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A highway patrol plane flies 3 miles above a level, straight road at a steady 120mph. The pilot sees an oncoming car and with radar determines that at the instant the line of sight distance from plane to car is 5 miles, the line of sight distance is decreasing at the rate of 160mph. Find the car's speed along the highway.

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What's confusing me is that they mention the rate of the plane. I know you would have to use the pythagorean theorem and incorporate the differentials, but it looks like you would have two different rates along the x axis. – Justin Brown Dec 16 '12 at 1:37

Let the horizontal position of the plane and car be $x_p,x_c$ respectively. Then we have the line of sight distance (in miles) at time $t$ is $l(t) = \sqrt{(x_p(t)-x_c(t))^2 + 3^2}$. Let $t=0$ be the moment the pilot takes the line of sight speed reading, then we have $l(0) = 5 = \sqrt{(x_p(0)-x_c(0))^2 + 3^2}$, hence we have by (Pythagoras) $x_p(0)-x_c(0) = \pm 4$. Since the problem says 'oncoming car' we may presume that, in fact, $x_p(0)-x_c(0) = -4$.

We are also given $\frac{d l(0)}{dt} = -160$. The formula for $l$ gives $\frac{d l(t)}{dt} = \frac{x_p(t)-x_c(t)}{l(t)} (\frac{d x_c(t)}{dt} - \frac{d x_p(t)}{dt})$. Plugging in values gives: $\frac{d l(0)}{dt} = -160= \frac{x_p(0)-x_c(0)}{l(0)} (\frac{d x_c(0)}{dt} - \frac{d x_p(0)}{dt}) = \frac{-4}{5} ( 120 -\frac{d x_p(0)}{dt})$, from which you can figure out the required answer.

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