An airplane is initially traveling at \( 119 \mathrm{~m} / \mathrm{s} \) when begins to accelerate at \( 25.6 \mathrm{~m} / \mathrm{s}^{2} \) in order to reach \( 233 \mathrm{~m} / \mathrm{s} \). How far does the plane travel during this time?

Given: - Initial velocity of the airplane, \( u = 119 \, \text{m/s} \) - Acceleration of the airplane, \( a = 25.6 \, \text{m/s}^2 \) - Final velocity of the airplane, \( v = 233 \, \text{m/s} \) We can use the kinematic equation to find the distance traveled by the airplane during this time: \[ v^2 = u^2 + 2as \] Substitute the given values into the equation: \[ 233^2 = 119^2 + 2 \times 25.6 \times s \] Now, we can solve for the distance \( s \) traveled by the airplane. Solve the equation by following steps: - step0: Solve for \(s\): \(233^{2}=119^{2}+2\times 25.6s\) - step1: Simplify: \(233^{2}=14161+51.2s\) - step2: Swap the sides: \(14161+51.2s=233^{2}\) - step3: Calculate: \(14161+51.2s=54289\) - step4: Move the constant to the right side: \(51.2s=54289-14161\) - step5: Subtract the numbers: \(51.2s=40128\) - step6: Divide both sides: \(\frac{51.2s}{51.2}=\frac{40128}{51.2}\) - step7: Divide the numbers: \(s=\frac{3135}{4}\) The distance traveled by the airplane during this time is \( 783.75 \, \text{m} \).