City S and City T are 384 km apart. At 3.49 p.m., Jack is travelling at a uniform speed left City S for City T while Zeph set off from City T to City S along the same road at a uniform speed, which was 12 km/h slower than that of Jack. The two met at 5.49 p.m.
- Find the speed of Jack.
- If Zeph continued to travel at the same speed, how long would it take for him to reach his destination after the two met? Express your answer in mixed number.
(a)
Time taken for Jack and Zeph to travel from 3.49 p.m. to 5.49 p.m. = 2 h
Average speed of Jack and Zeph
= 384 ÷ 2
= 192 km/h
Jack's speed
= (192 + 12) ÷ 2
= 204 ÷ 2
= 102 km/h
(b)
Zeph's speed
= 102 - 12
= 90 km/h
Distance that Zeph travelled in 2 h
= 2 x 90
= 180 km
Remaining distance that Zeph needed to travel
= 384 - 180
= 204 km
Time that Zeph needed to reach his destination after the two met
= 204 ÷ 90
= 2
2490 = 2
415 h
Answer(s): (a) 102 km/h; (b) 2
415 h