v(k-1)/2));
      end
      %b)
      if x(k-1)>275%when you pass the 50mph speed limit sign
          a(k)=-(F_drag/m);%acceleraton from the car is zero and the drag force is decelerating the car
          v(k)= v(k-1)+ ((a(k)+a(k-1))/2)*dt;%new velocity
      end
  end
    %c)
    if x(k-1)>1000 && v(k)<=SpeedLimit %when you pass the police car
        disp('keep coasting');%if the velocity os less than or equal to the speed limit
        v(k);
        speed=v(k)/.4474;
        disp('speed in mph');
        speed%displays velocity right after you pass the police car
    else
        disp('pull over');%if velocity is greater than the speed limit
        v(k);
        speed=v(k)/.4474;
        speed
    end
    figure;%plots
    subplot(3,1, 1)
    plot(t, x)%time vs. position plot
    xlabel('time(s)');
    ylabel('position(m)');
    subplot(3,1, 2)
    plot(t, v)%time vs. velocity
    xlabel('time(s)');
    ylabel('velocity(m/s)');
    subplot(3,1, 3)
    plot(t, a)%time vs. acceleration plot
    xlabel('time(s)');
    ylabel('acceleration');

MATLAB: I need a help with the project!!

v is a vector. what do you expect v(1.1) to return? Where will the 1.1 element of that vector be found? (Between the 1st and 2nd elements, I assume, but that is not how MATLAB does indexing.)

Oh, and you spelled "height" incorrectly. :)

I'd also wonder why you pass in v and t into the function as arguments? You create both of those variables inside the function.

Instead, you should pass in h as an argument, then you would never need to use that silly input statement.

Finally, you return y as an argument from this function. But you never create y as a variable. Perhaps a better choice of arguments for the function would have been

function [v,t]=freefall(h)

now, you would use the function like this:

[v,t]=freefall(h);

and completely drop out that silly input statement.

Best Answer

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