City F and City G is 468 km apart. Caden left City F for City G at 10.00 a.m. travelling at an average speed of 78 km/h. Justin left City F later than Caden and caught up with him at 12.00 p.m. Justin was travelling at a speed of 104 km/h.

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

(a)
From 10.00 a.m. to 12.00 p.m. = 2 h

Distance that Caden covered when Justin caught up with Caden
= 2 x 78
= 156 km

Time that Justin took to travel 156 km
= 156 ÷ 104
= 1¹₂ h

1 h = 60 min
¹₂ h = ¹₂ x 60 = 30 min
1¹₂ h = 1 h 30 min

Time that Justin left City F:
1 h 30 min before 12.00 p.m. = 10.30 a.m.

(b)
Remaining distance that they had to cover to reach City G
= 468 - 156
= 312 km

Duration that Justin had to travel before reaching City G
= 312 ÷ 104
= 3 h

Duration that Caden had to travel before reaching City G
= 312 ÷ 78
= 4 h

Duration that Caden took to arrive later than Justin in City G
= 4 - 3
= 1 h

Answer(s): (a) 10.30 a.m.; (b) 1 h