A fleet of ships is on a straight course at a speed of one knot. It is guarded by a cruiser that travels at two knots.
C. The fleet is in a straight line one nautical mile long but is not single file. Instead the motion is at an angle of θ from being abreast. The cruiser starts at one end and continuously motors to the other and back again. How long does one cycle take?
D. The fleet is in an equilateral triangle formation, one nautical mile on a side. The cruiser hems close to the sides as it cycles around this triangle. How long does one cycle take if:
i. One side of this triangle is perpendicular to the direction of travel?
ii. One side of this triangle is parallel to the direction of travel?
Try to give exact answers.
As in the previous problem I will assume the fleet is traveling North.
Part C
Let T be the angle measured clockwise between the fleet north vector and the cruiser's vector relative to the fleet. Let V be the speed of the cruiser relative to the fleet. The north vector is (0,1) and the vector along the fleet is (V*sin T, V*cos T). Their sum is the two knot speed of the cruiser. The speed along the line can then be found as a function of T:
abs[(0,1) + (V*sin T, V*cos T)] = 2
(V*sin T)^2 + (1+V*cos T)^2 = 2^2
V^2 + (2*cos T)V - 3 = 0
V = -cos T + sqrt[(cos T)^2 + 3] (the negative root is extraneous)
The reciprocal of V is the time it takes for the cruiser to travel 1 mile along one side of the line:
1/(-cos T + sqrt[(cos T)^2 + 3])
= (cos T + sqrt[(cos T)^2 + 3])/3
For the other side add 180 to the angle T:
(cos (T+180) + sqrt[(cos (T+180))^2 + 3])/3
= (-cos T + sqrt[(cos T)^2 + 3])/3
Then the sum of the times for the two sides of the line is (cos T + sqrt[(cos T)^2 + 3])/3 + (-cos T + sqrt[(cos T)^2 + 3])/3
= (2/3)*sqrt[(cos T)^2 + 3]
Part D
The one sided formula can be used three times, once for each side of the triangle.
side time = (cos T + sqrt[(cos T)^2 + 3])/3
For part 1 use T = {30,150, 270} OR {90,210,330}. Both options yield the same sum, I'll show the first option:
T=30 -> sqrt(3)/6+sqrt(15)/6
T=150 -> -sqrt(3)/6+sqrt(15)/6
T=270 -> sqrt(3)/3
Total time = (sqrt(3)/6+sqrt(15)/6) + (-sqrt(3)/6+sqrt(15)/6) + sqrt(3)/3
= sqrt(3)/3 + sqrt(15)/3 = 1.86834 hours or 1 hour, 52 minutes, 6 econds.
Similarly for part 2 use T = {0,120,240} OR {60,180,300}. Both options yield the same sum, I'll show the first option:
T=0 -> 1
T=120 -> -1/6 + sqrt(13)/6
T=240 -> -1/6 + sqrt(13)/6
Total time = 1 + (-1/6 + sqrt(13)/6) + (-1/6 + sqrt(13)/6)
= 2/3 + sqrt(13)/3 = 1.86852 hours or 1 hour, 52 minutes, 7 seconds.
Edited on December 16, 2015, 11:07 pm
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