City J and City K is 308 km apart. Caden left City J for City K at 8.00 a.m. travelling at an average speed of 77 km/h. Justin left City J later than Caden and caught up with him at 10.00 a.m. Justin was travelling at a speed of 84 km/h.

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

(a)
From 8.00 a.m. to 10.00 a.m. = 2 h

Distance that Caden covered when Justin caught up with Caden
= 2 x 77
= 154 km

Time that Justin took to travel 154 km
= 154 ÷ 84
= 1⁵₆ h

1 h = 60 min
⁵₆ h = ⁵₆ x 60 = 50 min
1⁵₆ h = 1 h 50 min

Time that Justin left City J:
1 h 50 min before 10.00 a.m. = 8.10 a.m.

(b)
Remaining distance that they had to cover to reach City K
= 308 - 154
= 154 km

Duration that Justin had to travel before reaching City K
= 154 ÷ 84
= 1⁵₆ h

Duration that Caden had to travel before reaching City K
= 154 ÷ 77
= 2 h

Duration that Caden took to arrive later than Justin in City K
= 2 - 1⁵₆
= ¹₆ h

Answer(s): (a) 8.10 a.m.; (b) ¹₆ h