City T and City U is 462 km apart. Sean left City T for City U at 9.00 a.m. travelling at an average speed of 77 km/h. Peter left City T later than Sean and caught up with him at 11.00 a.m. Peter was travelling at a speed of 88 km/h.
- At what time did Peter leave City T?
- How much later did Sean arrive in City U than Peter? Express your answer in h as a fraction or mixed number.
(a)
From 9.00 a.m. to 11.00 a.m. = 2 h
Distance that Sean covered when Peter caught up with Sean
= 2 x 77
= 154 km
Time that Peter took to travel 154 km
= 154 ÷ 88
= 1
34 h
1 h = 60 min
34 h =
34 x 60 = 45 min
1
34 h = 1 h 45 min
Time that Peter left City T:
1 h 45 min before 11.00 a.m. = 9.15 a.m.
(b)
Remaining distance that they had to cover to reach City U
= 462 - 154
= 308 km
Duration that Peter had to travel before reaching City U
= 308 ÷ 88
= 3
12 h
Duration that Sean had to travel before reaching City U
= 308 ÷ 77
= 4 h
Duration that Sean took to arrive later than Peter in City U
= 4 - 3
12=
12 h
Answer(s): (a) 9.15 a.m.; (b)
12 h
