1. **State the problem:** Two carts A and B are connected by a rope 39 ft long passing over a pulley P. The pulley P is 12 ft above point Q on the horizontal surface. Cart A is moving away from Q at 2 ft/s, and we want to find how fast cart B is moving toward Q when cart A is 5 ft from Q. 2. **Define variables:** Let $x$ be the distance from cart A to Q, and $y$ be the distance from cart B to Q. The rope length constraint relates $x$, $y$, and the vertical distance 12 ft. 3. **Express the rope length:** The rope length consists of the segment from cart A to pulley P, and from pulley P to cart B. Using the Pythagorean theorem for each side: $$\sqrt{x^2 + 12^2} + \sqrt{y^2 + 12^2} = 39$$ 4. **Differentiate with respect to time $t$:** $$\frac{d}{dt}\left(\sqrt{x^2 + 144}\right) + \frac{d}{dt}\left(\sqrt{y^2 + 144}\right) = 0$$ Using chain rule: $$\frac{x}{\sqrt{x^2 + 144}} \frac{dx}{dt} + \frac{y}{\sqrt{y^2 + 144}} \frac{dy}{dt} = 0$$ 5. **Known values:** - $x = 5$ ft - $\frac{dx}{dt} = 2$ ft/s (cart A moving away from Q) 6. **Find $y$ when $x=5$:** $$\sqrt{5^2 + 144} + \sqrt{y^2 + 144} = 39$$ $$\sqrt{25 + 144} + \sqrt{y^2 + 144} = 39$$ $$\sqrt{169} + \sqrt{y^2 + 144} = 39$$ $$13 + \sqrt{y^2 + 144} = 39$$ $$\sqrt{y^2 + 144} = 26$$ $$y^2 + 144 = 26^2 = 676$$ $$y^2 = 676 - 144 = 532$$ $$y = \sqrt{532} \approx 23.05 \text{ ft}$$ 7. **Substitute values into differentiated equation:** $$\frac{5}{13} \times 2 + \frac{23.05}{26} \frac{dy}{dt} = 0$$ Calculate first term: $$\frac{5}{13} \times 2 = \frac{10}{13} \approx 0.7692$$ So: $$0.7692 + 0.8865 \frac{dy}{dt} = 0$$ 8. **Solve for $\frac{dy}{dt}$:** $$\frac{dy}{dt} = -\frac{0.7692}{0.8865} \approx -0.868$$ The negative sign means cart B is moving toward Q. 9. **Final answer:** Cart B is moving toward Q at approximately **0.87 ft/s**.