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Protecting the fleet Parts C. & D. (Posted on 2015-12-15) Difficulty: 3 of 5
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.

No Solution Yet Submitted by Jer    
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Solution solution | Comment 1 of 2
C. In part B, theta was 0° so we could use the Pythagorean theorem. Now we have a triangle to solve given Angle-Side-Side, as we know two sides of a parallelogram of velocities and need one of the diagonals. Fortunately, going in one direction we need the acute case and in the other direction the obtuse case.

If v is the velocity of the boat relative to the fleet,

2^2 = v^2 + 1^2 - 2*v*1*cos(theta+90°)
v^2 - 2*cos(theta+90°)*v + 1 - 4 = 0
v^2 - 2*cos(theta+90°)*v - 3 = 0

v = (2*cos(theta+90°) +/- sqrt(4(cos(theta+90°)^2) + 12)) / 2

and in one direction the v uses the + and in the other the -.

Call the velocities v1 and v2. 

Actually we need the speeds rather than the velocities. This was glossed over in my answer to part B, but here it needs to be explicit. In the left/right legs of part B, theta was zero. And cos(90°) is zero, to the formula here comes out to +sqrt(3) and -sqrt(3), but we don't want time to cancel in opposite directions.

So let s1 = |v1| and s2 = |v2|.

The total time is 1/s1 + 1/s2.

D. i. Each side of the triangle is traversed only once in each cycle, unlike cases A and C. However, the diagonal legs match symmetrically so we could consider them as one diagonal traversed forward and backward, so we use the formula from part C, with theta equal to 60°.

v1 = -sqrt(3) + sqrt(12 + 3) = sqrt(15) - sqrt(3)
v2 = -sqrt(3) - sqrt(15)

s1 = sqrt(15) - sqrt(3)
s2 = sqrt(3) + sqrt(15)

t = 1/s1 + 1/s2 + 1/sqrt(3)

s1 and s2 come out to about 1.07046626931927 and 2.80251707688815 respectively

so  t ~= 1.86834471792544 hours.

  Posted by Charlie on 2015-12-15 18:58:34
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