Boat Speed and Resistance: Solving a Proportion Question in 140 Minutes"

[avcireis]

Homework Statement

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 traveled from B to A towards the current with 9F and the time taken when traveled 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.

 

[phinds]

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

 

[avcireis]

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

 

[pongo38]

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

 

[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

 

[pongo38]

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

 

[haruspex]

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

 

[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.

 

[avcireis]

ok then let's see,

While going from A to B, the boat has the speed of F_current+F_boat. Against this speed there is a relative resistance of α(F_current+F_boat). 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 F_current+9F_boat, resistance is α(F_current+9F_boat). 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 F_boat-F_current, and when going with 9F it is 9F_boat-F_current. The resistance against them will be α(F_boat-F_current) and α(9F_boat-F_current)

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

 

[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.

 

[avcireis]

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.

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 that's why i expressed it as "speed"

 

[haruspex]

I didn't say that it was.

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

 

[haruspex]

While going from A to B, the boat has the speed of F_current+F_boat.

Let's not confuse forces with speeds. How about V_current+V_boat? That's correct if V_boat is the speed relative to the water.

Against this speed there is a relative resistance of α(F_current+F_boat).

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 = αV_slowboat² etc.

 

[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 squaredSo,

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

 

[haruspex]

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

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 (V_boat=0)!

 

[avcireis]

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 (V_boat=0)!

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.

 

[haruspex]

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.

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

 

[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

 

[haruspex]

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.

In that case the force the boat's engine has to produce to maintain that speed is αV_boat². Why do you keep writing α(V_boat+V_water)²?

And resistance is related with the boat's sum speed against the water squared

I don't know what you mean by "the boat's sum speed against the water". I understand V_boat = "the boat's speed against the water", but I don't know why you insert the word "sum".

 

[avcireis]

In that case the force the boat's engine has to produce to maintain that speed is αV_boat². Why do you keep writing α(V_boat+V_water)²?

I understand V_boat = "the boat's speed against the water", but I don't know why you insert the word "sum".

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:

$α (V_{engine}+V_{current})^2$

 

[haruspex]

If there's two velocities at the same direction, don't you add them up?

Of course. So we have, when they're in the same direction:
speed of boat relative to water = v_boat
speed of water relative to bank = v_current
speed of boat relative to bank = v_boat+v_current
force boat engine has to produce, F = αv_boat²
To see that it cannot be α(v_boat+v_current)², consider turning off the engine. Now v_boat=0 and F=0.

 

[avcireis]

I think I understood thanks to you :) but still I haven't got any clue how to solve the problem.

 

[haruspex]

still I haven't got any clue how to solve the problem.

I'll recap the question, reworded slightly, so that it's on this page:

A boat travels from A to B with the current 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 relative velocity squared. What would be the ratio between the time taken when traveled from B to A towards the current with 9F and the time taken when traveled from B to A towards the current with F?

Let the current speed be w and the original boat's speed relative to the water be v (avoiding all that painful subscripting). Let the distance AB be d.
We have F = αv². What equation can you write relating w, v and d?
What will the boat's relative speed be when the engine's thrust is 9F? What equation connects that with w and d?

 

[avcireis]

I think we can cancel out distance since it never changes.

$140 (a v^2 + w) = 60 (a (9 v)^2 + w)$

 

[haruspex]

I think we can cancel out distance since it never changes.

Agreed.

$140 (a v^2 + w) = 60 (a (9 v)^2 + w)$

That's a very strange equation. a v^2 is a force, w is a speed. You can't add a force to a speed.

 

[avcireis]

Ok I'll write it again.

$140 (v + w) = 60 (9 v + w)$

$7 (v+w) = 3 (9 v + w)$

$7 v+7 w = 27 v + 3 w$

$4 w = 20 v$
$w= 5 v$

But here, something goes wrong, while going from B to A, the speed is v minus w. And since w is 5 times v, it seems impossible.

 

[haruspex]

$140 (v + w) = 60 (9 v + w)$

No, it's 9 times the force. So how many times the speed is it?

 

[avcireis]

Oh, right. So here I try again:

$7 ( v + w ) = 3 ( 3 v + w )$
$2w=v$

So the time taken when going with F

${\frac{140 (2w+w)}{2w-w}} = 420$

And when going with 9F

${\frac{140 (2w+w)}{6w-w}} = 84$

The answer is 5 thank you for your help :)