An aeroplane flying at a constant speed, parallel to the horizontal ground, $\sqrt{3} \, km$ above it, is observed at an elevation of $60^{\circ}$ from a point on the ground. If, after five seconds, its elevation from the same point, is $30^{\circ}$, then the speed (in km/hr) of the aeroplane, is :

Question

The position-time graphs for two students A and B returning from school to their homes are shown in the figure.

A. A lives closer to the school
B. B lives closer to the school
C. A takes lesser time to reach home
D. A travels faster than B
E. B travels faster than A
Choose the correct answer from the options given below :

Answer 1

To solve this problem, we need to determine the speed of an aeroplane given the change in the angle of elevation over a period of time. Let's break this down step by step.

Step 1: Understand the Geometry

The aeroplane is flying parallel to the ground at a height of \sqrt{3} \, \text{km}. Initially, it is observed at an angle of elevation of 60^\circ, and then after 5 seconds, the angle changes to 30^\circ from the same observation point on the ground.

Step 2: Use Trigonometric Ratios

We use the tangent function which relates the angle of elevation, the height of the plane, and the horizontal distance from the observer to the point on the ground directly below the plane.

Initially, at 60^\circ:

\[ \tan 60^\circ = \frac{\text{Height of the plane}}{\text{Initial horizontal distance (d1)}} \]

We know that \tan 60^\circ = \sqrt{3}. Thus,

\[ \sqrt{3} = \frac{\sqrt{3} \, \text{km}}{\text{d1}} \] Solving for \text{d1}, we find: \[ \text{d1} = 1 \, \text{km} \]

After 5 seconds, at 30^\circ:

\[ \tan 30^\circ = \frac{\text{Height of the plane}}{\text{Final horizontal distance (d2)}} \]

We know \tan 30^\circ = \frac{1}{\sqrt{3}}. Thus,

\[ \frac{1}{\sqrt{3}} = \frac{\sqrt{3} \, \text{km}}{\text{d2}} \] Solving for \text{d2}, we find: \[ \text{d2} = 3 \, \text{km} \]

Step 3: Calculate the Distance and Speed

The horizontal distance traveled by the plane in 5 seconds is:

\[ \text{Distance} = \text{d2} - \text{d1} = 3 \, \text{km} - 1 \, \text{km} = 2 \, \text{km} \]

The speed of the plane is then:

\[ \text{Speed} = \frac{\text{Distance}}{\text{Time}} = \frac{2 \, \text{km}}{5 \, \text{seconds}} \] To convert this speed into km/hr, multiply by 3600 (since 1 hour = 3600 seconds): \[ \text{Speed} = \frac{2}{5} \times 3600 = 1440 \, \text{km/hr} \]

Conclusion: The speed of the aeroplane is 1440 km/hr. Therefore, the correct answer is 1440.

Answer 2