City S and City T is 372 km apart. Luke left City S for City T at 10.00 a.m. travelling at an average speed of 62 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 93 km/h.
- At what time did Eric leave City S?
- How much later did Luke arrive in City T than Eric? Express your answer in h as a fraction or mixed number.
(a)
From 10.00 a.m. to 2.00 p.m. = 4 h
Distance that Luke covered when Eric caught up with Luke
= 4 x 62
= 248 km
Time that Eric took to travel 248 km
= 248 ÷ 93
= 2
23 h
1 h = 60 min
23 h =
23 x 60 = 40 min
2
23 h = 2 h 40 min
Time that Eric left City S:
2 h 40 min before 2.00 p.m. = 11.20 a.m.
(b)
Remaining distance that they had to cover to reach City T
= 372 - 248
= 124 km
Duration that Eric had to travel before reaching City T
= 124 ÷ 93
= 1
13 h
Duration that Luke had to travel before reaching City T
= 124 ÷ 62
= 2 h
Duration that Luke took to arrive later than Eric in City T
= 2 - 1
13=
23 h
Answer(s): (a) 11.20 a.m.; (b)
23 h