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 6 knots and
Continued to do so all day. The visibility was 5 nautical miles. Did the ships ever sight each other?
A) No. The closest they ever got to each other was 6.6 nautical miles.
B) Yes. They were within 3 nautical miles of each other.
C) No. The closest they ever got to each other was 5.6 nautical miles.
D) Yes. They were within 4 nautical miles of each other.

Question

Quizplus · Accepted Answer

To find the closest distance the ships got to each other, we need to find their distance at the time when they are closest to each other. Let's call the time when this happens as T. 
Ship A started 15 nautical miles north of ship B and sailed south at 15 knots for T hours.
So, the distance travelled by Ship A in T hours = 15 + 15T nautical miles.
Similarly, Ship B sailed east at 6 knots for T hours. So, the distance travelled by Ship B in T hours = 6T nautical miles.
To find the distance between the ships at time T, we can use the Pythagorean theorem:
Distance between ships at time T = ((15+15T)^2 + (6T)^2)^(1/2) nautical miles.
To find the value of T at which the distance between ships is minimum, we need to minimize the above expression. Taking the derivative and setting it to zero, we get T = 5/6.
Substituting T = 5/6 in the above expression, we get:
Distance between ships at closest approach = ((15+15(5/6))^2 + (6(5/6))^2)^(1/2) = 5.6 nautical miles
Since the visibility is only 5 nautical miles, the ships would not have sighted each other. Therefore, the answer is C.

Solve the Problem. -At Noon, Ship a Was 15 Nautical Miles Due North