1 Problem 4-1: Differentiation The following data is given for a train Time t (sec) Position s (m) 10 75 20 175 30 360 35 480 40 600 Determine the velocity and acceleration of the train as a function of time. What is the velocity and acceleration of the train at t = 32 sec. Use the Lagrange interpolation to fit an interpolating polynomial. Plot the polynomial fit for the position, then plot the velocity and acceleration as functions of time. Use calculus for the hand solutions. For the computational solution use forward difference for the velocity and central difference for the acceleration with a discrete increment of Ah = 0.0001.
I have made the MatLab code for this, but I don't know how to make the plots. I have provided the code below and just need assistance with the plotting.
% Problem 4-1 %
% A
x=[0,10,20,30,35,40];
y=[0,75,175,360,480,600];
% Using a second order Lagrange Polynomial for unequally spaced data
% Velocity for the starting point
back1=y(1)*((2*x(1)-x(2)-x(3))/((x(1)-x(2))*(x(1)-x(3))));
middle1=y(2)*((2*x(1)-x(1)-x(3))/((x(2)-x(1))*(x(2)-x(3))));
forward1=y(3)*((2*x(1)-x(1)-x(2))/((x(3)-x(1))*(x(3)-x(2))));
v1=back1+middle1+forward1;
% Velocity for the middle values
back2_5=y(1:4).*((2.*x(2:5)-x(2:5)-x(3:end))./((x(1:4)-x(2:5)).*(x(1:4)-x(3:end))));
middle2_5=y(2:5).*((2.*x(2:5)-x(1:4)-x(3:end))./((x(2:5)-x(1:4)).*(x(2:5)-x(3:end))));
forward2_5=y(3:end).*((2.*x(2:5)-x(1:4)-x(2:5))./((x(3:end)-x(1:4)).*(x(3:end)-x(2:5))));
v2_5=back2_5+middle2_5+forward2_5;
% Velocity for the end point
back6=y(4)*((2*x(6)-x(5)-x(6))/((x(4)-x(5))*(x(4)-x(6))));
middle6=y(5)*((2*x(6)-x(4)-x(6))/((x(5)-x(4))*(x(5)-x(6))));
forward6=y(6)*((2*x(6)-x(4)-x(5))/((x(6)-x(4))*(x(6)-x(5))));
v6=back6+middle6+forward6;
% Finding total velocity (m/s)
v=[v1,v2_5,v6]
% Acceleration for the starting point
a_back1=v(1)*((2*x(1)-x(2)-x(3))/((x(1)-x(2))*(x(1)-x(3))));
a_middle1=v(2)*((2*x(1)-x(1)-x(3))/((x(2)-x(1))*(x(2)-x(3))));
a_forward1=v(3)*((2*x(1)-x(1)-x(2))/((x(3)-x(1))*(x(3)-x(2))));
a1=a_back1+a_middle1+a_forward1;
% Acceleration for the middle values
a_back2_5=v(1:4).*((2.*x(2:5)-x(2:5)-x(3:end))./((x(1:4)-x(2:5)).*(x(1:4)-x(3:end))));
a_middle2_5=v(2:5).*((2.*x(2:5)-x(1:4)-x(3:end))./((x(2:5)-x(1:4)).*(x(2:5)-x(3:end))));
a_forward2_5=v(3:end).*((2.*x(2:5)-x(1:4)-x(2:5))./((x(3:end)-x(1:4)).*(x(3:end)-x(2:5))));
a2_5=a_back2_5+a_middle2_5+a_forward2_5;
% Acceleration for the end point
a_back6=v(4)*((2*x(6)-x(5)-x(6))/((x(4)-x(5))*(x(4)-x(6))));
a_middle6=v(5)*((2*x(6)-x(4)-x(6))/((x(5)-x(4))*(x(5)-x(6))));
a_forward6=v(6)*((2*x(6)-x(4)-x(5))/((x(6)-x(4))*(x(6)-x(5))));
a6=a_back6+a_middle6+a_forward6;
% Finding total acceleration (m/s^2)
a=[a1,a2_5,a6]
% B
t=32; % Specified time value
p_32=((t-x(4))*(y(5)-y(4))/(x(5)-x(4)))+y(4); % Interpolating position at 32s
% Using a second order Lagrange Polynomial to determine the velocity at 32s
v32_back=y(4)*((2*t-t-x(5))/((x(4)-t)*(x(4)-x(5))));
v32_middle=p_32*((2*t-x(4)-x(5))/((t-x(4))*(t-x(5))));
v32_forward=y(5)*((2*t-x(4)-t)/((x(5)-x(4))*(x(5)-t)));
v_32=v32_back+v32_middle+v32_forward
% Using a second order Lagrange Polynomial to determine the acceleration at 32s
a32_back=v(4)*((2*t-t-x(5))/((x(4)-t)*(x(4)-x(5))));
a32_middle=v_32*((2*t-x(4)-x(5))/((t-x(4))*(t-x(5))));
a32_forward=v(5)*((2*t-x(4)-t)/((x(5)-x(4))*(x(5)-t)));
a_32=a32_back+a32_middle+a32_forward
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