At 9 a.m., Riordan started from City R and travelled towards City S and his speed remained constant throughout. At 10 a.m., Tina started her journey from City R towards City S at an average speed of 88 km/h. Tina overtook Riordan at 1 p.m. After overtaking, Tina carried on her journey at the same average speed and reached at City S at 6:15 p.m.
- Find Riordan's average speed in km/h.
- What is the distance between the two cities?
(a)
From 10 a.m. to 1 p.m. = 3 h
Distance that Tina had travelled until Tina overtook Riordan
= 3 x 88
= 264 km
From 9 a.m. to 1 p.m. = 4 h
Riordan's average speed
= 264 ÷ 4
= 66 km/h
(b)
From 10 a.m. to 6:15 p.m. = 8 h 15 min
1 h = 60 min
8 h 15 min = 8
15 60
h = 8
14 h
Distance between the two cities
= 8
14 x 88
=
334 x 88
= 726 km
Answer(s): (a) 66 km/h; (b) 726 km