an airplane is initially traveling at 119 m/s when begins to accelerate at 25.6 m/s² in order to reach 233…

Explanation:

Step1: Identify the relevant kinematic equation

We use $v^{2}=v_{0}^{2}+2a\Delta x$, where $v$ is final - velocity, $v_{0}$ is initial - velocity, $a$ is acceleration and $\Delta x$ is displacement.

Step2: Rearrange the equation for $\Delta x$

$\Delta x=\frac{v^{2}-v_{0}^{2}}{2a}$

Step3: Substitute the given values

Given $v_{0} = 119\ m/s$, $v = 233\ m/s$, and $a=25.6\ m/s^{2}$. $\Delta x=\frac{(233)^{2}-(119)^{2}}{2\times25.6}$ First, calculate $(233)^{2}=54289$ and $(119)^{2}=14161$. Then, $(233)^{2}-(119)^{2}=54289 - 14161=40128$. And $2\times25.6 = 51.2$. So, $\Delta x=\frac{40128}{51.2}=784\ m$.

Answer:

$784$