Cricket Puzzle

Some boys are playing cricket. Wasim is at the striker' s end of the pitch and
Rachit is at the non-striker' s end. Wasim hits the ball far into the outfield and the
batsmen start running. Each of them runs at a constant speed, though their speeds are
not necessarily equal. They first cross each other at a distance of 9 ft from the striker' s
end while running the first run. They immediately turn back for the second run and cross each other at a distance of 3 ft from the non-striker' s end while running for the
second run. After this, they turn back for the third run and so on. What is the total
distance (In feet) Rachit has run when they cross each other for the third time?

Let Wasim and Rachit are running at a speed of 'S1','S2' repectively.
Assume that the length of the pitch is 'x'

When they crossed each other for 1st time,
Wasim covered a distance of 9 feet and
Rachit covered (x-9) feet.
Since both are running at their own individual constant speeds, the distance covered by them is proportional to their speed

S2/S1 = (x-9)/9 --- (1)

When they crossed each other for 2nd time,
Wasim covered a distance of (x+3) feet and
Rachit covered (2x-3) feet.

S2/S1 = (2x-3)/(x+3) --- (2)

Solving both the equation we get x = 24 feet.

From the explanation given in the above post I am taking 'y' as the distance from the non striker end where Wasim and Rachit crossed fo the 3 rd time.
Since wasim is still trying to complete his 2nd run he covered a distance of (x+y) feet
Rachit already completed his 2nd run and going to complete his 3rd run he covered a distace of (2x+y) feet

S2/S1 = (2x+y)/(x+y) ..... (3)

Solving the equations and substituting the value of 'x' we get 60 feet as the answer.