To find the final velocity of the airplane, we can use one of the kinematic equations. The equation that relates initial velocity, acceleration, distance, and final velocity is:

$v_f^2 = v_i^2 + 2ad$

Where:

• $v_f$ is the final velocity,
• $v_i$ is the initial velocity (21.59 m/s),
• $a$ is the acceleration (4.00 m/s$^2$),
• $d$ is the distance traveled (599.62 m).

Plugging in the values:

$v_f^2 = (21.59)^2 + 2 \times 4.00 \times 599.62$

Now, calculate step by step:

1. Square the initial velocity: $21.59^2 = 466.9281 \text{ m}^2/\text{s}^2$

2. Multiply acceleration by the distance: $2 \times 4.00 \times 599.62 = 4796.96 \text{ m}^2/\text{s}^2$

3. Add the two results: $v_f^2 = 466.9281 + 4796.96 = 5263.8881 \text{ m}^2/\text{s}^2$

4. Finally, take the square root to find the final velocity: $v_f = \sqrt{5263.8881} \approx 72.53 \text{ m/s}$

So, the final velocity of the airplane is approximately 72.53 m/s.

Would you like any more details or have any questions? Here are some related questions you might find interesting:

1. How long does it take the airplane to reach this final velocity?
2. What is the total time taken for the airplane to travel the 599.62 meters?
3. If the airplane continued to accelerate at the same rate, what would its velocity be after an additional 200 meters?
4. How much distance would the airplane cover if it started from rest under the same acceleration until reaching 72.53 m/s?
5. What would be the final velocity if the airplane only accelerated over 400 meters?

Tip: When dealing with kinematic equations, make sure to carefully identify which quantities are given and which equation relates them effectively.