City Z and City A is 252 km apart. Peter left City Z for City A at 7.00 a.m. travelling at an average speed of 63 km/h. Eric left City Z later than Peter and caught up with him at 9.00 a.m. Eric was travelling at a speed of 72 km/h.

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

(a)
From 7.00 a.m. to 9.00 a.m. = 2 h

Distance that Peter covered when Eric caught up with Peter
= 2 x 63
= 126 km

Time that Eric took to travel 126 km
= 126 ÷ 72
= 1³₄ h

1 h = 60 min
³₄ h = ³₄ x 60 = 45 min
1³₄ h = 1 h 45 min

Time that Eric left City Z:
1 h 45 min before 9.00 a.m. = 7.15 a.m.

(b)
Remaining distance that they had to cover to reach City A
= 252 - 126
= 126 km

Duration that Eric had to travel before reaching City A
= 126 ÷ 72
= 1³₄ h

Duration that Peter had to travel before reaching City A
= 126 ÷ 63
= 2 h

Duration that Peter took to arrive later than Eric in City A
= 2 - 1³₄
= ¹₄ h

Answer(s): (a) 7.15 a.m.; (b) ¹₄ h