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
[tex]\begin{gathered} v^2_f=v^2_1+2ax \\ where \\ a\text{ is the acceleration} \\ \text{ x is the distance} \end{gathered}[/tex]replace
[tex]\begin{gathered} v^2_f=v^2_1+2ax \\ v^2_f=(0)^2+2\cdot15\frac{m}{s^2}\cdot(3.7\cdot1000m) \\ v^2_f=111000 \\ v_f=\sqrt[]{111000} \\ v_f=333.166\text{ }\frac{m}{s} \end{gathered}[/tex]so, the final velocity is
333.16 meters per second ( wow¡, that's fast)
I hope this helps you