Differnetial Equations question Group Projects for chapter 3 D Aircraft Guidance

Question

Differnetial Equations question Group Projects for chapter 3 D Aircraft Guidance in a crosswind An aircraft flying under the guidence of nondirectional beacon (a fixed radio transmitter, abbreviated NDB) moves so that its longitudinal axis always points torward the beacon (see figure 3.19) A pilot sets torward an NBD from a point at which the wind is at right angles to the initial direction af the aircraft; the wind maintains this direction. Assume that the wind speed and the speed of the aircraft through the air (its "airspeed") remain constant. (Keep in mind that the latter is different from the aircraft's speed with respect to the ground.) (a) Locate the flight in the xy-plane, placing the start of the trip at (2,0) and the destination at (0,0). Set up the differential equation describing the aircrafts path over the ground. [Hint: dy/dx = (dy/dt)/(dx/dt).] (b) Make an appropriate substitution and solve this equation. [Hint: See Section 2.6.] (c) Use the fact that x = 2 and y = 0 at t = 0 to determine the appropriate value of the arbitrary constant in the solution set. (d) Solve to get y explicitly in terms of x. Write your solution in terms of a hyperbolic function. below is what I have so far.... Place the NDP at the origin and the airplane initially at location {2,0}. Since the plane is always flying directly toward the origin we can make an angle q at the origin which points at the plane. Let v be the plane's velocity through the air. The wind has no effect on the x component of the plane's velocity which is -v cos(q), the minus being necessary because the plane is flying in the direction of decreasing x as the time t advances. Ifw is the speed of the wind, then the plane's y component of velocity is w-vsin(q).Then the following equations must be true due to definitions of cosine and sine: dx/dt = -vcos(theta) = -(vy)/(sqrt(x^2+y^2)) dy/dt = w-vsin(theta) = w-(vy)/(sqrt(x^2+y^2)) Dividing the second equation by the first we obtain: dy/dx = (vy-w(sqrt(x^2+y^2)))/(vx) You can now solve this differential equation to obtain the following solution with the starting condition that y[2] = 0. I believe the portion above is correct but we are stuck on the rest.

Explanation / Answer

Place the NDP at the origin and theairplane initially at location {2,0}. Since the plane is alwaysflying directly toward the origin we can make an angleq at the originwhich points at the plane. Let v be the plane's velocitythrough the air. The wind has no effect on the x componentof the plane's velocity which is -v cos(q), the minus beingnecessary because the plane is flying in the direction ofdecreasing x as the time t advances. Ifw is the speed of the wind, then the plane's ycomponent of velocity is w-vsin(q).Then the following equations must be true due to definitions ofcosine and sine: Dividing the secondequation by the first we obtain: You can now solve thisdifferential equation to obtain the following solution with thestarting condition that y[2] = 0.

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