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Applied Maths Question

  • 05-12-2012 5:24pm
    #1
    Registered Users, Registered Users 2 Posts: 275 ✭✭


    can anyone tell me what to do in these : At 15.00 hours an aeroplane is 120km due east of a helicopter. The plane is flying in a direction west 30' south at a constant speed of 800km/h. The Helicopter has a max speed of 40km/h.

    Find (a) the course for the closest aproach.
    (b) the time that this event occurs.

    Also another question:
    A man swims at 4ms^-1 in still water. He swims across a river of width 120 metres. The river flows with a constant speed of 5ms^-1 parallel to the straight banks.
    He wishes to cross by the shortest path.
    Find a the direction he should take
    b the time taken to cross the river by the shortest path.

    Thanks

    Just another question if two objects are to intercept what relationship is there ?


Comments

  • Registered Users, Registered Users 2 Posts: 5,143 ✭✭✭locum-motion


    Is this homework?

    You're not allowed to get others to do your homework on this forum.


  • Registered Users, Registered Users 2 Posts: 5,141 ✭✭✭Yakuza


    We can't just do the problems for you here, but I can indicate the method I'd take:

    In both those problems, you need to split up the velocity into east-west (x) and north-south(y) components.
    In the first problem, (by the way, I'm assuming the helicopter is hovering over one spot), if the plane is on a heading of west / 30deg south, what is the speed of the plane in the x direction? And the y direction? (use trigonometry).

    Once you know these two speeds, you need to be able to express the distance in the x and y directions of the plane from the helicopter in terms of time t.

    When you have an expression for these, then the total distance between the helo and the plane is the square root of the x-direction and y-direction distances squared.

    To minimise this, differentiate it with respect to t (use the chain rule) and set it equal to zero (the second derivative is positive, indicating a minimum). Solve for t.

    Then, plug t into the equation for distance you got above and this will be the answer to the first part, adding t (in hours) to 1500 hours will give you the time it happens at.

    The second problem is has the same principle, you have to express the distance travelled in the x- and y-directions in terms of the angle [Latex]\theta[/Latex]he starts swimming at. Minimise this (again with calculus). Once you know what the angle is, you can divide the width of the river by the speed in the y direction to get the time.


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