Question 19

Explanation

Geometric Analysis of Plane Movement

This problem can be modeled using basic trigonometry on right-angled triangles. We have two different plane positions relative to the stationary ATC tower.

Given:

• Plane altitude (opposite side) remains constant: $h = 3000\text{ m}$.
• Initial elevation angle: $\theta_1 = 30^\circ$.
• Final elevation angle: $\theta_2 = 45^\circ$.
• Time interval: $t = 30\text{ s}$.

The plane is moving towards the tower because the elevation angle increases from $30^\circ$ to $45^\circ$. We need to find the horizontal distance of the plane from the tower at both positions.

Calculating Horizontal Distances

The relationship between altitude, horizontal distance ($x$), and elevation angle ($\theta$) is given by:

1. Initial Distance ($x_1$) when $\theta_1 = 30^\circ$:

   $$x_1 = \frac{3000}{\tan(30^\circ)} = \frac{3000}{\frac{1}{\sqrt{3}}} = 3000\sqrt{3}\text{ m}$$

2. Final Distance ($x_2$) when $\theta_2 = 45^\circ$:

   $$x_2 = \frac{3000}{\tan(45^\circ)} = \frac{3000}{1} = 3000\text{ m}$$

Calculating Plane Speed

The distance traveled by the plane ($\Delta x$) is the difference between the initial and final horizontal distances:

Speed ($v$) is defined as distance traveled divided by time:

Thus, the estimated speed of the plane is $100(\sqrt{3} - 1)\text{ m/s}$.