Calculating Relative Velocity of Two Boats in a River

[Aristarchus_]

Homework Statement

A boat B starts on one side of the river and goeses up to the middle. The river is $20m$ wide, and the speed of the stream is $0.5m/s$. The boat can move with a speed of $1m/s$ in still water. I am supposed to show that the boat uses $11.5$ seconds to come to the middle. Boat is traveling perpendicular to the river

Relevant Equations

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I thought I could calculate the relative velocity by Pythagorean theorem there $\sqrt{1^2+0.5^2}$
I could divide 10 meters by this value and obtain 8.95 seconds, but this is not the correct answer. Any suggestions?

 

[malawi_glenn]

Are the boats traveling perpendicular to the stream?

 

[Aristarchus_]

Are the boats traveling perpendicular to the stream?
The distance traveled is not 10 m

Boats are traveling perpendicular to the stream,yes. How so? I thought half of the way would be 10m

 

[malawi_glenn]

Boats are traveling perpendicular to the stream,yes. How so? I thought half of the way would be 10m

I have a somewhat hard time to imagine their path
Are the boats also supposed to travel perpendicular to the shore?
Then you must figure out what angle the boats must steer so that they meet on the middle of the straight line joining their initial positions. Then the distance is 10m for each boat.

 

[Aristarchus_]

Turns out they must steer 26.5 degrees. What does this imply and how to calculate the time it takes them to come halfway? The diagonal distance is then 11.17 meters...

 

[malawi_glenn]

Turns out they must steer 26.5 degrees. What does this imply and how to calculate the time it takes them to come halfway?

How did you calculate that? (its wrong)
Can you figure out the boats velocity perpendicular to the shore with that information?

 

[erobz]

Turns out they must steer 26.5 degrees. What does this imply and how to calculate the time it takes them to come halfway?

Ask Yourself: What is their component of velocity in the direction of travel, and how does the distance they travel relate to it?

 

[Aristarchus_]

How did you calculate that? (its wrong)
Can you figure out the boats velocity perpendicular to the shore with that information?

Wasn't the velocity already determined by using the Pythagorean theorem? Or am I mistaken

 

[malawi_glenn]

Wasn't the velocity already determined by using the Pythagorean theorem? Or am I mistaken

perhaps this picture will help

v_w : velocity of water
v_B : velocity of boat
You know three things here and can calcualte the fourth

 

[Aristarchus_]

I thought V_w = V_bx in this situation. I am looking for V_B, right?

 

[malawi_glenn]

I thought V_w = V_bx in this situation.

yes

I am looking for V_B, right?

Nope that one you know!

 

[Aristarchus_]

I am a little confused. It seems to me that I only know two values, the perpendicular speed of the boat and that of the stream... But then what would my objective here be? That is, which vector?

 

[Aristarchus_]

I thought about something. By the Pythagorean theorem, we also calculate the angle between two vectors Ten using that angle, we calculate the diagonal distance, (considering the perpendicular distance of 10m). That turns out to be 11.2m, then divide by the initial speed and we get the approximate value in seconds. Can it be done in this way?

 

[TSny]

There are three velocity vectors of interest:

The velocity of the boat relative to the shore $\vec V_{B/S}$

The velocity of the boat relative to the water $\vec V_{B/W}$

The velocity of the water relative to the shore: $\vec V_{W/S}$

For two of the three vectors, you know their magnitudes. For two of the three vectors, you know their directions.

Write a vector equation that relates these three vectors.

Draw a vector diagram that represents the equation.

Apply geometry/trig to the diagram to solve for unknowns.

 

[kuruman]

Homework Statement:: Two boats, A and B, start on one side of the river $\dots$

I assume this means that the boats start on opposite sides of the river because only then the problem makes sense. To travel perpendicular to the stream, each boat must have an upstream velocity component matching the downstream current. What's left over is the cross-stream component that takes each boat to the middle.

