Hello Beverly,
The first thing I would do is draw a diagram representing the scenario:
$T$ represents the trooper's position, $C$ represents the position of Amanda's car on I-95, $x$ represents the distance Amanda has traveled since passing the trooper, and $h$ represents the distance between Amanda and the trooper. All distances are in feet.
Now, we are ultimately being asked to find $$\frac{dx}{dt}$$, and we are given $$\left.\frac{dh}{dt}\right|_{h=150}=72\,\frac{\text{ft}}{\text{s}}$$.
What we need then, is a relationship between $x$ and $h$, and fortunately, we have what we need via Pythagoras:
$$x^2+90^2=h^2$$
Now, implicitly differentiating with respect to time $t$, we find:
$$2x\frac{dx}{dt}=2h\frac{dh}{dt}$$
and solving for $$\frac{dx}{dt}$$, we find:
$$\frac{dx}{dt}=\frac{h}{x}\frac{dh}{dt}$$
Using the Pythagorean relation, we find:
$$x(h)=\sqrt{h^2-90^2}$$
and so we have:
$$\frac{dx}{dt}=\frac{h}{\sqrt{h^2-90^2}}\frac{dh}{dt}$$
and finally, we may compute:
$$\left.\frac{dx}{dt} \right|_{h=150}=\frac{150}{\sqrt{150^2-90^2}}\left.\frac{dh}{dt} \right|_{h=150}=\frac{5}{4}\cdot72\, \frac{\text{ft}}{\text{s}}=90\, \frac{\text{ft}}{\text{s}}$$
To Beverly and any other guests viewing this topic, I invite and encourage you to post other related rates problems in our http://www.mathhelpboards.com/f10/ forum.
Best Regards,
Mark.