Speed and Distance

Solution:
Let the distance between A and B be d.
Let Jack’s speed be x km/hr and Jill’s speed be y km/hr.
Jack starts 1 hour earlier that is at 6AM.

In 1 hour he travels 1*x km or x km.
So distance between them at 7AM is d-x km.
At this time Jill is starting from B.
So when they meet let us say at point p, the time taken is (d – x)/(x+y) which is 9 – 7 or 2 hours.
Or (d-x)/(x+y) =2.
Or d = 3x+2y.
Now after they meet Jack is going on and reaching B.
Also the distance from point P to B is the distance Jill travels in 2 hours which is 2y.
So Jack covers this distance in 2y/x hours.
In this 2y/x hours distance traveled by Jill in the direction of A is y*2y/x = 2y^2/x.

So now after Jack reaches B distance between Jack and Jill is 2y + 2y^2/x. which is 2y(1+y/x).

So time in which Jack catches up with Jill when both are traveling in the direction of A is
{2y(1+y/x)}/(x-y).

Time at which this happens is 11AM.

They are meeting for the first time at 9AM. After this both Jack and Jill are traveling for 2y/x hours before Jack reaches B and after that they are meeting after {2y(1+y/x)}/(x-y) hours.

Or 2y/x + {2y*(1+y/x)}/(x-y) = 11- 9 = 2.

On solving we get x = 3y.

Or d is 9y + 2y is 11y.

Time taken by Jill to cover d is 11y/y is 11 hours.
This is after 7AM .
So Jill reaches A at 6PM.