Need help with this word problem, not sure how to complete this question.
A cop is trying to catch drivers who speed on the highway. She finds a long stretch of the highway. She parks her car behind some bushes, 400 metres away from the highway. There is a traffic sign at the point of the road closest to her car, and there is a phone by the road 600 metres away from the traffic sign.
(a) The cop points her radar gun at a car and learns that, as the car is passing by the phone, the distance between the car and the cop is increasing at a rate of 80 km/h. The speed limit is 120 km/h. Can she fine the driver?
(b) Why doesn’t the cop point her radar gun at cars as they pass by the traffic sign rather than as they pass by the phone?
Given the information in the problem, we know that the relationship between the distance from the officer to the car is \begin{equation} h^{2} = (0.6 - x(t))^{2} + 400^{2} \end{equation} where x is the distance the car has traveled since passing the phone.
Allow $h^{2} = f(x(t))$ Differentiating, using the chain rule, gives: \begin{equation} \frac{df}{dt} = \frac{df}{dx}\frac{dx}{dt} \end{equation} Plugging in the information we have, we come to the conclusion, \begin{equation} 80 = (-1.2 + 2x) * \frac{dx}{dt} \end{equation}
$\frac{dx}{dt}$ denotes the speed of the vehicle. Now lets assume that the car has reached the traffic sign. That implies x = 0. Therefore, $\frac{dx}{dt} = 66.6666...\frac{km}{h}$ which is not higher than $120\frac{km}{h}$
As for the second part of your problem, if the speed is evaluated at the traffic sign, the rate of change of the distance between the officer and the car would be $0\frac{km}{h}$, and the officer wouldn't have enough money to pay them billllzzzzzzz