Hello guys,
Please I need someone to help me solve this. I am still battling with it.

Three swimmers must swim from AA to BB and back. First, the first starts, after 5 seconds - the second, after another 5 seconds - the third. Some point CC, located between points AA and BB, all swimmers passed at the same time (until that time, none of them had visited BB). The third swimmer, having reached BB and turning back, meets the second one 9 m from and the first one - 15 m from BB. Find the speed of the third swimmer if the ABAB distance is 55 m.

Timijo:
Hello guys,
Please I need someone to help me solve this. I am still battling with it.
Three swimmers must swim from AA to BB and back. First, the first starts, after 5 seconds - the second, after another 5 seconds - the third. Some point CC, located between points AA and BB, all swimmers passed at the same time (until that time, none of them had visited BB). The third swimmer, having reached BB and turning back, meets the second one 9 m from and the first one - 15 m from BB. Find the speed of the third swimmer if the ABAB distance is 55 m.

Son, this is some bad ass question you've got here and I doubt if there exist an answer to this, as there are inadequate values to work with.
Even if there were an answer, it's definitely going to be a relative answer. That is, perhaps the speed of third swimmer relative to speed of second/first swimmer. E.g 2x speed of first swimmer, instead of a specific speed like 5m/s.
No specific time duration was given meaning the swimmers could have completed this task for any time length. Could have even taken them a whole year to swim to point BB.

The third swimmer, having reached BB and turning back, meets the second one 9m from (9m from where? From the sky or where?) and the first one - 15 m from BB. Find the speed of the third swimmer if the ABAB distance is 55 m.

@bolded areas. You don't make typo errors in this kind of sensitive question.
You typed '9m from', but failed to link it to anything.
Then what is ABAB? You mean AA to BB?

Some point CC, located between points AA and BB, all swimmers passed at the same time (until that time, none of them had visited BB).

Son, this information is useless. We're not dumb, we know they would definitely meet at a particular point. So how do will apply this info when no value was given as regards the distance or time taken to reach point CC, nor was this info linked to anywhere else in the question. It's just an idle info. In fact, it's more of a confusion than information.

Tell your lecturer who gave you this 'mission impossible' of a question to shove it back to wherever he brought it out from. Don't end up in a coma trying hard to solve this unsolvable question. My two cents.

light099:

Son, this is some bad ass question you've got here and I doubt if there exist an answer to this, as there are inadequate values to work with.
Even if there were an answer, it's definitely going to be a relative answer. That is, perhaps the speed of third swimmer relative to speed of second/first swimmer. E.g 2x speed of first swimmer, instead of a specific speed like 5m/s.
No specific time duration was given meaning the swimmers could have completed this task for any time length. Could have even taken them a whole year to swim to point BB.


@bolded areas. You don't make typo errors in this kind of sensitive question.
You typed '9m from', but failed to link it to anything.
Then what is ABAB? You mean AA to BB?


Son, this information is useless. We're not dumb, we know they would definitely meet at a particular point. So how do will apply this info when no value was given as regards the distance or time taken to reach point CC, nor was this info linked to anywhere else in the question. It's just an idle info. In fact, it's more of a confusion than information.

Tell your lecturer who gave you this 'mission impossible' of a question to shove it back to wherever he brought it out from. Don't end up in a coma trying hard to solve this unsolvable question. My two cents.

There is a solution to this question, it's just the way the question was coined that has an issue...The topic I believe falls under-speed, time and distance word problem.

hillsway:
There is a solution to this question, it's just the way the question was coined that has an issue...The topic I believe falls under-speed, time and distance word problem.

Alright but this isn't about belief or claim, it's a mathematics question. Simple calculations and answer is what is needed.
Since you said it has a solution, kindly show us the solution and workings. Only then can I agree with you.
Give the answer and I'll agree with your claim.

Was able to establish some form of ratio between each speed and the time taken from the first part of the equation. Using time t + 10, t + 5 and t as time taken for the 1st, 2nd and 3rd runner to get to same distance AA to CC, I established a ratio for all three speeds.

