City T and City U is 260 km apart. Caden left City T for City U at 9.00 a.m. travelling at an average speed of 65 km/h. Fred left City T later than Caden and caught up with him at 11.00 a.m. Fred was travelling at a speed of 104 km/h.

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

(a)
From 9.00 a.m. to 11.00 a.m. = 2 h

Distance that Caden covered when Fred caught up with Caden
= 2 x 65
= 130 km

Time that Fred took to travel 130 km
= 130 ÷ 104
= 1¹₄ h

1 h = 60 min
¹₄ h = ¹₄ x 60 = 15 min
1¹₄ h = 1 h 15 min

Time that Fred left City T:
1 h 15 min before 11.00 a.m. = 9.45 a.m.

(b)
Remaining distance that they had to cover to reach City U
= 260 - 130
= 130 km

Duration that Fred had to travel before reaching City U
= 130 ÷ 104
= 1¹₄ h

Duration that Caden had to travel before reaching City U
= 130 ÷ 65
= 2 h

Duration that Caden took to arrive later than Fred in City U
= 2 - 1¹₄
= ³₄ h

Answer(s): (a) 9.45 a.m.; (b) ³₄ h