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Relative velocity of flight

  1. May 7, 2004 #1
    Hi there,
    i am a cambridge student studying in the first year of O levels. Though relative velocity isn't in my syllabus, ihave inerest in gaining knowledge about it. I studied the topic, but was unable to do some of the exercise questions. One question which i foind difficult is as follows:-
    A plane flis in a straight line from A to B, where AB=580 km & the bearing of B from A is N 50(degrees)E. The speed of the plane in still air is 300km/h. Given that the wind is blowing from the direction N 40(degrees)W and that the flight takes 2 hours, calculate the speed of the wind and the course that should be set for the return journey, if the velocity of the wind remains unchanged.
    Please help me out to find the solution :rolleyes:
  2. jcsd
  3. May 7, 2004 #2


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    There are several different ways of doing this, depending on your "level of sophistication".

    Here is the most basic way:
    Set up a "coordinate system" with N upward, E to the right. A will be at the origin of the coordinate system. Draw a line at an angle of 50 degrees to the East of the N line and mark B 580 km along it (That's the hard part- you need a really BIG piece of paper! :) ). That represents the line of flight.
    Now draw a line at 40 degrees E of the S line (the wind was coming FROM 40 degrees W of N- it is blowing TO 40 degrees E of S). Unfortunately we don't know how long to make it since we don't know the speed of the wind. For now, just imagine it extended indefinitely.
    From B use compasses set at 600 km (the distance the plane would fly at 300 km in two hours) to strike an arc intersecting your second line (call it C). Draw the line segment BC to form a triangle. The length of the line from A to C, divided by 2 (for the 2 hours) is the speed of the wind!

    What? You can't draw and measure it accurately enough? You want to be able to calculate the wind speed? No problem.

    You now have a triangle drawn on which you know the lengths of two sides and one angle. You know the length of AB is 580 and the length of BC is 600. You know that angle AB makes with N is 50 degrees and the angle AC makes with S is 40 degrees so angle BAC is 180- 50- 40= 90 degrees. Hey, you've got a right triangle here! I was all set to use the "cosine law" to find the third side but that's not necessary: use the Pythagorean theorem (the cosine law "c2= a2[/sup]+ b2- 2ab cos(C) is a generalization of the Pythagorean theorem that applies to general triangles. If this angle were not a right angle, we could use that). Here, the hypotenuse has length 600 and one leg has length 580. The length of the other leg is given by a2= 6002-5802= 23600 so a= 154 km (rounded to the nearest kilometer). Since that is for two hours, the wind speed is 154/2= 77 km/h.

    Now turn it around: coming back, the airplane is still flying at 300 km/h, the wind is still blowing at 77 km/h in the same direction so we still have our line AC except that we should now have it up at B: draw BD parallel to AC and 77t km long. That's right: 77t- t is the time for the flight in hours and we don't know how long the flight will be. The third leg of our triangle, representing the airplane, will be 300t km long. AB is still 50 degrees E of N and BD is 130 degrees E of N (or, same thing, 50 degrees S of E). Since angle BAC was 90 degrees and BD is parallel to AC, angle ABD is also a right angle and we can, again, use the Pythagorean theorem. Now our two legs are of length 580 and 77t while the hypotenuse has length 300t. 5802+ (77t)2= (300t)2 so 336400+ 5929t2[/sup 90000t2 or 84071t2= 336400. t2= 4.00137 and that means t= 2 hours.

    Hey, that's the same as the flight from A to B! If the wind is "against" us in one direction shouldn't it be "helping" in the other- so that one way is shorter than the other? Have we gotten confused and redone the original problem? I confess that I stopped and recalculated everything! The point is, again, that right triangle! The wind is blowing at right angles to the line from A to B, the line of flight- coming back it is still at right angles. The two times are exactly the same.
  4. May 8, 2004 #3
    Thanks for ur answer, u just reduced a problem from me.
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