Answer:
[tex]1650\:\mathrm{m}[/tex]
Explanation:
We can use the following kinematics equations to solve this problem:
[tex]v_f=v_i+at,\\{v_f}^2={v_i}^2+2a\Delta x[/tex].
Using the first one to solve for acceleration:
[tex]132=88+a(15),\\15a=44,\\a=\frac{44}{15}=2.9\bar{3}\:\mathrm{m/s^2}[/tex].
Now we can use the second equation to solve for the distance travelled by the airplane:
[tex]132^2=88^2+2\cdot2.9\bar{3}\cdot \Delta x,\\\Delta x= \frac{9680}{2\cdot2.9\bar{3}},\\\Delta x =\fbox{$ 1650\:\mathrm{m}$}[/tex](three significant figures).