I'm not speeding, officer!

The Mean Value Theorem as applied to the velocity of this function states that the average value of the velocity equals the instantaneous velocity of the function at some point, as long as the function is continuous and differentiable. If we take the car's position as our function, we can use the velocity to apply the Mean Value Theorem. The car's average velocity was equal to 156 miles/2 hours = 78 mph. Therefore, at some point, the car had to be traveling at least at that speed, if not more, meaning that the car was definitely speeding in the distance between the two police cars.

Let us assume for the sake of contradiction that the car's velocity was never greater than 60 miles per hour. Then, integrating the car's velocity over the two hours between $10:30 \text{AM}$ and $12:30 \text{PM}$ , we find that the car's displacement is never greater than $(2 \text{hours})(60 \frac{\text{miles}}{\text{hour}}) = 120 \text{miles}$ . However, we are given that the car's displacement is 156 miles, which is clearly greater than 120 miles. Therefore, our initial assumption is false and the car at one point was speeding.

Since in two hours, the car traveled 156 miles, the average speed in these 2 hours is 156/2= 78 miles. This is higher than the speed limit. Therefore, at some point, the car was speeding between the two police cars, though it may be at more than 1 point.