Achiles and a tortoise (not Zeno's Paradox)

[Avathacis]

Homework Statement

When Achiles decided to race with the tortoise, it had already walked 64 meters. After Achiles lowered the distance between him and the tortoise 8 times, he realized that he was 15 times faster than his opponent. Achiles stopped the race. How far did Achiles run before he stopped the race? They ran in one direction.

Homework Equations

s = v*t
t = s/v
v = s/t
(v - velocity, t - time, s - distance)

The Attempt at a Solution

This looks like a physics problem, but there isn't a lot of physics in this. Anyway, it seems that when Achiles stopped the tortoise had run:

64+v_tt_t (where bottom t marks tortoise's characteristics)

Achiles stopped at:

64+v_tt_t/8

Achiles speed is: 15v_t

And I'm totally lost now :(.

 

[Mentallic]

On the first leg, achilles ran 64 metres and the tortoise (running at 1/15 the speed of achilles) would've ran 1/15 of that distance, thus 64/15 further than the 64 metre mark.

On the second leg, achilles had to run that 64/15 metres and the tortoise again going at 1/15 of the speed would've achieved 1/15 of that distance, thus (64/15)/15=64/15²

etc.

 

[Avathacis]

On the first leg, achilles ran 64 metres and the tortoise (running at 1/15 the speed of achilles) would've ran 1/15 of that distance, thus 64/15 further than the 64 metre mark.

On the second leg, achilles had to run that 64/15 metres and the tortoise again going at 1/15 of the speed would've achieved 1/15 of that distance, thus (64/15)/15=64/15²

etc.

I'm probably misunderstanding you, but this isn't Zeno's Paradox. Achilles probably won't even reach 64 meters before he stops the race.

 

[Mentallic]

I assumed that it is like Zeno's paradox, where Achilles runs up to the last point he saw the tortoise and counts that as one leg, then keeps getting closer by this method, but stops after 8 legs rather than doing this ad infinitum. But I guess not.

After Achiles lowered the distance between him and the tortoise 8 times

Can you please be more clear as to what this means then?

 

[Avathacis]

I assumed that it is like Zeno's paradox, where Achilles runs up to the last point he saw the tortoise and counts that as one leg, then keeps getting closer by this method, but stops after 8 legs rather than doing this ad infinitum. But I guess not.Can you please be more clear as to what this means then?

Yeah, sure.

Let's assume there's a guy and his girlfriend. The guy notices his girlfriend who is 160 meters away from the guy. When the guy decreases the distance between them by 8 times (meaning his girlfriend is now 20 (160/8=20) meters away from him) the girlfriend notices him.

If it's still unclear i'll try to draw.

 

[willem2]

let $v$ be the speed of the tortoise. The speed of
Achilles is $15 v$

after a time t, the tortoise has run ..... meters, and
Achilles has run ... meters, so their distance is .....

this distance should equal the initial distance divided by 8.

This happens for t = .... when Achilles has run ...

 

[Mentallic]

The guy notices his girlfriend who is 160 meters away from the guy. When the guy decreases the distance between them by 8 times (meaning his girlfriend is now 20 (160/8=20) meters away from him) the girlfriend notices him.

Ahh ok I see, the only difference is that they close the distance between each other. This isn't a race at all!

Let's make a quick logical calculation first. If we didn't consider how many times they closed the gap between each other and just let them run at each other, they will meet at the 60 metre mark, or in other words the tortoise would have run 4 metres since the tortoise runs at some speed, and achilles runs 15 times faster so they are closing the gap between each other at 16x this velocity, and the distance is 64 metres so 64/16=4 and thus 4 metres from the tortoise's starting position.

Ok so rather than using numbers, I'll use arbitrary constants which will hopefully give you a better understanding of the pattern that is going on.

• The tortoises speed is unknown and after some time he will travel some distance, let that be x.
• Achilles will be running at some constant multiplier of this speed, in this case 15, so let m denote the multiplier, thus he will run a distance of mx in the same time.
• Their distance apart will be denoted as s.
• The ratio of how much they close the gap each time will be given n, thus in this case we have n=8 since we are tightening the gap by 1/8.
• the distance they travel will each travel will be given the variable d.

Ok so on the first run, achilles is at the starting point and his distance traveled is mx in the positive direction. The tortoise will cover a distance of x in the negative direction starting at the point s. They will meet at a distance of s/n apart.
Thus we have:
$$d=mx$$
$$d=(s-x)-\frac{s}{n}$$

Solving this for d gives

$$d=s\left(\frac{m}{m+1}\right)\left(\frac{n-1}{n}\right)$$

EDIT: stop here, the rest is answering my misinterpretation of the question.