Then with the second information, I related the identical time taken for 3rd swimmer to get to 64m and 2nd swimmer at 46m. As well as the identical time taken for 3rd swimmer to get to 70m and 1st swimmer at 40m. The aim was to establish a ratio between the 3rd swimmer speed and each of the other two.

Since the speed is constant, I used the ratio from the first information (S3/S2) and the ratio from the second (still S3/S2) information to estimate the the time in the first instance t = 115/9.

Anyway, I landed erroneously at 2.5m/s. Don't ask me how, but I'm calling it a night.

DesChyko:
Was able to establish some form of ratio between each speed and the time taken from the first part of the equation. Using time t + 10, t + 5 and t as time taken for the 1st, 2nd and 3rd runner to get to same distance AA to CC, I established a ratio for all three speeds.

Then with the second information, I related the identical time taken for 3rd swimmer to get to 64m and 2nd swimmer at 46m. As well as the identical time taken for 3rd swimmer to get to 70m and 1st swimmer at 40m. The aim was to establish a ratio between the 3rd swimmer speed and each of the other two.

Since the speed is constant, I used the ratio from the first information (S3/S2) and the ratio from the second (still S3/S2) information to estimate the the time in the first instance t = 115/9.

Anyway, I landed erroneously at 2.5m/s. Don't ask me how, but I'm calling it a night.

Very cool job. Nice try. At least you're better than the other guy making claims with no answer to back it up.
Except that it could also be 4.5m/s, no way to verity the answer.
It's an invalid question.

light099:


Very cool job. Nice try. At least you're better than the other guy making claims with no answer to back it up.
Except that it could also be 4.5m/s, no way to verity the answer.
It's an invalid question.

Yeah. I feel there's a link. Just can't see it cleanly yet.

Wish I fit speak una language

Timijo:
Hello guys,
Please I need someone to help me solve this. I am still battling with it.

Three swimmers must swim from AA to BB and back. First, the first starts, after 5 seconds - the second, after another 5 seconds - the third. Some point CC, located between points AA and BB, all swimmers passed at the same time (until that time, none of them had visited BB). The third swimmer, having reached BB and turning back, meets the second one 9 m from and the first one - 15 m from BB. Find the speed of the third swimmer if the ABAB distance is 55 m.

The 3rd swimmer must be a dolphin.....Good luck to whomever gets the answer

Slynation:


The 3rd swimmer must be a dolphin.....Good luck to whomever gets the answer

Timijo:
Hello guys,
Please I need someone to help me solve this. I am still battling with it.

Three swimmers must swim from AA to BB and back. First, the first starts, after 5 seconds - the second, after another 5 seconds - the third. Some point CC, located between points AA and BB, all swimmers passed at the same time (until that time, none of them had visited BB). The third swimmer, having reached BB and turning back, meets the second one 9 m from and the first one - 15 m from BB. Find the speed of the third swimmer if the ABAB distance is 55 m.

do i look like Wole SOLVENka?

Op bewarn o oo ooo

anyway,
i laugh in a Fulani way

....

Martinez39s:
@Timijo

[1] We shall presume that each swimmer swam with a uniform velocity. Let's call the uniform velocities of first, second and third swimmer V₁ , V₂ , & V₃ respectively.

[2] Let t be time reading of the wrist watch of a neutral observer which started counting from 0 s when the first swimmer started his race. Let T₁ , T₂ , & T₃ respectively represent the time readings of the watches of first, second, and third swimmer which all started reading from 0 s when their respective owners started swimming.

[3] We have from [2] and the question the following equations:

[4] Let t_c represent the time reading on the neutral observer's watch when all swimmer's met at CC. From the question, you have

[5] Let t_b represent the time reading on the neutral observer's watch when the third swimmer got to BB. From the question, it follows that

[6] Solving the equations in [5] and [4] simultaneously (you will have to do this on your own), we discover the following

NB: when you solve simultaneously and get t_b = 4 t_c , you will, after the appropriate substitution, get the following quadratic equation 12t_c² – 375t_c + 150 = 0.

