1. **State the problem:** Car A moves due East at 10 m/s, and car B moves South 25° East at 20 m/s. The initial distance between them is 40 km. We want to find how fast the distance between them is increasing after one hour. 2. **Convert speeds to consistent units:** Since the distance is in km and time in hours, convert speeds from m/s to km/hr: $$10\ \text{m/s} = 10 \times 3.6 = 36\ \text{km/hr}$$ $$20\ \text{m/s} = 20 \times 3.6 = 72\ \text{km/hr}$$ 3. **Set up coordinate system and variables:** Let car A start at origin moving east, so its position after time $t$ hours is: $$x_A = 36t, \quad y_A = 0$$ Car B moves South 25° East, so its velocity components are: $$v_{Bx} = 72 \cos 25^\circ, \quad v_{By} = -72 \sin 25^\circ$$ Position of B after time $t$: $$x_B = 72t \cos 25^\circ, \quad y_B = -72t \sin 25^\circ$$ 4. **Distance between cars at time $t$:** $$D(t) = \sqrt{(x_A - x_B)^2 + (y_A - y_B)^2}$$ Initial distance $D(0) = 40$ km. 5. **Calculate $D(t)$ and differentiate:** $$D(t)^2 = (36t - 72t \cos 25^\circ)^2 + (0 + 72t \sin 25^\circ)^2$$ Simplify: $$D(t)^2 = t^2 \left[(36 - 72 \cos 25^\circ)^2 + (72 \sin 25^\circ)^2\right]$$ 6. **Find $D(1)$:** Calculate inside the bracket: $$a = 36 - 72 \cos 25^\circ$$ $$b = 72 \sin 25^\circ$$ Using $\cos 25^\circ \approx 0.9063$, $\sin 25^\circ \approx 0.4226$: $$a = 36 - 72 \times 0.9063 = 36 - 65.25 = -29.25$$ $$b = 72 \times 0.4226 = 30.43$$ Then: $$D(1)^2 = (-29.25)^2 + (30.43)^2 = 855.56 + 925.98 = 1781.54$$ $$D(1) = \sqrt{1781.54} \approx 42.21\ \text{km}$$ 7. **Differentiate $D(t)^2$ with respect to $t$ to find rate of change:** $$2D(t) \frac{dD}{dt} = 2t \left[a^2 + b^2\right]$$ At $t=1$: $$2 \times 42.21 \times \frac{dD}{dt} = 2 \times 1 \times 1781.54$$ $$\frac{dD}{dt} = \frac{1781.54}{42.21} \approx 42.21\ \text{km/hr}$$ **Final answer:** The distance between the cars is increasing at approximately **42.21 km/hr** after one hour.