Step-by-step explanation
an aircraft stars from the rest and accelerates at 15 m/s^2 until it takes off the end of the trip,what was the speed at the end of the 3.7 km way
use

replace
![\begin{gathered} v^2_f=v^2_1+2ax \\ v^2_f=(0)^2+2\cdot15(m)/(s^2)\cdot(3.7\cdot1000m) \\ v^2_f=111000 \\ v_f=\sqrt[]{111000} \\ v_f=333.166\text{ }(m)/(s) \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/9l6b399t96ib1a3idjra7qnd0dh3p7id2s.png)
so, the final velocity is
333.16 meters per second ( wow¡, that's fast)
I hope this helps you