To know at what time the two person will meet (when each person travelling in opposite directioon), we need to know what is the distance between them at the starting point and their speed of travel.
In the question, Ryan starts to turn back at 7:35pm, this will be our base of starting point. So we need to know at 7:35pm, what is the distance between Keith and Ryan.
We have two points A and B, so we need to find out, at 7:35pm:
- where is Keith, i.e. how many km has Keith travelled from A.
- where is Ryan, i.e. how many km has Ryan travelled from A.
Then only we can calculate the difference to find out the distance of Keith from Ryan at 7:35pm.
The formula we need to calculate distance given speed and time is:
distance travelled = time travelled x speed of travel
Keith travel from A towards B at the speed of 90 km/h.
30 min later at 2:35pm Keith distance from A is 0.5hr x 90km/hr = 45 km from A
Ryan starts driving from Town A to Town B, at 2:35pm at a constant speed of 120 km/hr.
At 7:35pm, the distance that Ryan travelled would be (7:35 - 2:35) hr x 120 km/hr = 5 hr x 120km/hr = 600 km.
At 7:35pm Keith would have travelled for (7:35pm - 2:05) = 5.5 hr , so his distance from A would be 5.5 hr x 90 km/hr = 495 km
From the above analysis, we can calculate the distance between Keith and Ryan (the distance from point C to point D) at 7:35 pm is 600 km - 495 km = 105 km
So, we have the base of the starting point when Ryan starts to turn back.
Distance between them can be illustrated with point C and point D above.
At 7:35pm, when Keith travel from C towards D at the speed of 90 km/hr, and Ryan travel from D towards C at the speed of 85 km/hr, at what time they will meet?
Let t be the time they meet, so the formula is
Keith's speed x t + Ryan's speed x t = distance from C to D
90 km/hr x t hr + 85 km/hr x t hr = 105 km
90t + 85t = 105
175 t = 105
t = 105/175 = 0.6 hr = 36 min
Thus, they will meet 0.6 hr later from 7:35pm, which is 7:35 pm + 36 min = 8:11 pm
The answer to the question, at 8:11 pm Ryan will pass Keith again.
