City V and City W is 567 km apart. Oscar left City V for City W at 7.00 a.m. travelling at an average speed of 63 km/h. Eric left City V later than Oscar and caught up with him at 11.00 a.m. Eric was travelling at a speed of 112 km/h.

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

(a)
From 7.00 a.m. to 11.00 a.m. = 4 h

Distance that Oscar covered when Eric caught up with Oscar
= 4 x 63
= 252 km

Time that Eric took to travel 252 km
= 252 ÷ 112
= 2¹₄ h

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

Time that Eric left City V:
2 h 15 min before 11.00 a.m. = 8.45 a.m.

(b)
Remaining distance that they had to cover to reach City W
= 567 - 252
= 315 km

Duration that Eric had to travel before reaching City W
= 315 ÷ 112
= 2¹³₁₆ h

Duration that Oscar had to travel before reaching City W
= 315 ÷ 63
= 5 h

Duration that Oscar took to arrive later than Eric in City W
= 5 - 2¹³₁₆
= 2³₁₆ h

Answer(s): (a) 8.45 a.m.; (b) 2³₁₆ h