Asked by Anonymous
A freighter, streaming on course 140„a at 20 knots, is 40 nautical miles N20„aE of a submarine with a cruising speed of 25 knots. Find the course to be set by the sub to overtake the freighter in the least amount of time, and find this minimum time.
Answers
Answered by
Steve
Let's place the submarine at (0,0) to start.
The freighter is 40 miles at N20°E, placing it at (13.68,37.59)
The freighter's heading is 140°, so at time t its position is
(fx,fy) = (13.68 + 20t*cos(140°), 37.59 + 20t*sin(140°))
(fx,fy) = (13.68-15.32t , 37.59 + 12.86t)
The sub, at speed 25, sails at a heading of a°, so its position at time t is
(sx,sy) = (25t*cos(a),25t*sin(a))
Now we can figure the distance between the vessels:
d^2 = (fx-sx)^2 + (fy-sy)^2
The equation is in two variables, and is not subject to direct solution, but a little iteration will show that the two ships will meet at
time t = 4.15 hours if the sub's heading is 118.75° or N28.75°W.
Hmmm. It occurs to me that I have been using common trig angles, with 0° pointing due East. Nautical headings have 0° = due North, so you may have to fiddle with the angles to correct things.
The freighter is 40 miles at N20°E, placing it at (13.68,37.59)
The freighter's heading is 140°, so at time t its position is
(fx,fy) = (13.68 + 20t*cos(140°), 37.59 + 20t*sin(140°))
(fx,fy) = (13.68-15.32t , 37.59 + 12.86t)
The sub, at speed 25, sails at a heading of a°, so its position at time t is
(sx,sy) = (25t*cos(a),25t*sin(a))
Now we can figure the distance between the vessels:
d^2 = (fx-sx)^2 + (fy-sy)^2
The equation is in two variables, and is not subject to direct solution, but a little iteration will show that the two ships will meet at
time t = 4.15 hours if the sub's heading is 118.75° or N28.75°W.
Hmmm. It occurs to me that I have been using common trig angles, with 0° pointing due East. Nautical headings have 0° = due North, so you may have to fiddle with the angles to correct things.
Answered by
Steve
Thinking about things a bit more, I realized that a simple law of cosines setup would work.
(25t)^2 = 40^2 + (20t)^2 - 2*40*cos110°
625t^2 = 1600 + 400t^2 + 684t
225t^2 - 684t - 1600 = 0
t = 4.58 (a little different value)
20t = 91.6
25t = 114.5
sin(a)/91.6 = sin110°/114.5
sin(a) = .75175
a = 48.74°
add that to the 70° initial bearing to get 118.74°
(25t)^2 = 40^2 + (20t)^2 - 2*40*cos110°
625t^2 = 1600 + 400t^2 + 684t
225t^2 - 684t - 1600 = 0
t = 4.58 (a little different value)
20t = 91.6
25t = 114.5
sin(a)/91.6 = sin110°/114.5
sin(a) = .75175
a = 48.74°
add that to the 70° initial bearing to get 118.74°