Solve() in a for loop giving an Empty sym: 1-by-0 answer

I've been working on a code to calculate velocity and acceleration. The issue is with the last section called %%Solve for position at time at a=0 and v=0 and the third function. I'm trying to get an array of positions at the times listed. I keep getting Empty sym: 1-by-0 as the answer.

I'll copy the function codes I've written as well as the document combining the functions. Let me know if you can help.

First functions:

function [velocity,acceleration] = Velocity_Acceleration( position )
%Velocity_Acceleration calculates the first and second differencial of
%position
%   this takes a given equation and turns it into a diff equation
velocity = diff( position ); % this is the velocity of a given equation
acceleration = diff( position, 2);% this it the acceleration of a given equation
end

Second functions:

function [time] = SolveforTime(equation, y, t1)
%SolveforTime solves for a unknow (t) variable if the answer for the
%equation is known 
%   tvalue is the unknown solution in the equation
% equation is the given equation
% whattheequationisequalto is the number the equation is equal to
% t1 is the name of the constant to be identified aka tvalue
whattheequationisequalto = y==equation;
time = solve(whattheequationisequalto, t1);
end

Third functions:

function [time] = SolveforXifTimeisKnown(eqn,t,t1)
%SolveforTime solves for a unknow (x) variable if the time for the
%equation is known 
%   tvalue is the unknown solution in the equation
% equation is the given equation
% whattheequationisequalto is the number the equation is equal to
% t1 is the name of the constant to be identified aka tvalue
counter = 1;
t1 = t1;
% first for loop runs for the primes of 1000
% ai is the individual number (1000) being checked in the loop
for i = length(t1)
    ti = t1(i);
    eqns = [eqn, t==ti];
    t2 = solve(eqns,t);
    time(counter,: ) = [t2];
    counter = 1 + counter;
end
time
end

Main code:

% Determine (a) when the velocity is zero, (b) the position and total distance traveled when the acceleration is zero.
%
% If the motion of a particle is defined by the relation x = t^3 - 9*t^2 +
% 24*t - 8, where x and t are expressed in inches and seconds, respectively.
% 
% use this to take a simple position equation in terms of time to find
% velocity and acceleration
clear
clc
%% what is the given position equation and what velocity and acceleration 
% needing to find

% t represents time
syms t;

% insert any equation in terms of t to get the first (v) and second(a)
% differential equation of x

x = t^3 - 9*t^2 + 24*t - 8;
v1 = 0;
a1 = 0;
x0 = 0;


%% solve for Velocity and Acceleration equations
%
fprintf('If position is: %s \nThen\n\n', x);

% this calls the differential function
[v,a] = Velocity_Acceleration(x);

fprintf('velocity is: ');
disp(v);
fprintf('acceleration is: ');
disp(a);


%% part a
%
equation = v; % choose v or a for which one you want to find.
v3 = v1; % this is what v or a equation above equals :

% this solves for t if velocity, acceleration, or position is known
tvalue1 = SolveforTime(equation, v3, t)';
a2 = tvalue1(:,1);
b2 = tvalue1(:,2);
% I'm just displaying what part of the problem is, this is part a
disp('a)');
% this shows the numbers solved for. for this one it's time
fprintf(' If the velocity is 0, time = %g and %g.\n\n\n', a2, b2);


%% part b
%
equation = a; % choose v or a for which one you want to find.
y2 = a1; % this is what v or a equation above equals :

% this solves for t if velocity, acceleration, or position is known
tvalue2 = SolveforTime(equation, y2, t);

% I'm just displaying what part of the problem is, this is part a
disp('b)')
% this shows the numbers solved for. for this one it's time
fprintf(' If the accileration is 0, time = %g.\n\n', tvalue2)



%% Solve for position at time at a=0 and v=0
%
t1 =[a2, tvalue2, b2];
t2 = length(t1);
syms x5 [1 t2];
x1 = x5 == x;
x2 = SolveforXifTimeisKnown(x1,t,t1)';

Answer 1

[Walter Roberson]

x = t^3 - 9*t^2 + 24*t - 8;

x will be a single scalar symbolic expression

t1 =[a2, tvalue2, b2];

t1 will be at least 3 values

t2 = length(t1);
syms x5 [1 t2];
x1 = x5 == x;

so x1 will be a vector with one equation for each t1 value.

for i = length(t1)
    ti = t1(i);
    eqns = [eqn, t==ti];
    t2 = solve(eqns,t);

ti will be a specific value for t. In particular, it will be the last t1 value, because you used i = length(t1) rather than i = 1:length(t1)

t==ti will effectively constrain t to be a single specific value for the purpose of the equations.

Then you solve the augmented system of three equations of the form XVARIABLE = t^3 - 9*t^2 + 24*t - 8 with the same t^3 - 9*t^2 + 24*t - 8 being used for each of the cases, but different XVARIABLE for the three equations. And you add on the t == specific value. Then you ask to solve that array of four equations for t . There is of course, only one solution, which is that t must be the ti value, since you have the equation t==ti .

Remember that the name of the function indicates that you are solving for x not for t

There is a much much easier way to proceed if you want to solve for x for a vector of t1 values:

    x = subs(eqn, t, t1);

with no solve needed.