Now on the second run, achilles is at that point d and the tortoise is at the point d+s/n (since they're that far apart) and we follow the same process:

$$d=mx+s\left(\frac{m}{m+1}\right)\left(\frac{n-1}{n}\right)$$

$$d=s\left(\frac{m}{m+1}\right)\left(\frac{n-1}{n}\right)-x-\frac{(\frac{s}{n})}{n}$$
$$=s\left(\frac{m}{m+1}\right)\left(\frac{n-1}{n}\right)-x-\frac{s}{n^2}$$

Solving for d, that big long expression from the first run cancels ofcourse, and we get

$$d=\frac{1}{n}.s\left(\frac{m}{m+1}\right)\left(\frac{n-1}{n}\right)$$

So after two runs, we just add those two distances together. Do you notice the pattern of the distance traveled after each run?

 

[D H]

When Achiles decided to race with the tortoise, it had already walked 64 meters. After Achiles lowered the distance between him and the tortoise 8 times, he realized that he was 15 times faster than his opponent. Achiles stopped the race. How far did Achiles run before he stopped the race?

Terribly worded question! Is that the way the question is worded in your assignment, or is this your rephrasing of the question? The problem is the phrase "After Achiles[sic] lowered the distance between him and the tortoise 8 times". Does it mean the distance between Achilles and the tortoise has been

• Reduced by a factor of 8 (so the distance between them is now 8 meters),
• Halved 8 times (so the distance between is now 1/4 meter), or
• Something else?
  • 
    
    

    [Long calculation trying to make this into a Achilles v. tortoise series elided]

    That's the tortoise approach to solving this problem.
    
    

    Achiles stopped at:
    
    64+v_tt_t/8

    And that's the wrong way. You don't even know that Achilles ran 64 meters.
    
    You do have the tortoise's position correct:
    

    Anyway, it seems that when Achiles stopped the tortoise had run:
    
    64+v_tt_t (where bottom t marks tortoise's characteristics)

    So let's give that distance a label (I'm using capital T to denote the tortoise, thereby avoiding the confusion between tortoise and time):
    
    d_T = 64 + v_Tt
    
    In that time we know that Achilles ran
    
    d_A = v_At = 15v_Tt
    
    The distance between Achilles and the tortoise at some time t is
    
    d_T - d_A = (64 + v_Tt) - (15v_Tt) = 64 - 14v_Tt
    
    This has a given value of 8 meters (but see the note at the top of this post):
    
    d_A = 64 - 14v_Tt = 8
    
    This provides you with all the information needed to solve for d_A, the distance Achilles has run.

 

[Mentallic]

That's the tortoise approach to solving this problem.

I took a few more steps to make it clear for the OP to understand. Giving the constants an arbitrary value so the pattern can be easily spotted for example.

And by the way, if my solution is a tortoise's approach to the problem, then yours is a turtle's approach

 

[D H]

You used my approach eight times over.

You also appear to have made just the opposite interpretation I did of that troublesome phrase (see start of post #8). I assumed that it means the distance is reduced by a factor of 8 while you appear to assume the distance is halved 8 times. If that is the case, do not take this as a jibe aimed at you. It is a jibe aimed at a poorly-stated problem.

 

[Mentallic]

Oh god yes I severely misunderstood the problem. Somehow the entire Zeno's paradox triggered and I solved the problem whereby they close the gap between each other by a factor of 8... 8 times

I wish it weren't achilles and the tortoise (or at least took the title of the post more seriously).

 

[Avathacis]

I'm sorry for all that confusion. The problem was a translation from Lithuanian to English (but the translation is nearly word to word. I was confused too.). It has been reduced by a factor of 8, but that doesn't equal to 8 (or I'm wrong) since we have to take the distance turtle ran into account too?

Thanks for being so patient. I can realize that it was incredibly hard to understand the problem.

Nearly solved!

 

[D H]

I'm sorry for all that confusion. The problem was a translation from Lithuanian to English (but the translation is nearly word to word. I was confused too.). It has been reduced by a factor of 8, but that doesn't equal to 8 (or I'm wrong) since we have to take the distance turtle ran into account too?

OK. You are supposed to solve for the distance Achilles has run when the distance between Achilles and the tortoise has been reduced by a factor of 8. As 64 meters / 8 = 8 meters, you are looking for the point in time at which the distance between the two is 8 meters. The answer is not 56 meters because you do have to account for the distance the turtle has run. At the time that Achilles reaches the 56 meter mark, the tortoise will be some distance beyond the initial 64 meter mark. The distance between them will be more than 8 meters. Achilles has to run somewhere beyond the 56 meter mark to reduce the distance between him and the tortoise to 8 meters.

 

[Avathacis]

The distance between Achilles and the tortoise at some time t is

d_T - d_A = (64 + v_Tt) - (15v_Tt) = 64 - 14v_Tt

This has a given value of 8 meters (but see the note at the top of this post):

d_A = 64 - 14v_Tt = 8

I'm not sure if i completely understand this. As i understand that the distance between them is (at some point in time):

d_T - d_A = (64 + v_Tt) - (15v_Tt) = 64 - 14v_Tt = 8

but if
d_T - d_A = 64 - 14v_Tt
then

d_A = d_T - d_A? (Which is clearly impossible, unless the tortoise ran 0 meters, which is, again, not true).

