# “How long 'til we get there?” Road trip puzzle

Road trips can be fun, but they often appear to go slower the closer you get to your destination. I thought up this puzzle while on a recent trip. Thought it would be good food for thought. Curious about the different approaches to solving it.

Suppose you have D miles until you reach your destination. The rule is that the speed at which you travel is equal to the distance to your destination. So when you are 60 miles from your destination your speed must be 60 mph; 50 miles from destination, 50 mph; etc.

How long until you reach your destination?

EDIT: I'm pretty sure that the answer is infinity—you will never get to your destination because you will always be one hour away. I'm curious about how people come up with their solutions. So far, very entertaining.

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Reminds me of the Zeno's paradox. Very different though. –  Paresh Nov 27 '12 at 15:06
Note, since speed and distance use different units, this is heavily dependent on your units. –  Thomas Andrews Nov 27 '12 at 15:10
Luckily, infinity does not depend on units. –  Phira Nov 27 '12 at 15:10
You might find the ending of Billy Jonas's song "Wichita" to be of interest. Lyrics at billyjonas.com/index.php?page=songs&display=57 . –  Michael Lugo Nov 27 '12 at 15:48
This reminds me of a joke I use to begin my classes in differential equations. Somewhere in the USA (where AFAIK the road signs actually have text on them!) a driver sees a sign telling "SPEED LIMIT 60". He slows down to 60mph, and after some time sees a "SPEED LIMIT 50" sign. He slows down etc... Finally he sees a sign saying "WELCOME TO SPEED LIMIT!" –  mbork Nov 27 '12 at 22:46

You are always moving forward as long as the distance is non-zero. Of course, your speed keeps slowing down with distance. So your travel time keeps increasing and asymptotically approaches infinity as your distance asymptotically approaches zero. Others have said "always one hour away" ... to me that is "never" or "infinite time".

Of course, you would have run out of gas far before that and your fellow passengers would have mocked you into shame.

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Your speed is strictly decreasing, so at any time, you know that it takes you at least one hour which would be the arrival time at present speed. So, you cannot reach the destination which is always more than an hour away.

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...though you will eventually reach a point where you could get out of your car, take one step and you'd be there. –  BlueRaja - Danny Pflughoeft Nov 27 '12 at 18:10
Yes, you might even give the car a little push. –  Phira Nov 27 '12 at 18:28
As no distance shorter than the Planck length is meaningful, we could instead calculate how much time it takes to be within one Planck length of the destination. Then, we, for all practical purposes, did arrive. So you cannot say you never reach the destination :) –  vsz Nov 27 '12 at 18:40
As all cars should have a self-distance and should not be a point, and the positions of any objects should be measured from the middle, the front of the car will reach the final position first and you will accept the car "reachs the final position" , provided that the car has sufficient fuel. So no need to consider the issue about the Planck length. –  doraemonpaul Nov 28 '12 at 0:18
@doraemonpaul sure, but I feel like you said the same thing as vsz? –  jcolebrand Nov 28 '12 at 3:47

Put a car following the same rules but which is exactly in the middle between you and your destination, at $D/2$ miles from the destination. It is moving at half of your speed so it stays exactly at half the distance between you and your destination, at all times. So, supposing your journey ends, the other car will arrive exactly at the same time as you at the destination.

However you quickly realize that, since you follow the same rules, the total duration of your trip is the time it takes you to cover half the distance to your destination + the total duration of his trip. So you will arrive strictly after him.

So supposing that the journey ends gives a contradiction : you will stay on the road forever.

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I like the way you think. –  Jeremy Nov 27 '12 at 22:25
Until you hit the other car... –  AJMansfield Nov 28 '12 at 0:55

The existing answers are very good and don't require calculus; nevertheless it seems worthwhile to also show how this would be solved using calculus.

If your position at time $t$ is $x(t)$, your velocity (in appropriate units) is $\dot x(t)=-x(t)$. The general solution of this linear first-order ordinary differential equation is $x(t)=c\,\mathrm e^{-t}$, with arbitrary constant $c$. Thus you get arbitrarily close to your destination, but your speed decreases exponentially and you never reach the destination.

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You might want to add that your answer proves that we do reach the destination "for all practical purposes". –  Phira Nov 27 '12 at 15:12
If you put in a variable for "this is close enough", I'm thinking you could actually construct an equation describing the time it takes to travel any distance. –  bdares Nov 28 '12 at 10:40

The problem can be written as linear ODE of order 1:

Lets say that $s(t)$ is the driven distance, and $v(t)$ is the speed. Then we have $v(t)=s'(t)$. On the other hand, we have the relation $v(t)=60-s(t)$ and therefore $$s'(t)=60-s(t).$$ This is a linear ODE of first order with constant coefficients, and its solution is $$s(t)=60-60e^{-t}.$$ Now you want $s(t)=60$, so we have $$60=60-60e^{-t}$$ and therefore $$e^{-t}=0.$$ There is no value $t\in\mathbb R$ that fulfills this equation, only the limit $\lim_{t\rightarrow\infty}e^{-t}=0$.

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Set up DE $\dfrac{dx}{dt} = \dfrac{a - x}{T}$, where $a$ and $T$ are constants. Asymptotic behavior with respect to time.

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