Effacer les filtres
Effacer les filtres

Using MATLAB to do Numeric Calculus

2 vues (au cours des 30 derniers jours)
Spaceman
Spaceman le 22 Avr 2024
Commenté : Spaceman le 29 Avr 2024
So I am culminating all of my newfound knowledge to solve and show a simple calculus problem. However I am running into issues...
Given: A piecewise function where v(t)=70t @ t<=20s & 1596.2-9.81t @ 20<t<t_max <---(eq 1)
Also given: s(t)=35t^2+7620 @ t<=20s & 1596.2t-4.905t^2-8342 @ 20<t<=t_max <---(eq 2)
Find: If aircraft ignites its engines at t=0s and accelerates vertically, there is only enough fuel for an engine burn of 20 seconds. At which time will the ship become a projectile? Find value of t_max. Plot the crafts velocity vs. t, (eq 1). Plot the crafts altitude vs. t, (eq 2).
My Solution:
%% Finding Value of tmax
% Finding tmax and rounding it to the nearest whole number
%%
tmaxr=roots([-9.81, 1596.2]);
tmax=ceil(tmaxr);
%% Plotting Equation 1
% Creating Figure 1 from Equation 1
%%
t=0:0.1:tmax;
v=zeros(size(t));
for N=1:length(t)
if t(N)<=20
v(N)=70*t(N);
else
v(N)=1596.2-9.81*t(N); % Only valid for t > 20 seconds
end
end
figure(1)
plot(t,v)
xlabel('Time (s)')
ylabel('Velocity (m/s)')
title('Velocity vs. Time')
%% Plotting Equation 2
% Creating Figure 2 from Equation 2
%%
t=0:0.1:tmax;
s=zeros(size(t));
for N=1:length(t)
if t(N)<=20
s(N)=35.*t(N).^2+7620;
else
s(N)=1596.2.*t(N)-4.905.*t(N).^2-8342;
end
end
figure(2)
plot(t,s)
xlabel('Time (s)')
ylabel('Altitude (m)')
title('Position vs. Time')
Issue: So I have found t_max, plotted the velocity but it looks funky, and also tried to plot my altitide but not sure if it's faultless.
P.s. Sorry for the formatting, there's no easy way to type piecewise functions that I know of.

Réponse acceptée

Torsten
Torsten le 22 Avr 2024
Modifié(e) : Torsten le 22 Avr 2024
Your code is correct.
Here is a simpler way to define s and v. Just substitute tcrash by tmax.
tcrash = 1596.2/(2*4.905)+sqrt((1596.2/(2*4.905))^2-8342/4.905);
s = @(t)(35*t.^2+7620).*(t<=20)+(1596.2*t-4.905*t.^2-8342).*(t>20).*(t<=tcrash);
v = @(t)(70*t).*(t<=20)+ (1596.2-9.81*t).*(t>20).*(t<=tcrash);
t = 0:0.01:tcrash;
figure(1)
plot(t,s(t))
grid on
figure(2)
plot(t,v(t))
grid on
  6 commentaires
Sam Chak
Sam Chak le 24 Avr 2024
I suppose @Kyle intended to estimate the rate of change based on the simulated data points, rather than directly from the kinematic equation. Observing the curve, it appears that the aircraft (modeled as a point mass) is descending rapidly towards the ground. Is this behavior intentional and meant to be displayed?
Spaceman
Spaceman le 29 Avr 2024
I was examining the kinematic equations to plot the information and compare it to the derivative approximation of velocity, etc. I was then plotting the integral approximation and comparing it to the past plots. I got it all figured out, though. :)

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