Question

An aeroplane at an altitude of $1km$. flying horizontally at $800$ km/hours passes directly over. An observer. Find the rate at which it is approaching the observer when it is $1250$ meters away from him.

Answer 1

Let $A$ be the position of aeroplane and $B$ is the position of observer at time $t$

Given the altitude of the plane.

i.e.,$OA=1$km

Let $OB=x,$ the horizontal distance and $AB=y,$ the actual distance of the aeroplane from the observer at time $t$.

Then $\frac{dx}{dt}=800$km/hr is the rate at which the aeroplane is moving over an observer and then

$\frac{dx}{dt}=800$km/hr is the rate at which the aeroplane is moving over an observer and $\frac{dy}{dt}$ is the rate at which the aeroplane is approaching to the observer.

From the figure,

$y^2=x^2+1$ ......$\left(1\right)$

Differentiating w.r.t $t$, we get

$2y\frac{dy}{dt}=2x\frac{dx}{dt}+0$

$\therefore \frac{dy}{dt}=\frac{x}{y}.\frac{dx}{dt}$ .....$\left(2\right)$

When $y=1250$m$=\frac{1250}{1000}=\frac{5}{4}$km, then from $\left(1\right)$, we get

$\frac{25}{16}=x^2+1$

$\therefore x^2=\frac{25}{16}-1=\frac{9}{16}$

$\therefore x=\frac{3}{4}$

$\therefore \left(2\right)$ gives, $\frac{dy}{dt}=\frac{\left(\frac{3}{4}\right)}{\left(\frac{5}{4}\right)}\times 800=480$km/hr.

$\therefore$ The aeroplane is approaching to the observer at the rate of $480$km/hr.