At noon, ship A is 150 km west of ship B. Ship A is sailing east at 25 km/h and ship B is sailing north at 20 km/h. How fast (in km/hr) is the distance between the ships changing at 4:00 p.m.? (Round your answer to three decimal places.)

To solve this problem, we can use the Pythagorean theorem and the concept of related rates.

Step 1: Set up the problem Let's denote the distance of ship A from the point directly north of ship B as x (km), and the distance of ship B from that point as y (km). The distance between the two ships is z (km).

At noon, ship A is 150 km west of ship B, so x = 150 km. Ship A is sailing east at 25 km/h and ship B is sailing north at 20 km/h.

Step 2: Apply the Pythagorean theorem According to the Pythagorean theorem, we have z^2 = x^2 + y^2.

Step 3: Differentiate both sides with respect to time (t) We get 2z(dz/dt) = 2x(dx/dt) + 2y(dy/dt).

Step 4: Plug in the known values At 4:00 p.m., 4 hours have passed. So, x = 150 km - 25 km/h * 4 h = 50 km (since ship A is sailing east, reducing x), and y = 20 km/h * 4 h = 80 km.

We also need to find z at this time. Using the Pythagorean theorem, z = sqrt(x^2 + y^2) = sqrt((50 km)^2 + (80 km)^2) = 94.339 km (rounded to three decimal places).

Step 5: Solve for dz/dt Now we can substitute all the known values into the differentiated equation: 294.339(dz/dt) = 250(-25) + 28020.

Solving for dz/dt, we get dz/dt = [250(-25) + 28020] / (2*94.339) = 9.899 km/h (rounded to three decimal places).

So, the distance between the ships is changing at a rate of 9.899 km/h at 4:00 p.m.