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finding the distance traveled before speed is reduced to a certain velocity

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kayla Levin
kayla Levin on 27 Oct 2019
Commented: kayla Levin on 3 Nov 2019
so i am given mass of an airplane, v1 and v2, F= -5*v^2 -570000 and the equation mv(dv/dx)= -5*v^2 - 570000
and i have to find how far the airplane travels before its speed is reduced to v2 using the trapezoid rule
So far i have
function Q = trapz( a,b,n )
x= linspace (a,b,n+1);
h = (b-a)/n;
Q = 0;
for i= 1:n
Q = Q + .5*(func(x(i))+func(x(i+1)))*h;
end
end
and i have
function value = Force(v)
value = -5*v^2-570000;
I don't know if i am on the right path or what to do from here. Please help and thank you

  3 Comments

darova
darova on 28 Oct 2019
Can you attach simple scheme how is this looks like? What is force here (gravity or air resistance)?
kayla Levin
kayla Levin on 28 Oct 2019
The given question is:
A Boeing 727-200 airplane of mass m = 97000 kg lands at a speed of 93 m/s and applies its thrust reversers at t = 0. The force F that is applied to the airplane, as it decelerates, is given by F = -5v^2 - 570000, where v is the airplane's velocity. Using Newton's second law of motion and flow dynamics, the lrelationship between the velocity and the position x of the airplane can be written as
mv(dv/dx) = -5v^2 - 570000
where x is the distance measured from the location of the jet at t = 0
Determine how far the airplane travels before its speed is reduced to 40 m/s
I was thinking that the function of x is -5v^2 - 570000
i = 93:40
but then i am stuck because i have to use the trapezoid rule to find distanced travled which probably is b and a can start at 0....
Sorry, thank you
darova
darova on 28 Oct 2019
if F = -5v^2 - 570000 then m(dv/dt) = F
Looks like diff equation
You can use ode45 to solve it

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Accepted Answer

Asvin Kumar
Asvin Kumar on 1 Nov 2019
The original equation from your question is .
Rearranging the terms would give .
We would need to integrate the equation above as follows to obtain the distance travelled:
Try using the following code to obtain that. This is just a modified version of the code you had provided which uses the trapezoidal method:
trapz(93,40,1e3,0)
% a is initial point
% b is final point
% n-1 is the number of points in between a and b
% ic is the initial condition
function Q = trapz( a,b,n,ic )
x = linspace (a,b,n+1);
h = (b-a)/n;
Q = ic;
for i= 1:n
Q = Q + .5*(func(x(i))+func(x(i+1)))*h;
end
% % optimized code below
% tmp = func(x);
% Q = ic + h * ( sum(tmp)-0.5*(func(a)+func(b)) );
end
function val = func(v)
m = 97e3;
val = (m*v)./(-5*v.^2-57e4);
end
As darova had mentioned in their comment this can also be solved using ode45. Comparing with its documentation, the variable of differentation (t) in this case is 'v' while the variable being differentiated (y) is 'x'.
The code for that is:
[v,x] = ode45(@(t,y) 97e3*t/(-5*t.^2-57e4) ,[93 40],0);
plot(v,x)
Have a look at an example from the documentation solving a similar problem: https://www.mathworks.com/help/matlab/ref/ode45.html#bu3ugj4

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