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 :

enter image description here

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?

  • 1
    $\begingroup$ Don't you mean $\frac{dp}{dt}=120\text{ mph}$? (instead of $\frac{dx}{dt}=120\text{ mph}$) $\endgroup$
    – Isaac Ren
    May 16, 2020 at 12:46
  • $\begingroup$ @IsaacRen fixed, sorry for that. $\endgroup$
    – Kais Hasan
    May 16, 2020 at 12:57
  • 2
    $\begingroup$ Since the question is asking for how FAST the car is moving, there are asking for speed and not velocity. Notice that all the numbers they gave you were positive regardless of direction. So all you have to do is take the absolute value of your answer. $\endgroup$
    – BigBear
    May 16, 2020 at 13:06
  • $\begingroup$ @BigBear I must say, that's an extremely unsatisfying answer, but it seems to be the right one xD $\endgroup$
    – Isaac Ren
    May 16, 2020 at 13:11
  • $\begingroup$ @BigBear actually in their short solution(which is the reason why I don't know where is my mistake) they assume that dD/dt = -136 m.p.h which the same in my solution so I think its not like that. $\endgroup$
    – Kais Hasan
    May 16, 2020 at 13:33

1 Answer 1


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.

  • $\begingroup$ what does that mean? I don't understand. $\endgroup$
    – Kais Hasan
    May 16, 2020 at 13:34
  • $\begingroup$ @KaisHasan; does that help? $\endgroup$
    – JMP
    May 16, 2020 at 13:42
  • $\begingroup$ I understand it, @BigBear comment is the same, but as I said to him, why they assume that dD/dt = -136 m.p.h if they want the speed only? $\endgroup$
    – Kais Hasan
    May 16, 2020 at 13:48
  • $\begingroup$ If they gave +136, the answer would be different. $\endgroup$
    – JMP
    May 16, 2020 at 13:50

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