The office concluded that the driver exceeded the speed limit by 64.3mi/h.
Given that the second cop, who arrived 28 minutes later, was recorded traveling at 60 mph while the patrol officer was traveling 30 miles along the roadway.
According to the mean value theorem, There exists a value of x = c such that f'(c) = [f(b)-f(a)][b-a] if f(x) is continuous on the closed interval [a,b] and differentiable on the open interval (a,b).
This implies the following in "lay man's" terms:
the instantaneous speed (the speed at any given time) must be equal to his average speed
Since the driver is driving along the highway, we can assume his position to be continuous and differentiable
his average speed is defined as:
average speed=[x(b)-x(a)]÷[b-a] where x represents his position
average speed=[30 - 0]÷[(28÷60) - 0]
average speed=30÷0.466
average speed=64.3mi/hr
Since the time is given in minutes, we convert it in hours by dividing it by 60.
Therefore, by the MVT the police officers can determine that at some point in time (even though he was only driving 60mph at the second patrol officer's location) since his average speed was approximately 64.3 mi/hr there was a point in time during the 28 minutes that his speed exceeded 60 mph.
Learn more about mean value theorem from here brainly.com/question/3403246
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