NCERT Solutions

We need to calculate the aircraft's speed using its altitude, the time interval, and the subtended observation angle.

Concept: Trigonometry

The problem utilizes trigonometry and the definition of average speed. Assuming the aircraft flies at a constant altitude $h$ and the observation point is positioned such that the path is bisected by the vertical, the horizontal distance $d$ traveled is related to the angle $\theta$ by:

$$d = 2h \tan\left(\frac{\theta}{2}\right)$$

The speed $v$ is then given by the ratio of distance to time:

$$v = \frac{d}{\Delta t}$$

Given:

• Height of the aircraft, $h = 3400\,\text{m}$
• Time interval, $\Delta t = 10.0\,\text{s}$
• Subtended angle, $\theta = 30^{\circ}$

Solving:

1. Calculate the horizontal distance traveled by the aircraft

The distance $d$ is covered in $\Delta t = 10.0\,\text{s}$. Using the geometric relationship:

$$\begin{aligned} d &= 2 \cdot h \cdot \tan\left(\frac{30^{\circ}}{2}\right) \\ &= 2 \cdot 3400\,\text{m} \cdot \tan(15^{\circ}) \end{aligned}$$

Using the trigonometric identity $\tan(15^{\circ}) = 2 - \sqrt{3}$:

$$\begin{aligned} d &= 6800 \cdot (2 - \sqrt{3})\,\text{m} \\ &\approx 6800 \cdot 0.267949\,\text{m} \\ &\approx 1822.0545\,\text{m} \end{aligned}$$

2. Calculate the speed of the aircraft

$$\begin{aligned} v &= \frac{d}{\Delta t} \\ &= \frac{1822.0545\,\text{m}}{10.0\,\text{s}} \\ &= 182.20545\,\text{m/s} \end{aligned}$$

3. Convert to standard units

$$\begin{aligned} v_{\text{km/h}} &= 182.20545 \cdot 3.6 \\ &= 655.93962\,\text{km/h} \end{aligned}$$

Answer:

$$182.2\,\text{m/s}$$

or

$$655.9\,\text{km/h}$$