To find the final cruising velocity ($v_f$), we can use the kinematic equation:

$v_f^2 = v_i^2 + 2a \cdot d$

Where:

• $v_i = 21.0 \, \text{m/s}$ (initial velocity)
• $a = 3.00 \, \text{m/s}^2$ (acceleration)
• $d = 535 \, \text{m}$ (distance)

Substituting the values:

$v_f^2 = (21.0)^2 + 2 \cdot (3.00) \cdot (535)$ $v_f^2 = 441 + 3210$ $v_f^2 = 3651$

Taking the square root: $v_f = \sqrt{3651} \approx 60.4 \, \text{m/s}$

Final cruising velocity:

$\boxed{60.4 \, \text{m/s}}$

Would you like a step-by-step breakdown of the calculations or have related questions?

Here are 5 related questions:

1. How much time did the airplane take to reach this final velocity?
2. What is the kinetic energy difference between the initial and final velocities?
3. How would the result change if the acceleration was doubled?
4. How far would the airplane travel in another 10 seconds at the final velocity?
5. What is the significance of constant acceleration in solving such problems?

Tip: Always use consistent units (e.g., m/s, m, s) to avoid errors in kinematic equations.