I thought about something. By the Pythagorean theorem, we also calculate the angle between two vectors Ten using that angle, we calculate the diagonal distance, (considering the perpendicular distance of 10m). That turns out to be 11.2m, then divide by the initial speed and we get the approximate value in seconds. Can it be done in this way?

You don't need the angle, just the Pythagorean theorem. See my comments above.

 

[erobz]

The boat can move in any direction on the stream at $1 \rm{ \frac{m}{s}}$ relative to the water. However, the boat has to use up some of that velocity fighting the current in order to move straight across the river. I suspect you need to recheck that answer of pointing $26.5^{\circ}$ from horizontal.

 

[Steve4Physics]

There are some issues with the question.

Why two boats? It sounds like they might be intended to start from different positions and/or at different times and/or with different directions. The question should relate to this. But there is no suggestion of why 2 boats is relevant.

Edit. Apologies. The 11.5 is a time in seconds. I realized too late. Therefore I've deleted the following text:
What does ‘the boats use $11.5$’ mean? You haven’t included a unit so we can’t tell to what sort of quantity ‘11.5’ refers. It could be a distance in metres - but a boat doesn’t ‘use’ a distance. It could be an amount of fuel if you know the rate of fuel-consumption.

Sounds like you need to go back to the original question and check you have given it completely and accurately.

 

[malawi_glenn]

I am a little confused. It seems to me that I only know two values, the perpendicular speed of the boat and that of the stream... But then what would my objective here be? That is, which vector?

The problem said something about what speed the boat has when the water is still.
Given my picture, can you figure out which of the two things VB and VBy this number is referring to?

And yes you do not actually need to know the angle, but it is nice to visualize how the boat need to steer

 

[erobz]

There are some issues with the question.

Why two boats? It sounds like they might be intended to start from different positions and/or at different times and/or with different directions. The question should relate to this. But there is no suggestion of why 2 boats is relevant.

What does ‘the boats use $11.5$’ mean? You haven’t included a unit so we can’t tell to what sort of quantity ‘11.5’ refers. It could be a distance in metres - but a boat doesn’t ‘use’ a distance. It could be an amount of fuel if you know the rate of fuel-consumption.

Sounds like you need to go back to the original question and check you have given it completely and accurately.

This too!

Enclose LAteX embedded in a paragraph as $Latex$, if you want it to be a standalone equation use $$ Latex $$

See the latex guide at the bottom of the post ( when making a new post )

 

[kuruman]

Why two boats? It sounds like they might be intended to start from different positions and/or at different times and/or with different directions. The question should relate to this. But there is no suggestion of why 2 boats is relevant.

Yes, that confused me too at first. The boats must start on opposite sides directly across each other. "Start on one side" probably means that each boat starts on its own side. There might be a translation problem here.

 

[Aristarchus_]

I assume this means that the boats start on opposite sides of the river because only then the problem makes sense. To travel perpendicular to the stream, each boat must have an upstream velocity component matching the downstream current. What's left over is the cross-stream component that takes each boat to the middle.

You don't need the angle, just the Pythagorean theorem. See my comments above.

Both boats start from the same side but think about one boat for simplicity...

 

[malawi_glenn]

Both boats start from the same side but think about one boat for simplicity...

Then what is the need of two boats?

 

[kuruman]

Both boats start from the same side but think about one boat for simplicity...

One is simpler than two. Just calculate how long it takes the boat to travel to the middle of the river moving perpendicular to the current. See my explanation in #15.

 

[Aristarchus_]

This too!

Enclose LAteX embedded in a paragraph as $Latex$, if you want it to be a standalone equation use $$ Latex $$

Seconds, 11.5 seconds. I am losing my mind over LaTex, as I am used to SE and using "$"

 

[Aristarchus_]

There are some issues with the question.

Why two boats? It sounds like they might be intended to start from different positions and/or at different times and/or with different directions. The question should relate to this. But there is no suggestion of why 2 boats is relevant.

What does ‘the boats use 11.511.5’ mean? You haven’t included a unit so we can’t tell to what sort of quantity ‘11.5’ refers. It could be a distance in metres - but a boat doesn’t ‘use’ a distance. It could be an amount of fuel if you know the rate of fuel-consumption.

