City S and City T is 384 km apart. Dylan left City S for City T at 9.00 a.m. travelling at an average speed of 64 km/h. Nick left City S later than Dylan and caught up with him at 12.00 p.m. Nick was travelling at a speed of 72 km/h.

1. At what time did Nick leave City S?
2. How much later did Dylan arrive in City T than Nick? Express your answer in h as a fraction or mixed number.

(a)
From 9.00 a.m. to 12.00 p.m. = 3 h

Distance that Dylan covered when Nick caught up with Dylan
= 3 x 64
= 192 km

Time that Nick took to travel 192 km
= 192 ÷ 72
= 2²₃ h

1 h = 60 min
²₃ h = ²₃ x 60 = 40 min
2²₃ h = 2 h 40 min

Time that Nick left City S:
2 h 40 min before 12.00 p.m. = 9.20 a.m.

(b)
Remaining distance that they had to cover to reach City T
= 384 - 192
= 192 km

Duration that Nick had to travel before reaching City T
= 192 ÷ 72
= 2²₃ h

Duration that Dylan had to travel before reaching City T
= 192 ÷ 64
= 3 h

Duration that Dylan took to arrive later than Nick in City T
= 3 - 2²₃
= ¹₃ h

Answer(s): (a) 9.20 a.m.; (b) ¹₃ h