Question 9.1: Distance traveled by a decelerating airplane. A Boeing 737-2......
Distance traveled by a decelerating airplane.
A Boeing 737-200 airplane of mass m = 97000 kg lands at a speed of 93 m/s (about 181 knots) and applies its thrust reversers at t = 0. The force F that is applied to the airplane, as it decelerates, is given by F = -5ν² – 570000, where ν is the airplane’s velocity. Using Newton’s second law of motion and flow dynamics, the relationship between the velocity and the position x of the airplane can be written as:
m \nu \frac{d\nu}{d x}=-5 \nu^{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 (about 78 knots) by using the composite trapezoidal method to evaluate the integral resulting from the governing differential equation.
Even though the governing equation is an ODE, it can be expressed as an integral in this case. This is done by separating the variables such that the speed ν appears on one side of the equation and x appears on the other.
\frac{97000 \nu d \nu}{\left(-5 \nu^{2}-570000\right)}=d x
Next, both sides are integrated. For x the limits of integration are from 0 to an arbitrary location x, and for ν the limits are from 93 m/s to 40 m/s.
\int_{0}^{x} d x=-\int_{93}^{40} \frac{97000 \nu}{\left(5 \nu^{2}+570000\right)} d \nu=\int_{40}^{93} \frac{97000 \nu}{\left(5 \nu^{2}+570000\right)} d \nu (9.14)
The objective of this example is to show how the definite integral on the right-hand side of the equation can be determined numerically using the composite trapezoidal method. In this problem, however, the integration can also be carried out analytically. For comparison, the integration is done both ways.
Analytical Integration
The integration can be carried out analytically by using substitution. By substituting z = 5u² + 570000, the integration can be performed to obtain the value x = 574.1494 m.
Script File
Numerical Integration
To carry out the numerical integration, the following user-defined function, named trapezoidal, is created.
Program 9-1: Function file, integration trapezoidal method.
function I = trapezoidal(Fun,a,b,N)
% trapezoidal numerically integrate using the composite trapezoidal method.
% Input Variables:
% Fun Name for the function to be integrated.
% (Fun is assumed to be written with element-by-element calculations.)
% a Lower limit of integration.
% b Upper limit of integration.
% N Number of subintervals.
% Output Variable:
% I Value of the integral.
h = (b-a)/N;
x = a:h:b;
F = Fun(x);
I =h* (F (1) + F (N+ 1)) /2 + h*sum(F (2 :N));
Calculate the value of the integral according to Eq. (9.13).
I(f) \approx \frac{h}{2}[f(a)+f(b)]+h \sum\limits_{i=2}^N f\left(x_i\right) (9.13)
The function trapezoidal is used next in the Command Window to determine the value of the integral in Eq. (9.14).
To examine the effect of the number of subintervals on the result, the function is used three times using N = 10,100, and 1000 . The display in the Command Window is:
>> format long g
>> Vel =@ (v) 97000*v./(5*v.^2+570000);
>> distance = trapezoidal(Vel,40,93,10)
distance =
574.085485133712
>> distance = trapezoidal(Vel,40,93,100)
distance =
574.148773931409
>> distance = trapezoidal(Vel,40,93,1000)
distance =
574.149406775129
As expected, the results show that the integral is evaluated more accurately as the number of subintervals is increased. When N = 1000, the answer is the same as that calculated analytically to four decimal places.