Chapter 3: Derivatives - Section 3.8 - Related Rates - Exercises 3.8 - Page 164: 42

$80~mi/h$

Step 1. See figure; assume the car's speed is $v$ and its x-coordinate is $x$ Step 2. At the time of measurement, $x^2+3^2=s^2$, where $s$ is the hypotenuse. When $s=5, x=4$ Step 3. Differentiate to get $2xx'=2ss'$, since $x=4, x'=-(v+120), s=5$ and $\frac{ds}{dt}=-160mi/h$, we have $-2(4)(v+120)=2(5)(-160)$ which gives $v+120=200$ and $v=80mi/h$