An aeroplane at an height of 1 km is flying horizontally at 800 km/hr passes directly over an observer .find the rate at which it is approaching the observer when it is 1250 m away from him
h = 1000
x = sqrt (1250^2 - 1000^2)
= 750 meters horizontal
r = distance (hypotenuse)
r^2 = h^2 + x^2
2 r dr/dt = 0 + 2 x dx/dt
so
dr/dt = (x/r) dx/dt
= (750/1250)(800)
Ans is 480 km per second..
To find the rate at which the airplane is approaching the observer, we can use the concept of related rates. Let's define the following variables:
- Let "A" represent the airplane.
- Let "O" represent the observer.
- Let "d" represent the distance between the airplane "A" and the observer "O".
- Let "h" represent the height of the airplane above the ground.
According to the problem, the airplane is flying horizontally, which means it is not changing its height. So, the rate at which the height changes, dh/dt, is zero.
We are given the following information:
- The height of the airplane above the ground, h, is 1 km (or 1000 meters).
- The speed of the airplane, dx/dt (horizontal component), is 800 km/hr (or 800,000 meters/hr).
We want to find the rate at which the airplane is approaching the observer, dd/dt, when it is 1250 meters away from the observer.
We can start by writing the equation for the relationship between the variables:
d^2 = h^2 + x^2
Differentiating both sides of the equation with respect to time (t), we get:
2d * dd/dt = 2h * dh/dt + 2x * dx/dt
Since dh/dt is zero (as the height does not change), the equation simplifies to:
2d * dd/dt = 2x * dx/dt
Now, let's substitute the given values into the equation:
- d = 1250 meters
- h = 1000 meters
- dx/dt = 800,000 meters/hr
Now we can solve for dd/dt:
2 * 1250 * dd/dt = 2 * 1250 * 800,000
Simplifying the equation:
2500 * dd/dt = 2,000,000
Dividing both sides by 2500:
dd/dt = 2,000,000 / 2500
Calculating the rate at which the airplane is approaching the observer:
dd/dt = 800 meters/hr
Therefore, the rate at which the airplane is approaching the observer when it is 1250 meters away from him is 800 meters/hr.
To find the rate at which the airplane is approaching the observer, we can use the concept of similar triangles and differentiation.
Let's consider the situation when the airplane is 1250 meters away from the observer. We can draw a right triangle, where the observer is at one vertex, the airplane is at the top vertex, and the base of the triangle represents the horizontal distance between them.
The vertical side of the triangle represents the height of the airplane above the observer, which is 1 km (or 1000 m). The hypotenuse of the triangle represents the distance between the observer and the airplane, which is 1250 m.
To find the rate at which the airplane is approaching the observer, we need to differentiate the distance equation with respect to time. The distance at any time can be represented by the Pythagorean theorem:
distance^2 = horizontal distance^2 + vertical distance^2
Differentiating both sides of the equation with respect to time will give us:
(2 * distance) * (rate of change of distance) = (2 * horizontal distance) * (rate of change of horizontal distance) + (2 * vertical distance) * (rate of change of vertical distance)
Since the observer is not moving, the rate of change of horizontal distance is the same as the velocity of the airplane. Additionally, the rate of change of vertical distance is zero since the height of the airplane is constant.
Let's plug in the given values: distance = 1250 m, vertical distance = 1000 m, and velocity of the airplane = 800 km/hr.
(2 * 1250) * (rate of change of distance) = (2 * horizontal distance) * (800 km/hr) + (2 * 1000) * (0)
Simplifying the equation:
2500 * (rate of change of distance) = (2 * 1250) * (800 km/hr)
2500 * (rate of change of distance) = 2000 * 1250 km/hr
Dividing both sides by 2500:
rate of change of distance = (2000 * 1250 km/hr) / 2500
rate of change of distance = 1000 km/hr
Therefore, when the airplane is 1250 m away from the observer, it is approaching the observer at a rate of 1000 km/hr.