How fast is the car traveling along the highway? MIT 18.01 OCW final exam The following question appeared in the MIT 18.01 single variable calculus OCW final exam:

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
  $\sqrt{5} = 2.2$ .)

I'm going to explain how I solve it :
first I sketch this :

where $p$ is the plane and $c$ is the car.
We know the following things: 
$\frac{dp}{dt} = 120 \ \ m.p.h,$
$\frac{dD}{dt} = -136 \ \ m.p.h,$
and using Pythagorean theorem we can find that $x = \frac{\sqrt{5}}{2} \approx 1.1 \ \ m$
And $\frac{dx}{dt} = \frac{dc}{dt}  - \frac{dp}{dt}$ because there are two things which affect the distance x, the first one is the speed of the car (and the sign is positive because if the car is moving forward its speed should be positive and that should increase x and the opposite for the plane ).
Using Pythagorean theorem again we have:
$D^2 = x^2 + 1^2$.
We take the derivative with respect to t so:
$\implies 2D\frac{dD}{dt} = 2x\frac{dx}{dt}$
$\implies (2)(1.5)(-136) = (2)(1.1)\frac{dx}{dt}$
$\implies \frac{dx}{dt} \approx -185.45 \  \  m.p.h.$
And we know that $\frac{dx}{dt} = \frac{dc}{dt} - \frac{dp}{dt}.$
So $ \ \ \implies \frac{dc}{dt} = -65.45 \  \ m.p.h.$
But the answer in the solution pdf is $65.45 \  \ m.p.h$
Can anyone tell where the wrong in my solution?

The car is travelling the wrong way down the freeway, so the speed is $65.45$ as MIT state (and they only ask for the speed), but the direction of the car is towards the aeroplane (because your maths looks valid to me), which I assume is unusual.
