1. **State the problem:** We are given two ships, A and B, initially 150 km apart horizontally at noon. Ship A sails south at 20 km/h and ship B sails north at 40 km/h. We want to find the rate at which the distance between the ships is changing at 4:00 pm, i.e., after 4 hours. 2. **Set variables:** Let $t$ be the time in hours after noon. - Ship A's southward distance after $t$ hours is $20t$ km. - Ship B's northward distance after $t$ hours is $40t$ km. 3. **Express the distance between the ships at time $t$:** - The horizontal distance between ships remains constant at 150 km. - The total vertical separation is the sum of distances traveled by both ships: $20t + 40t = 60t$ km. Thus, the distance $s$ between ships is given by Pythagoras theorem: $$s = \sqrt{150^2 + (60t)^2} = \sqrt{22500 + 3600t^2}$$ 4. **Find the rate of change of distance $s$ with respect to time $t$: ** Differentiate both sides with respect to $t$: $$\frac{ds}{dt} = \frac{1}{2\sqrt{22500 + 3600t^2}} \cdot 7200t = \frac{7200t}{2s} = \frac{3600t}{s}$$ 5. **Evaluate at $t=4$ hours:** Calculate $s$ at $t=4$: $$s = \sqrt{22500 + 3600 \times 16} = \sqrt{22500 + 57600} = \sqrt{80100} \approx 283.07\text{ km}$$ Calculate $$\frac{ds}{dt}$$: $$\frac{ds}{dt} = \frac{3600 \times 4}{283.07} \approx \frac{14400}{283.07} \approx 50.87\text{ km/h}$$ **Final answer:** The distance between the ships is increasing at approximately $50.87$ km/h at 4:00 pm.