Calculus word problems
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
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Rob
Updated on October 27, 2022Comments

Rob about 1 year
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?

Simon about 8 yearsHave you drawn a diagram? Always start by drawing a diagram. Also, try to convert units so that they match with each other.

Rob about 8 yearsyes, I have drawn a diagram but I don't know what to do after that.

Simon about 8 yearsSince the police officer is facing the highway, and cars are moving on the highway, which as we can now deduce is orthogonal to the line of sight of the officer, we essentially have a triangle. Why? Because the distance between the officer and the car forms the hypotenuse of a right triangle. You also know what the rate of change of that hypotenuse is, 80km/h. Getting any ideas?

Rob about 8 yearsOk, so how would i complete the question from that point?

Simon about 8 yearsCan you think of a relationship that relates the hypotenuse to two other sides of a triangle?

Rob about 8 yearsNo, I'm not really sure

Simon about 8 yearsHow about Pythagoras's theorem? Use that to relate the distance between the officer to the other sides of the "triangle"

André Nicolas about 8 yearsb) At the sign, the distance between the car and the radar gun is changing at $0$ km/h, and she has a ticket quota to fill.

Rob about 8 yearsCould you show me the first few steps?

André Nicolas about 8 yearsLet $z=z(t)$ be the distance, in km, of the radar gun from the car. Let $x$ be the distance of the car from the sign. Then (picture) $z^2=x^2+(0.4)^2$. Differentiate with respect to $t$.

Simon about 8 yearsWe know that there is a triangle formed by the police officer's distance from the bushes to the highway, the distance between the police officer and the car, as well as the distance between the car to the traffic sign. Allow x to be the car's position at any time. This implies that the relationship between the police officer's distance to the car is $h^{2} = (0.4)^{2} + (0.6x)^{2}$. You now have a function relating distance to of a car, to the police officer. Note that you will have to implicitly differentiate.

Simon about 8 years@AndréNicolas Correct me if I'm wrong, but I believe that the distance between the car and the sign is $0.6  x$, because x is the distance the car has traveled since it passed the phone.

André Nicolas about 8 years@Simon: We are using a different meaning for $x$. You will be evaluating at $x=0$. I will be evaluating at $x=0.6$. Same final result.

Simon about 8 years@AndréNicolas You're right. To finish up this problem, differentiate implicitly, $h^{2}$ and note that $x(t)$, that is, x is a function of t, so there should be a $\frac{dx}{dt}$ term in there
