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Math Help - Direction of aeroplane query

  1. #1
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    Direction of aeroplane query

    Hi, could someone please help me on this as I'm a bit stuck...

    An aeroplane flies in a straight line from city A (300, -200) to city B at (-100, 600).

    (positions are given with reference to a Cartesian coordinate system whose x and y axes point due East and Due North respectively. Distance is in km.)

    (1) Find the equation of the line'

    I have m = rise/run = 800/-400 = -2

    Then by applying equation y-y1 =m(x-x1) with m = -2 and (x1, y1) = (300, -200) I get:
    y - (-200) = -2(x-300)

    y = -2x + 400
    (2) Find the direction of travel of the aeroplane, as a bearing, with the angle in degrees correct to one decimal place

    I think I dont quite get this one...I've tried tan = -2 and get N 334.9 E

    (3) Find the distance between cities A and B to the nearest km

    Here I have done (-100-300)squared = (600-(-200))squared = 480000
    Then apply the square root to this final figure get 693km

    Then after landing at city B, the aeroplane flies in a straight line in the direction of S 23 W, to city C, before finally flying in a straight line in the direction of N 54 E back to city A.

    (4) What is the distance between cities A and C, to the nearest km?


    Please if any one could give me a few pointers I would be really grateful.

    With regard to the aeroplane's flight from city A to city B. During this flight, it flies within range of an air traffic control centre at position (300,0)

    (5) Find the parametric equations for the line of flight of the aeroplane. Your equations should be in terms of the parameter t , and should be such that the aeroplane is at city A when t = 0 and at city B when t= 1

    This is what I have done.

    The line of the slope is -2 and passes through (x1, y1) = (300, 0)

    The parametric expressions are:

    x = t + x1, y = mt + y1 leading to x = t + 300, y = -2t
    (6) Write down an expression, in terms of t, for the square of the distance between the air traffic control centre and the aeroplane at the point with parameter t on the line of the flight of the aeroplane. Simplify your answer.

    (the distance required here is the horizontal distance i.e. the distance between the air traffic control centre and the point on the ground immediately below the aeroplane.

    (7) Using the answer to (6) above, and the method of completing the square, to determine the distance, to the nearest km, between the air traffic control centre and the aeroplane at the point on the line of flight of the aeroplane where it is closest to the air traffic control centre.

    Many thanks
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  2. #2
    Banned
    Joined
    Sep 2008
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    for the second one you want to take the inverse tangent of the change in y over the change in x.... keeping in mind you may get 2 possible solutions, just think about what direction its goin in should be straightforward. just take 180 - arctan(2) .

    for the third problem draw out your vectors on a piece of paper, I don't know what this notation N 54 E means, but whatever it means you can draw a picture of it, label the angels of the triangle, and use some trig to find the distances involved.


    so if your air traffic control center is at some point (x0,y0) and your plane is at the point (x(t), y(t)), then the distance between them, say L(t) = sqrt( (x0-x(t))^2 + (y0-y(t))^2) ill let you do the 'simplification'
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