Sounds like you need to go back to the original question and check you have given it completely and accurately.

seconds, 11.5 seconds

 

[kuruman]

Seconds, 11.5 seconds. I am losing my mind over LaTex, as I am used to SE and using "$"

Here, at PF, we are rich and use $$ instead of just $.

 

[Aristarchus_]

I have restated the problem. Hopefully, it makes more sense now.

 

[Aristarchus_]

What vector component should refer to the halfway distance and how do I obtain 11.5 seconds?

 

[erobz]

I have restated the problem. Hopefully, it makes more sense now.

The latex still isn't fixed, and what do you mean it goes "up to the middle". I'm feeling like this problem is not what I was envisioning.

 

[Aristarchus_]

The latex still isn't fixed, and what do you mean it goes "up to the middle". I'm feeling like this problem is not what I was envisioning.

The middle of the river

 

[malawi_glenn]

What vector component should refer to the halfway distance and how do I obtain 11.5 seconds?

In the picture I gave, you said correctly that VBx = Vw
Which of the vectors VB and VBy do you think will give you what you need to calculate the time it takes to reach that halfway distance?
It was given in the problem that the boat can travel at 1 m/s when the water is still. Which of the vectors VB and VBy do you think has something to do with this information?

 

[Aristarchus_]

In the picture I gave, you said correctly that VBx = Vw
Which of the vectors VB and VBy do you think will give you what you need to calculate the time it takes to reach that halfway distance?

Well...Intuitively Vby but I thought that one could calculate the diagonal value since that is the direction which the boat is actually moving towards...

 

[malawi_glenn]

If the water is still, then VBx = 0
How would the picture i gave you look?

If the water is streaming with Vw = 1m/s
How would the picture I gave you look? Will the boat ever be able to get to the middle of the river then?

 

[Aristarchus_]

In the picture I gave, you said correctly that VBx = Vw
Which of the vectors VB and VBy do you think will give you what you need to calculate the time it takes to reach that halfway distance?
It was given in the problem that the boat can travel at 1 m/s when the water is still. Which of the vectors VB and VBy do you think has something to do with this information?

But would the speed then still be 1m/s, or the relative one, by the Pythagorean theorem?

 

[kuruman]

What vector component should refer to the halfway distance and how do I obtain 11.5 seconds?

Imagine a bridge over the river and a man walking on it so he keeps lined up with the boat until they both reach the middle of the river (10 m) at the same time. How fast should the person be moving for that to happen? Remember that the boat has a velocity vector of 1 m/s. That has two components as mentioned in post #15. One of them is 0.5 m/s to cancel the current so that boat can move straight across. What is the other one?

 

[malawi_glenn]

Well...Intuitively Vby but I thought that one could calculate the diagonal value since that is the direction which the boat is actually moving towards...

These things you have to be careful with. What does it mean that it is actually moving towards? You have several frames of reference here, the water frame, the boat frame, and the shore frame.

But would the speed then still be 1m/s, or the relative one, by the Pythagorean theorem?

Answer the questions I gave you carefully

Which of the vectors VB and VBy do you think will give you what you need to calculate the time it takes to reach that halfway distance?

 

[Aristarchus_]

 

[malawi_glenn]

View attachment 305265

Nope.

 

[Aristarchus_]

Nope.

Got it. Where is the mistake?

 

[malawi_glenn]

Got it. Where is the mistake?

The speed of the boat is 1m/s according to the problem.
the VBx and VBy are velocity components
if VBx = 0.5m/s and VBy = 1m/s then VB > 1m/s
Imagine again how the picture would look if the water was still vw = 0 and when the water is streaming with 1m/s.
At all times, VB must equal 1m/s, the boat can not move faster than this relative the water

 

[erobz]

Got it. Where is the mistake?

If the water was still the boat could only move at $1 \rm{\frac{m}{s}}$ in any direction.

