# A proportion question

1. Feb 8, 2013

### avcireis

1. The problem statement, all variables and given/known data

A boat travels from A to B in the current's direction with constant speed in 140 minutes. In this situation the boat has F pushing force. If the boat travels with 9F, it arrives B in 60 minutes. In every circumstance the water resistance is directly proportional with the boat's sum velocity's square. What would be the ratio between the time taken when travelled from B to A towards the current with 9F and the time taken when travelled from B to A towards the current with F?

Choices

A) 3
B) 10
C) 8
D) 5
E) 2

Note: Sorry if the question isn't clear enough, I had to translate. Please ask if you didn't understand.

2. Feb 8, 2013

### phinds

You need to show some attempt to solve it yourself. We don't just spoon-feed answers.

3. Feb 8, 2013

### avcireis

I have been working on the question for a week and neither me or my friends have an idea how to solve it.

4. Feb 8, 2013

### pongo38

Clue: Let R be the ratio required. Then...........?

5. Feb 8, 2013

### avcireis

R=t1/t2

where t1 is 9F, towards the water current from B to A

and t2 is F , twoards the water current, from B to A

6. Feb 9, 2013

### pongo38

What relationships between time, distance, speed and force do you think might be candidates for approaching this problem?

7. Feb 9, 2013

### haruspex

How, according to the statement, is the pushing force related to the speed relative to the water?

8. Feb 10, 2013

### pongo38

I didn't say that it was. I was encouraging you to think about possibilities you may have overlooked. What definitions or laws do you know that might apply here? One approach might be to see if you can solve the problem by assuming (incorrectly according to the question, but not that far removed) that resistance is proportional to speed to the power 1. If you can solve that, then the question might seem easier.

9. Feb 10, 2013

### avcireis

ok then lets see,

While going from A to B, the boat has the speed of Fcurrent+Fboat. Against this speed there is a relative resistance of α(Fcurrent+Fboat). And as well, 1>α>0. Becouse if it was 1, the boat wouldn't move, if it was zero there wouldn't be resistance. When the boat travels with 9F, speed is Fcurrent+9Fboat, resistance is α(Fcurrent+9Fboat). So,

$60 [F_{current}+9F_{boat}-α(F_{current}+9F_{boat})]$

equals to

$140 [F_{current}+F_{boat}-α(F_{current}+F_{boat})]$

So when simplificated,

$3 (1-α) (F_{current}+9F_{boat}) = 7 (1-α) (F_{current}+F_{boat})$

$3 (F_{current}+9F_{boat}) = 7 (F_{current}+F_{boat})$.

When going from B to A, the boat's speed will be Fboat-Fcurrent, and when going with 9F it is 9Fboat-Fcurrent. The resistance against them will be α(Fboat-Fcurrent) and α(9Fboat-Fcurrent)

So what I do from now on? Or am I going wrong?

10. Feb 10, 2013

### pongo38

quote "speed of Fcurrent+Fboat."
what are the units of speed? And what are the units of "Fcurrent+Fboat"? Have you mistakenly used F in that last expression? In the question there are some unknown proportionality constants. These can be eliminated by considering ratios, and I think that is what the question is getting at. F is a pushing force, presumably provided by an engine. You seem to have adopted F with suffixes as a symbol for speed. That is throwing me. Another constant that might be eliminated one way or the other is the constant distance, say 'd', between A and B.

11. Feb 10, 2013

### avcireis

The units are not given in the question so i can't make a comment about it. and if misunderstood, Fboat+Fcurrent is boat's net force but of course without the resistance. But its related to its speed directly thats why i expressed it as "speed"

12. Feb 10, 2013

### haruspex

My question was intended for avcireis. The force is related to the speed, in a way defined in the OP.

13. Feb 10, 2013

### haruspex

Let's not confuse forces with speeds. How about Vcurrent+Vboat? That's correct if Vboat is the speed relative to the water.
Wrong on a couple of counts. The force needed will only depend on the boat's relative speed, and it says it goes as the square, so we have F = αVslowboat2 etc.

14. Feb 10, 2013

### avcireis

But I think there are 3 things that effect the boat, and one of them is towards the boat

The current
The boat's engine etc.
The resistance related to the boat's sum speed squared

So,

Can the resistance be expressed like this: $α (V_{current}+V_{boat})^2$

15. Feb 10, 2013

### haruspex

If $V_{boat}$ is the speed of the boat relative to the water, the resistance must be $α V_{boat}^2$. If it were $α (V_{current}+V_{boat})^2$ then the boat would have to run its engine just to travel at the same speed as the current (Vboat=0)!

16. Feb 10, 2013

### avcireis

I don't know if I misunderstood the question, but I believe even if the boat haven't had an running engine, it should've the speed of current - resistance.

17. Feb 11, 2013

### haruspex

Please clarify: Is your Vboat variable the speed of the boat relative to the water or the speed of the boat relative to the bank?

18. Feb 11, 2013

### avcireis

I think it's relative to the water, if you mean resistance by "water".

By V_{boat} I meant the speed caused only by the boat's engine.
Current is constant
And resistance is related with the boat's sum speed against the water squared

19. Feb 11, 2013

### haruspex

In that case the force the boat's engine has to produce to maintain that speed is αVboat2. Why do you keep writing α(Vboat+Vwater)2?
I don't know what you mean by "the boat's sum speed against the water". I understand Vboat = "the boat's speed against the water", but I don't know why you insert the word "sum".

20. Feb 12, 2013

### avcireis

If there's two velocities at the same direction, don't you add them up? If there's someone walking with 2 km/h in a train with 200 km/h, doesn't that make the person going with 202 km/h? Let the train represent the current and person represent the boat. That's why I add them up and call it the boat's sum speed. But maybe it's better to express it like this:

$α (V_{engine}+V_{current})^2$