The airplane undergoes uniformly decelerated motion. Let the initial landing velocity be $v_0$ and the constant deceleration be $a$. The key kinematic equations are:

1. $v(t) = v_0 + at$
2. $x(t) = v_0 t + \frac{1}{2} a t^2$
3. $x(t) = \frac{v_0 + v(t)}{2} t$

The given conditions are: at $t_1 = 10$ s, the distance traveled is $x_1 = 450$ m, and the velocity is $v_1 = v_0/2$.

[Q1] Find the airplane's landing velocity. We use the kinematic equation relating distance, velocities, and time.

$$x_1 = \frac{v_0 + v_1}{2} t_1$$

Substituting $v_1 = v_0/2$:

$$x_1 = \frac{v_0 + v_0/2}{2} t_1 = \frac{3v_0}{4} t_1$$

Solving for the initial landing velocity $v_0$:

$$v_0 = \frac{4x_1}{3t_1} = \frac{4(450 \text{ m})}{3(10 \text{ s})} = 60 \text{ m/s}$$

The deceleration $a$ can be found from $v_1 = v_0 + at_1$:

$$a = \frac{v_1 - v_0}{t_1} = \frac{v_0/2 - v_0}{t_1} = -\frac{v_0}{2t_1} = -\frac{60 \text{ m/s}}{2(10 \text{ s})} = -3 \text{ m/s}^2$$

[Q2] How far is the airplane from the landing point $20$ s after landing? Let $t_2 = 20$ s. The position $x_2$ is given by the equation of motion:

$$x_2 = v_0 t_2 + \frac{1}{2} a t_2^2$$

Substituting the known values:

$$x_2 = (60 \text{ m/s})(20 \text{ s}) + \frac{1}{2}(-3 \text{ m/s}^2)(20 \text{ s})^2 = 1200 \text{ m} - 600 \text{ m} = 600 \text{ m}$$

[Q3] How long after landing does it take for the airplane to travel $540$ m? Let $x_3 = 540$ m. We must find the time $t_3$ using the position equation:

$$x_3 = v_0 t_3 + \frac{1}{2} a t_3^2$$ $$540 = 60 t_3 - \frac{3}{2} t_3^2$$

Rearranging into a standard quadratic form ($At^2+Bt+C=0$):

$$\frac{3}{2} t_3^2 - 60 t_3 + 540 = 0 \implies t_3^2 - 40 t_3 + 360 = 0$$

Using the quadratic formula, $t_3 = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$:

$$t_3 = \frac{40 \pm \sqrt{(-40)^2 - 4(1)(360)}}{2} = \frac{40 \pm \sqrt{1600 - 1440}}{2} = \frac{40 \pm \sqrt{160}}{2}$$ $$t_3 = \frac{40 \pm 4\sqrt{10}}{2} = (20 \pm 2\sqrt{10}) \text{ s}$$

As the airplane is decelerating, it will eventually stop. The second, larger time solution may be unphysical. From Q4, we find the stopping time is $20$ s. Since $20 + 2\sqrt{10} > 20$, we must choose the smaller root.

$$t_3 = (20 - 2\sqrt{10}) \text{ s}$$

[Q4] How long after landing does it take for the airplane to come to a complete stop? Let $t_{stop}$ be the time to stop, where the final velocity $v_f = 0$.

$$v_f = v_0 + a t_{stop}$$

Solving for $t_{stop}$:

$$t_{stop} = -\frac{v_0}{a} = -\frac{60 \text{ m/s}}{-3 \text{ m/s}^2} = 20 \text{ s}$$

[Q5] What is the total distance from the landing point to where the airplane stops? Let $x_{stop}$ be the stopping distance. This is the distance traveled in $t_{stop}=20$ s, which was already calculated in Q2. Alternatively, using the time-independent kinematic equation:

$$v_f^2 = v_0^2 + 2ax_{stop}$$

Solving for $x_{stop}$ with $v_f = 0$:

$$x_{stop} = -\frac{v_0^2}{2a} = -\frac{(60 \text{ m/s})^2}{2(-3 \text{ m/s}^2)} = \frac{3600}{6} \text{ m} = 600 \text{ m}$$