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I have to prove this by induction:

$n$ cars are travelling down a narrow one-way street. We know that:

  • The distance d between each two cars is the same.
  • The safe breaking distance b is the minimum distance between two cars that is needed for the second car to stop on time if the car in front suddenly breaks.
  • d < b

Prove by induction or refute: if the first car suddenly stops moving, all cars will stop moving. Before you do the induction state the property P you are using in the induction axiom.

So this is what I have until now and please tell me if it is right or wrong.

The property P: if $n=1$ stops, then $n=2$ will stop. (Is this property correct?)

We consider 2 cases:

Base case:

n=1 -> Since there is one only car on the road, that car will stop. (Should the base case start at 0, 1 or 2? )

Step case: (Supposed that the base case should start at 1)

As mentioned, for n=1 if the only car stops, then we can say all the cars have stopped.

Assume that all cars stop for n cars and then we prove that all cars stop for $n+1$. We create the set C of Cars in which are included all the cars in road. The moment the first car $(n=1)$ stops, the following cars have 2 ways to stop:

  1. $n+1$ car stops without touching the n car
  2. $n+1$ car stops by touching and having/or not an accident with n car

Any suggestion?

ben
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  • Not sure I understand. As the road is now blocked, of course all the cars have to stop. This would be true even if $d>>b$. In other words, what is the alternative to stopping? – lulu Sep 18 '15 at 20:09
  • yeah but if d>>b, there would be no chance they the cars would crash. if d<b, there is a probability that cars would crash cause they are close and outside the safe breaking distance. – ben Sep 18 '15 at 20:11
  • So the question ought to read "is it possible they will not crash?" That is very different. And have you given us enough information? Driver #2 appears doomed, but if, say, $2d>b$ then then all the others should see the lead accident. Of course, this depends on them getting the knowledge instantly. – lulu Sep 18 '15 at 20:18
  • i have given all the information I have. And I have to prove that "if the first car suddenly stops moving, all cars will stop moving." – ben Sep 18 '15 at 20:24
  • Well, I still feel that I don't understand. Based on what I think you are asking, the statement that you "have to prove" is false. As soon as $n$ is large enough so that $nd>b$ the $n^{th}$ driver has time to avoid a crash. Technically, he doesn't have to actually stop...he can keep decelerating forever. As I say, I doubt this is what the author of the problem had in mind. – lulu Sep 18 '15 at 20:32
  • i didn't say all cars have to crash, but since the distance is smaller than the safety zone, then cars may crash and then stop. But some other cars can not stop without crashin. The whole point is that cars stop since it is an one-way road. The matter is, how do I prove it? Did I do it correctly or not – ben Sep 18 '15 at 20:35
  • As I say, maybe it is a technical point, but a driver with ample braking room never has to stop. If you need to walk one mile, you can take one hour for the first half, one hour for the next quarter, one hour for the next eighth, and so on for ever. But if you exclude (admittedly unphysical) things like this, then I truly don't see the question. The road is blocked up ahead, so of course they all have to stop. I would ask for clarification. – lulu Sep 18 '15 at 20:40

1 Answers1

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When the first car stops, the second cannot stop on time, as $d < b$. Then the statement "all cars will stop moving" is false.

This holds whatever $n$, except for $n=1$.

  • why false? The following cars will stop, but maybe by crashing with the one ahead of them. I do not understand your last sentence. – ben Sep 18 '15 at 20:25
  • It suffices that one car crashes for "all" to be falsified. –  Sep 18 '15 at 20:26