I assume "first phase" refers to the first 6.00 seconds of the dash.
Since the sprinter starts from rest and accelerates uniformly, he covers a distance x after time t of
x = 1/2 at²
where a is his acceleration.
The sprinter's velocity v after time t is
v = at
so that if he attains a speed of 14.7 m/s after 6.00 s, he does so with acceleration a such that
14.7 m/s = a (6.00 s) ⇒ a = 2.45 m/s²
Then in the first 6.00 seconds, the sprinter travels
x = 1/2 (2.45 m/s²) (6.00 s)² = 44.1 m
