City S and City T is 252 km apart. Xavier left City S for City T at 10.00 a.m. travelling at an average speed of 63 km/h. Cole left City S later than Xavier and caught up with him at 12.00 p.m. Cole was travelling at a speed of 72 km/h.

1. At what time did Cole leave City S?
2. How much later did Xavier arrive in City T than Cole? 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 Xavier covered when Cole caught up with Xavier
= 2 x 63
= 126 km

Time that Cole 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 Cole left City S:
1 h 45 min before 12.00 p.m. = 10.15 a.m.

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

Duration that Cole had to travel before reaching City T
= 126 ÷ 72
= 1³₄ h

Duration that Xavier had to travel before reaching City T
= 126 ÷ 63
= 2 h

Duration that Xavier took to arrive later than Cole in City T
= 2 - 1³₄
= ¹₄ h

Answer(s): (a) 10.15 a.m.; (b) ¹₄ h