You did a nice job here, but I need to digest it.

Martinez39s:
@Timijo

[1] We shall presume that each swimmer swam with a uniform velocity. Let's call the uniform velocities of first, second and third swimmer V₁ , V₂ , & V₃ respectively.

[2] Let t be time reading of the wrist watch of a neutral observer which started counting from 0 s when the first swimmer started his race. Let T₁ , T₂ , & T₃ respectively represent the time readings of the watches of first, second, and third swimmer which all started reading from 0 s when their respective owners started swimming.

[3] We have from [2] and the question the following equations:

[4] Let t_c represent the time reading on the neutral observer's watch when all swimmer's met at CC. From the question, you have

[5] Let t_b represent the time reading on the neutral observer's watch when the third swimmer got to BB. From the question, it follows that

[6] Solving the equations in [5] and [4] simultaneously (you will have to do this on your own), we discover the following

NB: when you solve simultaneously and get t_b = 4 t_c , you will, after the appropriate substitution, get the following quadratic equation 12t_c² – 375t_c + 150 = 0.

Please, what are the 4th and 5th equations you referred to?

Martinez39s:
@Timijo
[5] Let t_b represent the time reading on the neutral observer's watch when the third swimmer got to BB. From the question, it follows that
40 = V1 t_b
46 = V2 (t_b – 5)
55 = V3 (t_b – 10)

I'm impressed, I must say.
This is a job well done. Smart and neat. Highly organised.

Check the part of your post I quoted here.
At time t_b, when S3 covered 55m, no one knows the distance covered by S1 and S2.
Rather, S3 covered 64m at the time S2 was at 46m.
While S3 covered 70m at the time S1 was at 40m.

The only place we can link them all up is point CC like you've rightly done.

DesChyko:


Yeah. I feel there's a link. Just can't see it cleanly yet.

You might not see it cleanly eventually. It's a crazy question. When you go back to your calculations, it's likely you're not able to follow the steps anymore or you discover you made a mistake in the calculation.
I doubt if it has an answer.

light099:


You might not see it cleanly eventually. It's a crazy question. When you go back to your calculations, it's likely you're not able to follow the steps anymore or you discover you made a mistake in the calculation.
I doubt if it has an answer.

I swear

Timijo:
Please, what are the 4th and 5th equations you referred to?

I never said 4th and 5th equations. I only said solve the equations in [4] and [5] simultaneously.

== The equations in [4] are:

S₁ (t_c ) = S₂ (t_c ) = S₃ (t_c )

V₁ t_c = V₂ (t_c – 5) = V₃ (t_c – 10)

== The equations in [5] are:

40 = V₁ t_b
46 = V₂ (t_b – 5)
55 = V₃ (t_b – 10)

Hope you get.

light099:


I'm impressed, I must say.
This is a job well done. Smart and neat. Highly organised.

Check the part of your post I quoted here.
At time t_b, when S3 covered 55m, no one knows the distance covered by S1 and S2.
Rather, S3 covered 64m at the time S2 was at 46m.
While S3 covered 70m at the time S1 was at 40m.

The only place we can link them all up is point CC like you've rightly done.

The first swimmer started first. The second swimmer started 5s later, and the third swimmer started another 5s later. Despite this, they all got to the same place CC at the same time on the neutral observer's wrist watch. What does this tell you? The third swimmer is swimming the fastest follow by the second and then the first. This means

V₃ > V₂ > V₁

The implication of this is that starting from CC (where the three swimmers are at the same point), the third swimmer reached BB before the first and second swimmers. When the third swimmer reached BB, the second swimmer still needed to travel 9m to reach BB and the first needed to travel 15m to reach BB.

Hence when the third swimmer had travelled 55m (from AA to BB), the second had travelled 46m and the third had travelled 40m.

