Let's set up the situation and variables to find your car's actual speed:

Variables

1. Let:

   • $x$ be the distance between your car and the point on the road closest to the police officer.
   • $y = 22 \, \text{m}$ be the constant distance between the road and the police officer.
   • $s$ be the distance between your car and the police officer (the hypotenuse of the triangle formed by $x$ and $y$).

2. The problem states:

   • $s = 60 \, \text{m}$ (distance from the police officer to your car at that instant).
   • $\frac{ds}{dt} = -$ (the rate at which this distance is decreasing, indicating the officer measures your distance to be getting closer).

Relationships and Goal

Since we want the actual speed of your car along the road, we're interested in $\frac{dx}{dt}$, the rate at which $x$ changes.

Using the Pythagorean theorem, we have: $s^2 = x^2 + y^2$

Step 1: Differentiate with Respect to Time

Differentiating both sides with respect to $t$: $2s \frac{ds}{dt} = 2x \frac{dx}{dt}$

Step 2: Solve for $\frac{dx}{dt}$

Dividing both sides by $2x$ gives: $\frac{dx}{dt} = \frac{s \frac{ds}{dt}}{x}$

Step 3: Substitute Values

Since $y = 22$ m and $s = 60$ m, we can find $x$ using the Pythagorean theorem: $x = \sqrt{s^2 - y^2} = \sqrt{60^2 - 22^2}$

Now, plug in these values to get the unsimplified expression for $\frac{dx}{dt}$: $\frac{dx}{dt} = \frac{60 \cdot \frac{ds}{dt}}{\sqrt{60^2 - 22^2}}$

This expression gives the speed of your vehicle along the road.