If I understand true bearings as relating to true north, then we have a diagram where
L is the light
A is the initial observation point
B is the second observation point
In 20 minutes, the ship has traveled 4 miles. AB=4
∠A = 60.3-37.3 = 23°
∠B = 180-(60.3-20) = 139.7°
∠L = 37.3-20 = 17.3°
Using the x-y plane, as opposed to compass bearings, if the first sighting was taken at (0,0), then
the ship's position is
s = tan29.7° x = 0.57x
the light is somewhere along the line
y = tan52.7° x = 1.31x
Now, after 20 minutes, s is 4 miles from (0,0), at (3.47,1.98)
So, the second sighting is somewhere along the line
y-1.98 = tan70° (x-3.47), or
y = 2.74(x-3.47)+1.98
y = 2.74x - 7.53
That places the light at (5.27,6.90)
At time t hours, we have the coordinates of s as
x = 12*cos29.7 t = 10.42t
y = 12*sin29.7 t = 5.95t
Now, the distance from the ship to the light at time t is
d^2 = (10.42t-5.27)^2 + (5.95t-6.90)^2
d^2 = 143.98t^2 - 191.94t + 75.39
When is that a minimum?
dd/dt = (143.98t-95.97)/d
Since d≠0, dd/dt=0 when t = .666, or 40 minutes from the first sighting.
Extra credit: how close is the nearest approach? 3.38 miles.
when is the ship 11 miles from the light? when
143.98t^2 - 191.94t + 75.39 = 11^2
t = 1.54 hours
a pilot entering a bay on course 60.3 degrees at the speed of 12 knots sees a light bearing 37.3 degrees true, and 20 mins later he sees it bearing 20 degrees true. if he keeps the same course and speed,(a)when will he be nearest to the lighthouse and how near (b) when will he be 11 miles from the lighthouse?
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