@Timijo

Martinez39s:
The first swimmer started first. The second swimmer started 5s later, and the third swimmer started another 5s later. Despite this, they all got to the same place CC at the same time on the neutral observer's wrist watch. What does this tell you? The third swimmer is swimming the fastest follow by the second and then the first. This means
V₃ > V₂ > V₁
The implication of this is that starting from CC (where the three swimmers are at the same point), the third swimmer reached BB before the first and second swimmers.

Sure it is very obvious.
It's the first thing to note when one reads the question. Third swimmer is the fastest, followed by second and first swimmer.

When the third swimmer reached BB, the second swimmer still needed to travel 9m to reach BB and the first needed to travel 15m to reach BB.
Hence when the third swimmer had travelled 55m (from AA to BB), the second had travelled 46m and the third had travelled 40m.
@Timijo

This is the point you misinterpreted the question.
Note from the original question:
1. "Three swimmers must swim from AA to BB and back."
They're going to and fro. Not just to BB. Their movement is from AA to BB, then back to AA.

2. "The third swimmer, having reached BB and turning back, meets the second one 9 m from BB."
We were not told what happened when the third swimmer got to point BB. We're only informed after he has gotten there and now turned back to AA.
If a car travels 55m and turns back and travels another 9m. What's is the total distance covered up to that point? 55m? Obviously not.
It would be 55m + the 9m after turning back = 64m.
So at the time (T1), second swimmer covered 46m, while third swimmer covered 64m.

3. and the first one - 15 m from BB.
At that time (T2), first swimmer covered 40m, while third swimmer has now covered 70m (55m + 15m).

_{@Timijo, I misinterpreted your question at first due to the fact that I rushed it. @light099 has pointed out my errors and I have corrected my workings. I present the solution to your question, below.}

[1] We shall presume that each swimmer swam with a uniform velocity. Let's call the uniform velocities of first, second and third swimmer V₁ , V₂ , & V₃ respectively. From the question, it is obvious that

V₃ > V₂ > V₁

[2] Let t be time reading of the wrist watch of a neutral observer which started counting from 0 s when the first swimmer started his race. Let T₁ , T₂ , & T₃ respectively represent the time readings of the watches of first, second, and third swimmer which all started reading from 0 s when their respective owners started swimming.

[3] We have from [2] and the question the following equations:

S₁ (t) = V₁ T₁ = V₁ t
S₂ (t) = V₂ T₂ = V₁ (t – 5)
S₃ (t) = V₃ T₃ = V₃ (t – 10)

Where S₁ , S₂ and S₃ are distances travelled by the first, second and third swimmers respectively.

[4] Let t_c represent the time reading on the neutral observer's watch when all swimmers met at CC. From the question, you have

S₁ (t_c ) = S₂ (t_c ) = S₃ (t_c )

V₁ t_c = V₂ (t_c – 5) = V₃ (t_c – 10)

It is clear that t_c > 10 s

[5] Let t₃₂ represent the time reading on the neutral observer's watch when the third swimmer, on his return, met the second swimmer. Also, let t₃₁ represent the time reading on the neutral observer's watch when the third swimmer, on his return, met the first swimmer. From the question, it follows that

64 = V₃ (t₃₂ – 10) ------(1)
46 = V₂ (t₃₂ – 5) ------(2)
70 = V₃ (t₃₁ – 10) ------(3)
40 = V₁ t₃₁ ------(4)

It is clear that t₃₂ > 10 s and t₃₁ > 10 s

[6] In [5] when you divide (1) by (2), you obtain this equation

35t₃₂ = 32t₃₁ + 30

[7] Using the equation in [6] on those in [5], you can obtain the following with appropriate substitution

40 = V₁ t₃₁
70 = V₃ (t₃₁ – 10)
1610 = V₂ (32t₃₁ – 145)

[8] Solving the equations in [7] and [4] simultaneously (you will have to do this on your own), we discover the following

• t₃₁ = 4 t_c
• t_c = 20 s
• t₃₁ = 80 s
• t₃₂ = 74 s
• V₃ = 1 m/s (YOUR ANSWER).
• V₂ = 0.666... m/s
• V₁ = 0.5 m/s

NB: when you solve simultaneously and get t₃₁ = 4 t_c , you will, after the appropriate substitution, get the following quadratic equation 12t_c² – 245t_c + 100 = 0.

