1. **State the problem:** Two ships start from the same point. The Queen of May sails due east at 10 knots, and the Blue Star sails 30° west of north at 15 knots. We want to find how fast the distance between them is increasing after 5 hours. 2. **Set up the problem:** Let the position of the Queen of May after time $t$ hours be $Q(t)$ and the Blue Star be $B(t)$. The Queen of May moves east at 10 knots, so $$Q(t) = (10t, 0).$$ The Blue Star moves 30° west of north at 15 knots. We decompose its velocity into components: - North component: $15 \cos 30^\circ = 15 \times \frac{\sqrt{3}}{2} = \frac{15\sqrt{3}}{2}$ - West component: $15 \sin 30^\circ = 15 \times \frac{1}{2} = 7.5$ Since west is negative x-direction and north is positive y-direction, $$B(t) = (-7.5t, \frac{15\sqrt{3}}{2} t).$$ 3. **Distance between ships:** The vector from Queen of May to Blue Star is $$D(t) = B(t) - Q(t) = (-7.5t - 10t, \frac{15\sqrt{3}}{2} t - 0) = (-17.5t, \frac{15\sqrt{3}}{2} t).$$ The distance between them is $$d(t) = \sqrt{(-17.5t)^2 + \left(\frac{15\sqrt{3}}{2} t\right)^2} = t \sqrt{17.5^2 + \left(\frac{15\sqrt{3}}{2}\right)^2}.$$ 4. **Calculate the magnitude inside the square root:** $$17.5^2 = 306.25,$$ $$\left(\frac{15\sqrt{3}}{2}\right)^2 = \frac{225 \times 3}{4} = \frac{675}{4} = 168.75,$$ So, $$\sqrt{306.25 + 168.75} = \sqrt{475}.$$ 5. **Distance function:** $$d(t) = t \sqrt{475}.$$ 6. **Rate of separation:** The rate of change of distance is the derivative $$\frac{dd}{dt} = \sqrt{475}.$$ 7. **Evaluate at $t=5$ hours:** Since the rate is constant, $$\frac{dd}{dt} = \sqrt{475} \approx 21.794.$$ **Final answer:** The ships are separating at approximately **21.794** knots after 5 hours.