City V and City W is 576 km apart. Will left City V for City W at 7.00 a.m. travelling at an average speed of 72 km/h. Justin left City V later than Will and caught up with him at 10.00 a.m. Justin was travelling at a speed of 96 km/h.

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

(a)
From 7.00 a.m. to 10.00 a.m. = 3 h

Distance that Will covered when Justin caught up with Will
= 3 x 72
= 216 km

Time that Justin took to travel 216 km
= 216 ÷ 96
= 2¹₄ h

1 h = 60 min
¹₄ h = ¹₄ x 60 = 15 min
2¹₄ h = 2 h 15 min

Time that Justin left City V:
2 h 15 min before 10.00 a.m. = 7.45 a.m.

(b)
Remaining distance that they had to cover to reach City W
= 576 - 216
= 360 km

Duration that Justin had to travel before reaching City W
= 360 ÷ 96
= 3³₄ h

Duration that Will had to travel before reaching City W
= 360 ÷ 72
= 5 h

Duration that Will took to arrive later than Justin in City W
= 5 - 3³₄
= 1¹₄ h

Answer(s): (a) 7.45 a.m.; (b) 1¹₄ h