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........

Step-by-step solution

Let the horizontal distance of the aeroplane from the observer be x km and the actual distance be s km. The altitude of the aeroplane is h = 1 km. From the right-angled triangle formed by the observer, the aeroplane, and the point on the ground directly below the aeroplane, we have: s^2 = x^2 + 1^2 Differentiating with respect to time t , we get: 2s ds dt = 2x dx dt ds dt = x s dx dt We are given the horizontal velocity dx dt = 600 km/hr. When the distance s = 1250 m = 1.25 km, the horizontal distance x is: x = s^2