Peter started driving from City E towards City F at 14 30 at an average speed of 60 km/h. Luis began driving from City E towards City F at 16 20 at an average speed of 110 km/h.

1. At what time did Luis pass Peter on the road?
2. 1¹₁₀ hours after passing Peter, Luis reached City F. At what time did Peter reach City F?

(a)
From 14 30 to 16 20 = 1 h 50 min

Time that Peter travelled before Luis started driving = 1 h 50 min

1 h = 60 min
1 h 50 min = 1⁵⁰₆₀ h = 1⁵₆ h

Distance that Peter travelled before Luis started driving
= 1⁵₆ x 60
= ¹¹₆ x 60
= 110 km

Difference between Luis's and Peter's speed
= 110 - 60
= 50 km/h

Time that Luis took to catch up with Peter
= 110 ÷ 50
= 2.2 h
= 2 h 12 min

2 h 12 min after 16 20 = 18 32

Time that Luis passed Peter on the road = 18 32

(b)
1¹₁₀ h = 1 h 6 min

Total time that Luis travelled
= 1 h 6 min + 2 h 12 min
= 3 h 18 min

3 h 18 min = 3¹⁸₆₀h = 3³₁₀h

Distance between City E and City F
= 3³₁₀ x 110
= 363 km

Time that Peter travelled from City E to City F
= 363 ÷ 60
= 6.05 h
= 6 h 3 min

6 h 3 min after 14 30 = 20 33

Time that Peter reached City F = 20 33

Answer: (a) 18 32; (b) 20 33