Question

An aeroplane at an altitude of 1 km is flying horizontally at 600 km / hr, passes directly over an observer. Then the rate at which it is approaching the observer when it is 1250 meters away from him is........

• A. \(360 km/hr\)
• B. \(430 km/hr\)
• C. \(600 km/hr\)
• D. \(250 km/hr\)

Hint:

Join the plane, the observer and the point below the plane in a right triangle and differentiate the Pythagoras relation.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
Let the plane be at horizontal distance \(x\) from the point above the observer and at height \(h=1\) km. The distance from the observer is \(s\), so \(s^2=x^2+1\).

Step 2: Differentiate
\[ 2s\frac{ds}{dt}=2x\frac{dx}{dt}\Rightarrow\frac{ds}{dt}=\frac xs\frac{dx}{dt} \]

Step 3: Values at the given moment
\(s=1250\) m \(=1.25\) km. Then \(x=\sqrt{1.25^2-1}=\sqrt{0.5625}=0.75\) km. The speed is \(\dfrac{dx}{dt}=-600\) km/h (distance decreasing).

Step 4: Calculate
\[ \frac{ds}{dt}=\frac{0.75}{1.25}\times(-600)=-360\ \text{km/h} \]
The plane is approaching the observer at 360 km/h, option (A).

Final Answer:
The distance to the observer shrinks at 360 km/h, option (A). \[ \boxed{360\ \text{km/hr}} \]