Martinez39s:
_{@Timijo , I misinterpreted your question at first due to the fact that I rushed it. @light099 has pointed out my errors and I have corrected my workings. I present the solution to your question below.}

[1] We shall presume that each swimmer swam with a uniform velocity. Let's call the uniform velocities of first, second and third swimmer V₁ , V₂ , & V₃ respectively. From the question, it is obvious that

[2] Let t be time reading of the wrist watch of a neutral observer which started counting from 0 s when the first swimmer started his race. Let T₁ , T₂ , & T₃ respectively represent the time readings of the watches of first, second, and third swimmer which all started reading from 0 s when their respective owners started swimming.

[3] We have from [2] and the question the following equations:

[4] Let t_c represent the time reading on the neutral observer's watch when all swimmers met at CC. From the question, you have

[5] Let t₃₂ represent the time reading on the neutral observer's watch when the third swimmer, on his return, met the second swimmer. Also, let t₃₁ represent the time reading on the neutral observer's watch when the third swimmer, on his return, met the first swimmer. From the question, it follows that

[6] In [5] when you divide (1) by (2), you obtain this equation

[7] Using the equation in [6] on those in [5], you can obtain the following with appropriate substitution

[8] Solving the equations in [7] and [4] simultaneously (you will have to do this on your own), we discover the following

NB: when you solve simultaneously and get t₃₁ = 4 t_c , you will, after the appropriate substitution, get the following quadratic equation 12t_c² – 245t_c + 100 = 0.

Well done and thank you.

Timijo:
Well done and thank you.

Did it tally with the answer in your textbook?

Martinez39s:
Did it tally with the answer in your textbook?

The question is not from textbook, it is a Russian Olympiad question.
From the solution you provided and compared to the the answer on the Russian Olympiad website, it is correct.

Martinez39s:

• t31 = 4 tc
• tc = 20 s
• t31 = 80 s
• t32 = 74 s
• V3 = 1 m/s (YOUR ANSWER).
• V2 = 0.666... m/s
• V1 = 0.5 m/s

Now, this is very cool.
Finally, I think this is the long awaited answer. I'm impressed.
You deleted the initial solution. Why? Can I have your initial answers?
Timijo, so far, this should be your answer you've been looking for right here.
Martinez39s you're the GOAT, far better than irrelevant Pessi and Penaldo combined.
If you can solve this, then you could also solve some world's vital problems.
If some Pessi and Penaldo earn a lot per week, why should you earn less when you can offer a lot more to the world? It is sheer deprivation.
The 'system' cheats you. And I won't be surprised if you're one of the so called fans who worship these clowns while in the real sense, they should be your fans worshipping you.
The 'system' has turned 'smart' people into psychological 'simps'. That is why a genius would happily worship and lick the feet of a clown. The weak have taken the earth.

If Bill Gate didn't rebel, he would have been a unpopular pauper today licking the feet of some Pessi and Penaldo.
Until every smart, intelligent mind out there wakes up in the same consciousness of Bill Gate, the weak would continue to rule the strong.
There is going to be a great rebellion someday, when the 'powerholds' of the world would wake up and take their places and all the clowns would be sent back to the trash where they all belong to. But how would that day come soon if bright minds continue to act dumb?

Jack273 ortega004 you can see from this thread that I don't seek to always win a conversation? That's even immature.
When I sincerely see that you're right, I agree.
In fact, I like it when someone else is right and can prove his/her points. I learn this way and it impresses me.
But how do you want me to agree with you when I've already made valid points to prove myself right, while you're obviously wrong and you can't even prove your points?

All I seek for in life is right and sincere answers. The right answers don't always have to come from me. They could come from someone else. It doesn't matter. As long as they're right answers, I'm sincere enough to agree.