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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?

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  • $\begingroup$ Welcome to MathSE. Please edit your question to show what you have attempted and explain where you are stuck so that you receive responses that address the specific difficulties you are encountering. This tutorial explains how to typeset mathematics on this site. $\endgroup$ – N. F. Taussig Nov 22 '19 at 16:04
  • $\begingroup$ logic and computations both seem correct, don't think of the velocity as something positive or negative but as same or opposite. If we assume the car is traveling along the x-axis, the plane will be traveling opposite to this direction as their relative distances (both x and y) are shrinking. $\endgroup$ – DubsVeer23 Nov 23 '19 at 6:21
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You defined $x$ as the horizontal distance between the airplane and car. As the airplane approaches the car, the distance shrinks, which makes $$\frac{dx}{dt}$$ negative.

The rate at which the horizontal distance shrinks depends on whether the airplane and car are heading in the same direction or opposite directions. If the airplane and car were heading in the same direction, the rate at which the horizontal distance shrinks would be the difference of the speeds of the airplane and car, which would be smaller than the speed of the airplane. Since $$\left|\frac{dx}{dt}\right| = 185.45~\text{mph}$$ exceeds the speed of the airplane, the airplane and car must be headed in opposite directions, so the rate at which the horizontal distance shrinks must equal the sum of the speeds of the airplane and car. Since $$\frac{dx}{dt} = -185.45~\text{mph}$$ the relative horizontal speed at which the airplane and car approach each other is $$\left|\frac{dx}{dt}\right| = 185.45~\text{mph}$$ Hence, $$185.45~\text{mph} = \left|\frac{dp}{dt}\right| + \left|\frac{dc}{dt}\right| = 120~\text{mph} + \left|\frac{dc}{dt}\right| \implies \left|\frac{dc}{dt}\right| = 65.45~\text{mph}$$ The reason the velocities of both the airplane and car are negative is that as the airplane and car approach each other, the distance between them shrinks.
$$\frac{dp}{dt} = -120~\text{mph}$$ since the speed of the plane causes the distance $x$ between the airplane and the car to decrease at a rate of $120~\text{mph}$. Similarly, $$\frac{dc}{dt} = -65.45~\text{mph}$$ since the speed of the car causes the distance $x$ between the airplane and the car to decrease at a rate of $65.45~\text{mph}$.

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The problem statement is ambiguous.

Assuming both the airplane and the car are moving along parallel, perfectly horizontal straight lines (one along the road, which is assumed to be level, the other parallel to and above the road) and that those parallel lines are exactly $1$ mile apart, the horizontal distance $x$ between the car and the airplane must be decreasing at the rate of about $182.46$ miles per hour. (Note: if you're going to give an answer to five digits, don't make it depend on a factor that is accurate to only two digits.)

And that is all we know from the problem statement about the relative motion of the car and the aircraft. We don't know whether the car is approaching the aircraft from far ahead or catching up to it from behind.

If it's approaching from far ahead, the car's speed is approximately $$ 182.46 - 120 = 62.46 \mathrm{mph}. $$

If it's catching up from behind, the car's speed is approximately $$ 182.46 + 120 = 302.46 \mathrm{mph}. $$

We can reasonably rule out the second possibility if the car is a stock Ford Escort, but not if it is a specially-modified Bugatti Chiron.

The best assumption to make when a problem like this is set, if you must give the answer as one number with no further explanation, is that whoever set the problem did not even consider the second possibility.

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