 

[Aristarchus_]

The speed of the boat is 1m/s according to the problem.
the VBx and VBy are velocity components
if VBx = 0.5m/s and VBy = 1m/s then VB > 1m/s
Imagine again how the picture would look if the water was still vw = 0 and when the water is streaming with 1m/s.
At all times, VB must equal 1m/s, the boat can not move faster than this relative the water

I get the correct answer by considering the diagonal as the speed of 1m/s which is a hypotenuse. Thus calculating the vertical component I get the right vertical speed...

 

[malawi_glenn]

I get the correct answer by considering the diagonal as the speed of 1m/s which is a hypotenuse. Thus calculating the vertical component I get the right vertical speed...

Great
Now, can you figure out why this gives the correct result and is just not due to some happy accident?

 

[Aristarchus_]

Great
Now, can you figure out why this gives the correct result and is just not due to some happy accident?

I will do so

 

[malawi_glenn]

hint: when we decompose a velocity vector like that, what is the hypotenuse representing?

Can you calculate the time it would take the boat to reach the middle if the speed of the water is
vw = 0m/s
vw=1m/s
vw = 0.75 m/s
good practice

 

[Aristarchus_]

hint: when we decompose a velocity vector like that, what is the hypotenuse representing?

Can you calculate the time it would take the boat to reach the middle if the speed of the water is
vw = 0m/s
vw=1m/s
vw = 0.75 m/s
good practice

Thank you, I will sleep over it, and come back in the morning, But what I roughly understood is that the perpendicular speed (component) is different from the diagonal speed of 1m/s. So 1m/s would in a way be a resultant speed from the effect of the stream, but not quite...

 

[malawi_glenn]

So 1m/s would in a way be a resultant speed from the effect of the stream, but not quite...

1 m/s is the speed of the boat relative to the water.
Its magnitude does not depend on VW.
But since you want the boat to travel straight, VBx needs to be equal in magnitude of VW but opposite direction.
This means that the boat has to steer. The VBy is then the resulting velocity perpendicular to the shore.

might be confusing to draw vectors like this when their magnitude is 0...
but it should give you an idea what's going on.
You can not change the magnitude of VB, only its direction and that will change VBx and VBy

Here is a great song you can listen to when you try my bonus questions

 

[erobz]

1 m/s is the speed of the boat relative to the water.
Its magnitude is does not depend on VW.
But since you want the boat to travel straight, VBx needs to be equal to VW but opposite direction.
This means that the boat has to steer. The VBy is then the resulting velocity perpendicular to the shore.

View attachment 305266
might be confusing to draw vectors like this when their magnitude is 0...
but it should give you an idea what's going on.
You can not change the magnitude of VB, only its direction and that will change VBx and VBy
View attachment 305267

Here is a great song you can listen to when you try my bonus questions

"Rolling on a River" was not covered in my physics lectures!

 

[Steve4Physics]

May be superfluous now, but here's a restatement of the problem (based on the origonal 2-boat version).

A 20m wide river runs east to west with a speed of 0.5m/s.

Boat A is on the south bank at point P.
Boat B is directly opposite on the north bank at point Q.

Each boat moves with a speed of 1m/s relative to the water.

The boats start together and move perpendicular to the river bank:
- Boat A moves due north, so it’s path is along line PQ;
- Boat B moves due south, so it’s path is along line QP.

Show that the boats collide after 11.5s (there are no casualties).

 

[Aristarchus_]

hint: when we decompose a velocity vector like that, what is the hypotenuse representing?

Can you calculate the time it would take the boat to reach the middle if the speed of the water is
vw = 0m/s
vw=1m/s
vw = 0.75 m/s
good practice

I have tried calculating with vw=1m/s, but now I see that it can be a little confusing drawing strictly vectors. As I do not know what happens if the speed of the stream is greater than or equal to the relative speed of the boat (In this case, Vw =>1m/s).

You see... I cannot use the same method, because the drawing is obviosly wrong. The hypothanuse cannot be equal in magnitute, considering speed as a vector quantity...