City S and City T is 560 km apart. Luke left City S for City T at 11.00 a.m. travelling at an average speed of 70 km/h. Eric left City S later than Luke and caught up with him at 2.00 p.m. Eric was travelling at a speed of 72 km/h.

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

(a)
From 11.00 a.m. to 2.00 p.m. = 3 h

Distance that Luke covered when Eric caught up with Luke
= 3 x 70
= 210 km

Time that Eric took to travel 210 km
= 210 ÷ 72
= 2¹¹₁₂ h

1 h = 60 min
¹¹₁₂ h = ¹¹₁₂ x 60 = 55 min
2¹¹₁₂ h = 2 h 55 min

Time that Eric left City S:
2 h 55 min before 2.00 p.m. = 11.05 a.m.

(b)
Remaining distance that they had to cover to reach City T
= 560 - 210
= 350 km

Duration that Eric had to travel before reaching City T
= 350 ÷ 72
= 4³¹₃₆ h

Duration that Luke had to travel before reaching City T
= 350 ÷ 70
= 5 h

Duration that Luke took to arrive later than Eric in City T
= 5 - 4³¹₃₆
= ⁵₃₆ h

Answer(s): (a) 11.05 a.m.; (b) ⁵₃₆ h