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A table representing data of velocities at different times of an Airbus A320 jetliner is given here.

The take off speed of the jetliner is, $u=80m/s$

The accelerations of the jetliner at same time interval is the same according to the table as shown below-

$a=\frac{23}{10}=\frac{46}{20}=2.3m/s^2$

So the acceleration of the jetliner is constant during take off.

We can determine the time when the wheels leave the ground by using the formula$v=u+at$

Consider the wheels leave the ground at time $t$

Therefore,

$$\begin{gathered} v=0+at \\ 80=2.3t \\ t=\frac{80}{2.3}s \\ t=34.78s \end{gathered}$$

For aborted takeoff, the runway must be three times the takeoff distance to takeoff safely.

The takeoff distance can be calculated by the equation $s=ut+\frac{1}{2}a{t}^2$

Initial speed of the jetliner is, $u=0$

Therefore, the distance of takeoff is,

$$\begin{gathered} s=0+\left(0.5脳2.3脳34.{78}^2\right)m \\ s=1391.1m \end{gathered}$$

Considering the safety of taking off, the runway should be, $$\begin{gathered} x\text{'}=\left(3脳1391.1\right)m \\ x\text{'}=4173.3m \\ x\text{'}=2.59mi \end{gathered}$$

Hence, it is not safe to takeoff in a $2.5mi$long runway as at least the runway should be$2.59mi$long.