We use the kinematic equation,
[tex]s= u t +\frac{1}{2}a t^2[/tex]
Here, s is the distance traveled by elevator u is the initial velocity and t is time.
As elevator accelerates uniformly from rest to a speed of 2.5 m/s in 12 s, so
[tex]a = \frac{v_{f}- v_{i} }{t} = \frac{2.5 m/s-0}{12 s} = 0.208 \ m/s^2[/tex]
Thus, the distance the elevator travels during 12 s time is
[tex]s= 0+\frac{1}{2} 0.208 \ m/s^2(12)^2 = 14.97 m \simeq 15 m[/tex]
