City A and City B is 540 km apart. Sean left City A for City B at 8.00 a.m. travelling at an average speed of 60 km/h. Wesley left City A later than Sean and caught up with him at 12.00 p.m. Wesley was travelling at a speed of 96 km/h.

1. At what time did Wesley leave City A?
2. How much later did Sean arrive in City B than Wesley? Express your answer in h as a fraction or mixed number.

(a)
From 8.00 a.m. to 12.00 p.m. = 4 h

Distance that Sean covered when Wesley caught up with Sean
= 4 x 60
= 240 km

Time that Wesley took to travel 240 km
= 240 ÷ 96
= 2¹₂ h

1 h = 60 min
¹₂ h = ¹₂ x 60 = 30 min
2¹₂ h = 2 h 30 min

Time that Wesley left City A:
2 h 30 min before 12.00 p.m. = 9.30 a.m.

(b)
Remaining distance that they had to cover to reach City B
= 540 - 240
= 300 km

Duration that Wesley had to travel before reaching City B
= 300 ÷ 96
= 3¹₈ h

Duration that Sean had to travel before reaching City B
= 300 ÷ 60
= 5 h

Duration that Sean took to arrive later than Wesley in City B
= 5 - 3¹₈
= 1⁷₈ h

Answer(s): (a) 9.30 a.m.; (b) 1⁷₈ h