At noon, ship A is 50 nautical miles due west of ship B. Ship A is sailing west at 19 knots and ship B is sailing north at 24 knots. How fast (in knots) is the distance between the ships changing at 3 PM? (Note: 1 knot is a speed of 1 nautical mile per hour.)

Answer 1

after 3hrs the distance travelled by A increases by 19*3=57 nautical miles.Distance by A equals 50+57=117nm distance by B equals 24*3=72nm.distance between them is sqrt(117^2+72^2) equals 137.4nm

Answer 2

To find the rate of increase, note that the distance x after t hours is given by

x^2 = (50+19t)^2 + (24t)^2
At 3 pm, t=3, and we have
x^2 = 117^2 + 72^2 = 16633
x = 137.4

Now we have to find the derivative:

2x dx/dt = 38(50+19t)+48(24t) = 1874t + 1900

So, at t=3, we have

2(137.4) dx/dt = 1874(3)+1900
dx/dt = 7522/274.8 = 27.4 knots