A highway patrol plane is flying $1$ mile above a long, straight road, with constant ground speed of $120$ m.p.h. Using radar, the pilot detects a car whose distance from the plane is $1.5$ miles and decreasing at a rate of $136$ m.p.h.
How fast is the car traveling along the highway? (Hint: You may give an exact answer, or use the fact that $\sqrt5 \approx 2.2$.)
I drew a triangle diagram depicting the position of the plane and car relative to the ground, where the height of the triangle is 1, the horizontal distance from the plane to the car is given by $x$, and the hypotenuse of the triangle $y$ is given by the Pythagorean theorem: $y = \sqrt{x^2+1}$.
At time $t = 0$, we know that $y = 1.5$ and $x= \sqrt{y^2-1} = \frac{\sqrt{5}}{2}$
By differentiating y, we get: $$\frac{dy}{dt} = \frac{2x}{2\sqrt{x^2+1}} \frac{dx}{dt} = \frac{x}{\sqrt{x^2+1}} \frac{dx}{dt} = -136 mph$$
Note that $\frac{dy}{dt}$ is negative since the distance between the car and the plane is decreasing at a rate of 136 mph.
When we subsitute $x= \frac{\sqrt{5}}{2}$, we get: $$\frac{dx}{dt} = \frac{\sqrt{x^2+1}}{x} *\left(-136 mph \right) = \frac{3}{\sqrt5} *\left(-136 mph \right) \approx \frac{-408}{2.2} \approx -185.45 mph$$
Given that we now know that $\frac{dx}{dt} $ is approximately equal to -185.45 mph and the velocity of the plane $\left( \frac{dp}{dt}\right)$ is given by 120 mph, what is the formula for the velocity of the car $\left( \frac{dc}{dt}\right)$?
What confuses me is the negative and positive signs of the velocities. For example, I got the following formula: $\frac{dc}{dt} = \frac{dp}{dt} -\frac{dx}{dt}$
However, this formula only produces the correct answer if $\frac{dp}{dt} = -120 mph$, making $\frac{dc}{dt} = 65.45 mph$
How do we know that the velocity of the plane is negative?