1. **Problem statement:** Alan cycled from P to Q (64 km) starting at 10:00, stopped for a break, then continued to arrive at 15:00. Brendan started at 12:00 from P to Q, passing Alan at 14:00. Colin cycled from Q to P starting at 10:00 and arrived at 15:00. 2. **(a) Time Alan stopped:** From the graph, Alan stopped at 11:30. 3. **(b) Alan's constant speed:** Alan cycled 24 km in 1.5 hours before stopping. Formula: $$S=\frac{D}{T}$$ Calculate speed: $$S=\frac{24}{1.5}=16\text{ km/h}$$ 4. **(c)(ii) Brendan's speed:** Brendan left at 12:00 and passed Alan at 14:00. At 14:00, Alan's distance: Alan cycled 24 km before break, then resumed at 16 km/h for 1.5 hours (from 12:30 to 14:00), so distance after break at 14:00: $$24 + 16 \times 1.5 = 24 + 24 = 48\text{ km}$$ Brendan cycled from 12:00 to 14:00 (2 hours) and covered 48 km. Brendan's speed: $$S=\frac{48}{2}=24\text{ km/h}$$ 5. **(d)(ii) Colin's speed:** Colin cycled 64 km from Q to P in 5 hours (10:00 to 15:00). Speed: $$S=\frac{64}{5}=12.8\text{ km/h}$$ 6. **(d)(ii) Time Colin and Brendan passed each other:** Let $t$ be hours after 10:00 when they meet. Brendan starts at 12:00 (2 hours after 10:00), so Brendan's travel time at meeting is $t-2$. Distance Brendan covers from P: $$24(t-2)$$ Distance Colin covers from Q: $$64 - 12.8t$$ They meet when distances sum to 64: $$24(t-2) + 64 - 12.8t = 64$$ Simplify: $$24t - 48 + 64 - 12.8t = 64$$ $$11.2t + 16 = 64$$ $$11.2t = 48$$ $$t = \frac{48}{11.2} = 4.2857\text{ hours}$$ Convert to time: $$10:00 + 4.2857\text{ hours} = 14:17\text{ (approx)}$$ **Final answers:** (a) Alan stopped at 11:30 am. (b) Alan's speed: 16 km/h. (c)(ii) Brendan's speed: 24 km/h. (d)(ii) Colin's speed: 12.8 km/h. (d)(ii) Colin and Brendan passed each other at approximately 14:17.