If they move at uniform velocity, determine their time of meeting & their distance from stop X.

From the given data, it is clear that X and Y are 8km apart.

Car A is travelling from X to Y in 8 mins $(1:44 – 1:36)$

Car B is travelling from Y to X in 10 mins $(1:46 – 1:36)$

Given:

$Ta = 8 min = 480 sec \\ Tb = 10 min = 600 sec \\ Dist = 8 km = 8000 m$

Solution:

$Va = \text {dist/Ta} = 8000/480 = 16.67 m/s \\ Vb = \text {dist/Tb } = 8000/600 = 13.33 m/s $

Let they meet at a distance D from stop X.

So, car A travels D metres and car B will travel (8000-D) metres. But both would have travelled for same time (t).

$Va = D/t \\ 16.67 = D/t .. (i) \\ Vb = (8000-D)/t \\ 13.33 = (8000-D)/t .. (ii) $

From (i) and (ii)

$D= 4445 m $ & $t = 267 sec$