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Homework Help: Optimisation help

  1. Aug 14, 2011 #1
    1. The problem statement, all variables and given/known data
    An enemy fighter jet has invaded friendly airspace and is travelling on a bearing of 40 degrees. At the time they scramble an interceptor from a nearby airbase, the enemy jet is 200km away and on a bearing of 10 degrees from the airbase. The enemy jet is travelling at 1200km/h. The interceptor can fly at 2500km/h. On what bearng should the
    interceptor fly in order to intercept the enemy in the shortest possible time, assuming that the interceptor accelerates immediately to 2500km/h after take off?

    2. Relevant equations
    cosine rule, sine rule

    3. The attempt at a solution
    I am puzzled alot by this problem. I really don't know how to optimise this problem. Please help!
  2. jcsd
  3. Aug 14, 2011 #2
    It seems to me that there is incomplete information to answer the question. You need to know what time the first observation (that the enemy was at 40 degrees) was made at. Or you need to know how far away it was when the observation was made. But, as is, there is no was to figure out the direction of the enemies flight, so there is no way to figure out how to intercept it.

    Are you sure there isn't a little more detail to the problem that you're leaving out?
  4. Aug 14, 2011 #3


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    This is a problem where a drawing is going to be useful. You'll want to make a graph with the positive x-axis pointing East and the positive y-axis pointing North. Bearings are measured as angles clockwise from due North.

    The initial time is the moment when the interceptor aircraft takes off. For convenience, put the intruder jet at the origin; it is traveling on a straight line pointing 40 degrees clockwise from the positive y-axis. This jet is moving at 1200 km/hr, so you will want to write linear functions x(t) and y(t) for its position, given that it is at the origin at t = 0 .

    Next, we need to locate the airbase where the interceptor starts from. It is 200 km. away from the origin along a line making a 10-degree angle with the y-axis (since the intruder is on a bearing of 10 degrees from there at t = 0 ). So the interceptor is starting from a point in the fourth quadrant; you will need to calculate those starting coordinates.

    Then we need to write linear functions X(t) and Y(t) for the interceptor's position as functions of time. We don't know what bearing we need yet, so we'll just call it (theta). We do know the interceptor's speed is 2500 km/kr . We then write the position functions using this speed, the unknown angle, and the starting coordinates.

    The two lines of flight of these aircraft are intended to intersect at some time t = T . So we want to set up x(T) = X(T) and y(T) = Y(T) , so the intruder and interceptor arrive at the same point at the same time. We will end up with a relationship between T , the time required to intercept the intruder jet, and (theta), the bearing angle used. This finally gives us a function for which we can find a critical point at which T is minimized.

    That's as far as I can take you at the level of suggestions. You'll get a bit of exercise with trigonometry in this problem...
  5. Aug 14, 2011 #4
    Ah, forgive my first post. The bearing in the first part is the direction of travel of the enemy, not the direction of the enemy relative to the airbase.

    I have a suggestion then. Suppose the interceptor intersects the enemies line of travel at the exact same time that the enemy reaches the intersection point. What do we know about the distances that the planes have flown? The ratio of the distances traveled must be the same as the ratio of their velocities since they are arriving at the same point at the same time. So you just need to find the point along the enemy's line of travel for which those two ratios are the same. Then, work out what the bearing from the airbase to that point is.
  6. Aug 15, 2011 #5


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    I beat on this a while longer and finally recognized what the "hint" is about. The difficulty with the approach I described is that it requires a careful arrangement to make sure that the x- and y-coordinates of the two aircraft match at the same time. That, however, bothered me because it should not be necessary to plan that in an optimization problem...

    Another approach is to use the Law of Cosines. You have a triangle with one side that is 200 km. long and make an angle 10º clockwise to due North, a second side at 40º clockwise to due North with a length 1200T km. (the intruder's flight path), where T is the time required for interception, and the third side has a length 2500T km. (the interceptor's flight path) making an unknown angle (theta) to due North. Using geometry, it can be shown that the angle opposite the side of length 2500T is 150º. The Law of Cosines will give an equation which can be rearranged into a quadratic equation in T with only one positive solution. (I'll just say that the time to interception is close to 0.14 hour.)

    The point is that the conditions of the problem give a unique answer, so the problem is over-constrained as an optimization problem. There would need to be another variable, such as the speed of the interceptor, to make this a matter for optimization (though I suspect the solution would still be to fly at maximum practicable speed).

    A way that you can see that this problem only has one answer not subject to optimization is to think of a circle centered on the airbase which is expanding at 2500 km/hr. The intruder is fleeing along its flight line at 1200 km/hr with a specified lead. Under these specifications, there is only one place on the line where the circle will overtake the retreating jet, so there is only one direction possible for the interceptor to take in order to reach that point. If the velocity of the interceptor were a variable, then there would be a range of interception points and a question of minimal time to be resolved.

    Returning to the problem as given, all three sides of the triangle can now be found and the Law of Sines employed to find the unknown angle, which can be shown geometrically to be (theta)-10º . (The bearing of the interceptor is close to 24º clockwise from North.) I leave you to fill in the detailed calculations...
    Last edited: Aug 15, 2011
  7. Aug 15, 2011 #6
    Your an absolute legend, that's cleared it up really well for me! thank you so much for taking the time to help me.
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