v 0 = 0, v to = v 0 + a t = a t
x = a t² /2
2500 = 1.3 t² / 2
1.3 t² = 5000
t² = 5000 : 1.3 = 3,846.15
t = √ 3,846.15 ≈ 62 s
v to = 1.3 m/s² · 62 s = 80.625 m / s > 75 m/s ( takeoff speed )
Answer: Yes, the plane will be able to use this runway safely.