Fast car

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1. Jun 20, 2015

Carlos Gouveia

A car goes from repose (0 mph) to 50 mph in, say, 30 seconds. Math tells us that there is an infinite amount of numbers between 0 and 50 (or between any two other numbers). Therefore, isn't it "obvious" or "intuitive" that it would take a car an infinite amount of time to go from 0 mph to 50 mph? Or from 0 mph to whatever other speed, be it smaller or bigger than 50 mph? However, as we all know, that's not what happens. What's the explanation for this (apparent) clash between the "theoretical" (so to speak) and the "real", "physical" worlds?

Any answer is welcome. Thank you.

Last edited: Jun 20, 2015
2. Jun 20, 2015

Orodruin

Staff Emeritus
Why do you think there is a paradox? (There is not.) The time interval during which you have a particular velocity has length zero. (Assuming a non-zero acceleration.) This is perfectly well described by standard calculus.

3. Jun 20, 2015

Carlos Gouveia

Orodruin, I'm not sure if I understood your explanation, but if the time interval during which we have a particular velocity has length zero, then shouldn't all cars accelerate at the same "rate", or go from 0 to x mph instantaneously? I know that from definition a = dv/dt but that doesn't make things better as "dt" is not exactly zero, it tends to zero. Sorry, you got me a bit confused. Thanks for your answer anyway.

4. Jun 20, 2015

Orodruin

Staff Emeritus
No. The time during which you have a particular velocity being zero does not mean that all cars accelerate at the same rate (and definitely not instantaneously). Acceleration determines how much time it will take the car to reach the velocity v+dv from the velocity v.

5. Jun 21, 2015

Josh S Thompson

dt is 0, your question is hard to explain I think you just need to accept it about continous numbers. dt is 0 because the limit tends to 0 but limits are hard to prove I think they are actually impossible to prove but they are the basis for calculus. It makes more sense to just think about numbers as countable pieces and then your "paradox" goes away.

6. Jun 21, 2015

jbriggs444

Essentially none of this is correct.

Neither dv nor dt is zero. They are notational devices and can be thought of as infinitesimals. If working with limits, they properly exist inside a quantifier and have no value at all outside the scope of the quantifier.

Limits do not "tend" to anything. The limit of a chosen function at a chosen point is (if it exists at all) a fixed and exact value. We might say that the function tends to the limit as its argument approaches that point. But that's not the same thing as saying that the limit tends to something.

Limits are rigorous. There is no particular difficulty "proving them".

Thinking of the real number line as a countable sequence of points, one after another, very closely spaced may resolve some paradoxes. But that approach has its own difficulties.

7. Jun 21, 2015

Josh S Thompson

What do you mean quantifier? How can you prove 1/x = 0 as x = infinte.

8. Jun 21, 2015

jbriggs444

A quantifier is a mathematical term used to refer to a statement of the form "for all <x in some set> <some statement about x>" or to a statement of the form "there exists a <y in some set> such that <some statement about y>". The "for all" version is a Universal Quantifier. The "there exists" version is an Existential Quantifier.

The definition of limit is cast in terms of a set of nested quantifiers. It is not a statement about infinite x. Traditionally the definition uses epsilon and delta as variables. In this case the statement would be: "For all epsilon greater than 0 there exists a delta greater than 0 such that for all x > delta and all x < -delta, the difference between 1/x and 0 is less than epsilon. That statement is provably true. Please attempt the proof.

The statement above is what is meant when a mathematician says that the limit as x approaches infinity of 1/x is zero. (Though my teachers would be rolling over in their graves if they heard me say "approaches infinity" rather than "increases without bound")

9. Jun 21, 2015

Josh S Thompson

I don't really understand these quatifiers to be honest. I think I understand what you are saying basically you can always pick a higher number and make the difference smaller than epsilon. But my understanding would have delta as the inverse of epsilon.

Thank you though I appreciate the knowledge

10. Jun 21, 2015

Josh S Thompson

What does that mean; for all epsilon greater than 0 there exists delta greater than 0.

Why do you need to say epsilon and delta. for all epsilon greater than 0 means all numbers greater than 0?
What is the relation between delta and epsilon?

11. Jun 21, 2015

A.T.

Your confusion has nothing to do with physics or movement. Any finite non-zero number can be split into infinitely many non-zero parts. That doesn't make the number infinite.

12. Jun 21, 2015

Carlos Gouveia

As far as I know, any finite non-zero number is unique. How come could it be "split into infinitely many non-zero parts"? I don't get it, sorry.

13. Jun 21, 2015

A.T.

I don't get what being unique has to do with it, sorry. You can represent a positive finite non-zero number as a sum of infinitely many positive non-zero numbers:

1 = 1/2 + 1/4 + 1/8 + ....

14. Jun 21, 2015

Carlos Gouveia

I understand your point now, but I don't think that that has much to do with my original post. What I meant to say was that in the interval between 0 and 50 mph there is an infinite amount of speeds. Thus, to go from 0 to 50 mph a car should go through all those speeds, and that would take an infinite amount of time. However, that's not what happens. Surely the fault is mine -- I don't think I made myself quite clear in the post that originated this thread.

15. Jun 21, 2015

A.T.

Why? How do you calculate the total time? My example shows that you can have infinitely many non-zero time intervals, and yet their sum is still finite.

16. Jun 21, 2015

jbriggs444

An infinite number of speeds in an infinite number of times. What's the problem?

17. Jun 21, 2015

Carlos Gouveia

Good point. Yes, the number of "times" is infinite as well. I like this answer.

18. Jun 21, 2015

jbriggs444

It may help to think of it as a game.

For any epsilon you choose that is greater than zero, I can then choose a delta greater than zero. Then you get to choose an x that is either greater than delta or less than negative delta.

What I'm claiming is that no matter what epsilon you choose and no matter what x you choose within those rules, my a choice for delta can ensure that 1/x < epsilon.

Last edited: Jun 21, 2015
19. Jun 21, 2015

DaveC426913

20. Jun 21, 2015

Carlos Gouveia

I got your point now. :-)