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.
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