Calculating Relative Velocity of Two Boats in a River

[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...

 

[malawi_glenn]

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).
View attachment 305287You 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...

What is the physical significance of this? :)

 

[Aristarchus_]

What is the physical significance of this? :)

Well, the first thing I thought about was that the boat would never move vertically upwards (in Vy direction) and would just be carried away by the stream, which is intuitively wrong. I have to think about this...

 

[Aristarchus_]

What is the physical significance of this? :)

The boat will first move a little in the vertical direction, after which it would eventually be carried away by the stream. So it will never reach the end of the river...But what distance could it reach? That is, what is the maximum distance it would traverse(in the vertical direction) with the speed of the water greater than its relative velocity?

 

[malawi_glenn]

Well, the first thing I thought about was that the boat would never move vertically upwards (in Vy direction) and would just be carried away by the stream, which is intuitively wrong. I have to think about this...

Why is it intuitively wrong? It make perfect sense. If the water stream speed is greater than the speed which the boat can travel with in the water, the boats can not travel along the line connecting their starting points! If the boats want to travel in a direction perpendicular to the shore, they would have to have a non zero VBx.
This becomes impossible:

 

[Aristarchus_]

Why is it intuitively wrong? It make perfect sense. If the water stream speed is greater than the speed which the boat can travel with in the water, the boats can not travel along the line connecting their starting points!
This becomes impossible
View attachment 305289

Right, but it will move some minimal distance in the y-direction before being carried away, or is this not so?

 

[malawi_glenn]

some minimal distance

zero

 

[Aristarchus_]

zero

Wow, who would've thought... Thank you very much for all the feedback

 

[malawi_glenn]

Wow, who would've thought... Thank you very much for all the feedback

its like dropping a ball at rest. As soon as you release it, it will obtain some velocity > 0 (time is not a discrete variable)

And in the case where VW = 0. The boats can use all their speed in the perpendicular direction and travel along that straight line connecting their original positions using minimal time 10s.

When VW = 1m/s, you can "think of it" as it will take an "infinite" time for the boats to meet at the center of the line connecting their original positions.

You should be able to make a function that give the time needed to meet at centre of that line, with VW as the input