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An aircraft passes over you with a velocity of 950 km/h at an altitude of 11000 meters. What is the angular velocity two minutes later?

**this problem is tricky because 2 minutes later the aircraft is no longer tangent to the circle.

  • physics -

    First do the conversion
    950 km/h = 263.9 m/s

    Measured from the instant the plane is overhead, the Y coordinate remains
    y = 11,000m,
    and the X coordinate is
    x = 263.9 t, if t is in seconds
    The angle from zenith direction (straight up) is
    A(t) = arctan (x(t)/y)

    What you want is dA/dt. It equals
    dA/dt = (dA/dx)*(dx/dt)

    = {1 /[1 + (x/y)^2]}*(1/y)*263.9

    Evaluate this at t = 120 s, when x = 31,668 m and y = 11,000 m. The answer will be in radians per second

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