Hello Gina,
First, let's draw a diagram:
View attachment 1445
The plane is at $P$, the radar station is at $R$, $h$ is the altitude of the plane (which is constant since its flight is said to be horizontal) and $x$ is the distance from the radar station to the point on the ground (or at the same level as the radar station) directly below the plane. $s$ is the distance from the plane the the radar station.
Using the Pythagorean theorem, we may state:
(1) $$x^2+h^2=s^2$$
Implicitly differentiating (1) with respect to time $t$, we find:
$$2x\frac{dx}{dt}=2s\frac{ds}{dt}$$
We are interested in solving for $$\frac{ds}{dt}$$ since we are asked to find the rate at which the distance from the plane to the station is increasing when it is 5 mi away from the station. So, we find:
$$\frac{ds}{dt}=\frac{x}{s}\frac{dx}{dt}$$
Now, we do not know $x$ but we know $h$ and $s$, and so solving (1) for $x$ (and taking the positive root since it represents a distance), we find:
$$x=\sqrt{s^2-h^2}$$
And so we have:
$$\frac{ds}{dt}=\frac{\sqrt{s^2-h^2}}{s}\frac{dx}{dt}$$
Now, the speed $v$ of the plane represents the time rate of change of $x$, hence:
$$v=\frac{dx}{dt}$$
and so we may write:
$$\frac{ds}{dt}=\frac{v\sqrt{s^2-h^2}}{s}$$
Now, we may plug in the given data:
$$v=470\frac{\text{mi}}{\text{hr}},\,s=5\text{ mi},\,h=1\text{ mi}$$
and we then find:
$$\frac{ds}{dt}=\frac{470\sqrt{5^2-1^2}}{5} \frac{\text{mi}}{\text{hr}}=188\sqrt{6}\frac{\text{mi}}{\text{hr}}\approx460.504071643238 \text{ mph}$$
And so we have found that the rate at which the distance from the plane to the station is increasing when it is 5 mi away from the station is about 461 mph.
