Solve the problem.

At noon, ship A was 15 nautical miles due north of ship B. Ship A was sailing south at 15 knots (nautical miles per hour; a nautical mile is 2000 yards) and continued to do so all day. Ship B was sailing east at and continued to do so all day. The visibility was 5 nautical miles. Did the ships ever sight each other?

Question

Solve the problem.

At noon, ship A was 15 nautical miles due north of ship B. Ship A was sailing south at 15 knots (nautical miles per hour; a nautical mile is 2000 yards) and continued to do so all day. Ship B was sailing east at  and continued to do so all day. The visibility was 5 nautical miles. Did the ships ever sight each other?

BlackTom AI · Accepted Answer

Let me restate the problem to ensure we’re analyzing the same scenario: At noon, A is 15 nautical miles due north of B. A sails south at 15 knots, while B sails east at 9 knots. Visibility is 5 nautical miles. We’re asked whether the ships ever come within sight of each other.

Option 1: No. The closest they ever got to each other was 7.7 nautical miles.
To evaluate this, model the positions as a function of time t (in hours) after noon. Let B be at the origin (0,0). Then A, starting 15 miles north of B, is at coordinates (0, 15 − 15t) since A moves south at 15 knots. B moves east at 9 knots, so B is at (9t, 0). The separation distance s(t) between the ships is
s(t) = sqrt[(9t − 0)^2 + (0 − (15 − 15t))^2] = sqrt[(9t)^2 + (15 − 15t)^2].
Expanding, s(t)^2 = 81t^2 + (15 − 15t)^2 = 81t^2 + 225 − 450t + 225t^2 = 306t^2 − 450t + 225.
To find the closest approach, minimize s(t). Since s(t) is minimized where s(t)^2 is minimized, take the derivative of s(t)^2 with respect to t and set it to zero:
d/dt [306t^2 − 450t + 225] = 612t − 450 = 0 ⇒ t = 450/612 ≈ 0.7353 hours ≈ 44.1 minutes.
Plugging back: s^2 ≈ 306(0.7353)^2 − 450(0.7353) + 225 ≈ 59.1, so s ≈ sqrt(59.1) ≈ 7.7 nautical miles.
Thus the minimum separation is about 7.7 nm, which is larger than the 5 nm visibility. So they never sight each other.

Option 2: Yes. They were within 3 nautical miles of each other.
This would require s(t) ≤ 3 at some time t ≥ 0. But the computed minimum is about 7.7 nm, which is well above 3, so this option is incorrect.

Option 3: No. The closest they ever got to each other was 8.7 nautical miles.
The math above gives a minimum distance near 7.7 nm, not 8.7 nm. Since 8.7 nm is farther than the actual minimum, this statement misstates the closest approach and is therefore incorrect.

Option 4: Yes. They were within 4 nautical miles of each other.
Again, the closest approach is about 7.7 nm, which is greater than 4 nm. Therefore, this option is false.

Across all options, the key steps were setting up the relative position, expressing the separation as a function of time, finding the time of closest approach by minimizing s(t)^2, and comparing the minimum distance to the visibility limit. The computed minimum distance of roughly 7.7 nm exceeds the 5 nm visibility, so sighting could not occur under these conditions.

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