Also, the last equation gives us 2 variables. How should i solve that without an equation system? Unless, changing t to t = s/v is correct.

 

[D H]

I'm not sure if i completely understand this. As i understand that the distance between them is (at some point in time):

d_T - d_A = (64 + v_Tt) - (15v_Tt) = 64 - 14v_Tt = 8

Correct. That d_T - d_A = (64 + v_Tt) - (15v_Tt) is obvious (I hope). The next step, d_T - d_A = 64 - 14v_Tt, follows immediately from this first step. Finally, that d_T - d_A = 8 is a given; this is the mathematical interpretation of the problem statement.

but if
d_T - d_A = 64 - 14v_Tt
then

d_A = d_T - d_A? (Which is clearly impossible, unless the tortoise ran 0 meters, which is, again, not true).

No! How did you arrive at that?

Also, the last equation gives us 2 variables. How should i solve that without an equation system? Unless, changing t to t = s/v is correct.

Yes, there are two unknowns here, v_T and t. However, you do not need to know v_T and t. All you need to know is there product, and the product of the two is something for which you can solve.

 

[Mentallic]

Wait a second. I went by the assumption that the tortoise and achilles are running towards each other (as per the boyfriend girlfriend analogy in post #5) but then D H is assuming the tortoise is running away from achilles. So which is it?

but if
d_T - d_A = 64 - 14v_Tt
then

d_A = d_T - d_A?

How did you get 64-14v_Tt=d_A ?

 

[Avathacis]

Correct. That d_T - d_A = (64 + v_Tt) - (15v_Tt) is obvious (I hope). The next step, d_T - d_A = 64 - 14v_Tt, follows immediately from this first step. Finally, that d_T - d_A = 8 is a given; this is the mathematical interpretation of the problem statement.No! How did you arrive at that?Yes, there are two unknowns here, v_T and t. However, you do not need to know v_T and t. All you need to know is there product, and the product of the two is something for which you can solve.

I arrived there because of this:

d_A = 64 - 14v_Tt = 8

Because it at least looks the same as d_T - d_A. But never mind that.

Anyway, food wins. I just finished eating and i managed to solve it while bringing the plate downstairs.

v_Tt = d_T

14d_T = 64 - 8

d_T = 4

d_A = 64 + 4 -8 = 60.

The answer is 60 meters!

Thanks a lot guys! You went through quite a lot :O. I need to learn to word problems properly.

@Mentallic, they were running away from each other. I never said that the girlfriend was running towards the guy. It was stated in the original problem as the last sentence.

 

[Mentallic]

I'm not enjoying all these unnecessary variables with subscripts...

If we let the distance traveled by the tortoise be x after some time, then achilles would travel a distance of 15x in the same time. Achilles is at the starting point and the tortoise is 64 metres away. They are running towards each other (an assumption I'm making since I'm not quite sure what the question is asking) and want to close the gap to 64/8=8 metres.

We just need to solve for x in these two equations then,

$$d=15x$$

$$d=(64-x)-8$$

To explain what these equations are, the d=15x is describing the distance achilles is covering. We are looking to solve for d since this is the distance achilles runs. The other equation describes the tortoise's movement such that he starts at 64 metres and moves towards achilles (or in the negative direction, thus -x). The -8 is as though the tortoise is 8 metres closer to achilles and thus will run into him (the graphs will intersect) at the same time that they would be 8 metres apart if they weren't 8 metres closer. We could obviously instead have made it d=15x+8 and d=64-x.

Hopefully this has cleared some light on the problem...

edit: ok so they were running away from each other. So that means I solved two other problems in this thread, without solving the actual problem you asked

 

[Avathacis]

I'm not enjoying all these unnecessary variables with subscripts...

If we let the distance traveled by the tortoise be x after some time, then achilles would travel a distance of 15x in the same time. Achilles is at the starting point and the tortoise is 64 metres away. They are running towards each other (an assumption I'm making since I'm not quite sure what the question is asking) and want to close the gap to 64/8=8 metres.

We just need to solve for x in these two equations then,

$$d=15x$$

$$d=(64-x)-8$$

To explain what these equations are, the d=15x is describing the distance achilles is covering. We are looking to solve for d since this is the distance achilles runs. The other equation describes the tortoise's movement such that he starts at 64 metres and moves towards achilles (or in the negative direction, thus -x). The -8 is as though the tortoise is 8 metres closer to achilles and thus will run into him (the graphs will intersect) at the same time that they would be 8 metres apart if they weren't 8 metres closer. We could obviously instead have made it d=15x+8 and d=64-x.

Hopefully this has cleared some light on the problem...

edit: ok so they were running away from each other. So that means I solved two other problems in this thread, without solving the actual problem you asked

Hey, any math i can get is of great use to me. Especially if it's something that hasn't been taught to me in school. Somehow things "outside" school seem more interesting to me.

Also the above is correct. It's just +x, not -x. It's also way faster.