Relative velocity in 2 dimensions

[LR5]

Homework Statement

A passenger walks from one side of a ferry to the other as it approaches a dock. If the passenger's velocity is 1.50m/s due north relative to the ferry, and 4.50m/s at an angle of 30.0 degrees west of north relative to the water, what are the direction and magnitude of the ferry's velocity relative to the water?

Relevant Equations

V ⃗13 = V ⃗12 + V ⃗23

(pw is person to water, pf is person to ferry, fw is ferry to water)
I get Vpw = Vpf + Vfw. Therefore Vfw = Vpw + Vpf, which is
Vfw = -Vwp + Vpf. I also have that the x component of Vpw is negative.
I think I'm getting confused with either the order of the formula or the negative negatives (or both) as my answers just don't make sense.

 

[PeroK]

Homework Statement:: A passenger walks from one side of a ferry to the other as it approaches a dock. If the passenger's velocity is 1.50m/s due north relative to the ferry, and 4.50m/s at an angle of 30.0 degrees west of north relative to the water, what are the direction and magnitude of the ferry's velocity relative to the water?
Relevant Equations:: V ⃗13 = V ⃗12 + V ⃗23

(pw is person to water, pf is person to ferry, fw is ferry to water)
I get Vpw = Vpf + Vfw. Therefore Vfw = Vpw + Vpf, which is
Vfw = -Vwp + Vpf. I also have that the x component of Vpw is negative.
I think I'm getting confused with either the order of the formula or the negative negatives (or both) as my answers just don't make sense.

First, have you drawn a diagram?

Second, look at what happens to the positions of things during ##1s##. First draw the ##1s## displacement of the ferry relative to the water; then show the ##1s## displacement of the person relative to the ferry. Then use the diagram you have drawn to calculate the answer.

 

[LR5]

I drew a diagram which is basically the person walking north (positive x axis) from the origin and the then for the person relative to the water a vector 30 degrees from the north axis (to the left of the y axis) also starting at the origin.
I am confused as to using the displacement diagrams you suggest, as I have no information for the ferry to water. Do you mean person to water? It indicates that the person travels a lot further relative to the water in 1s than the person relative to the ferry, but I'm not sure how to work it from there.

 

[David Lewis]

Sketch a velocity vector diagram. Based on your diagram write a velocity vector equation but place the variable you are looking for to the left of the equal sign.

 

[PeroK]

I drew a diagram which is basically the person walking north (positive x axis) from the origin and the then for the person relative to the water a vector 30 degrees from the north axis (to the left of the y axis) also starting at the origin.
I am confused as to using the displacement diagrams you suggest, as I have no information for the ferry to water. Do you mean person to water? It indicates that the person travels a lot further relative to the water in 1s than the person relative to the ferry, but I'm not sure how to work it from there.

Yes, I see you are given the person relative to the water. Apologies. So that is the resultant vector on your diagram. The ferry relative to the water becomes the unknown.

Another approach is to use the motion of the ferry relative to the person. And then follow the simpler method.

 

[LR5]

Sketch a velocity vector diagram. Based on your diagram write a velocity vector equation but place the variable you are looking for to the left of the equal sign.

This is what I have:
Vfw = Vpf + Vpw
= -Vfp + Vpw
Vpf = 0x = (1.5)y
Vpw = (-4.5 cos 30)x + (4.5 sin 30)y
Vfw = -3.90x + 0.75y
= 15.8 m/s at 10.9 degrees west of north

 

[PeroK]

View attachment 256540 This is what I have:

This is not right. The vector ##\vec v_{pw}## is the resultant vector. It is the sum of ##\vec v_{fw} + \vec v_{pf}##. Moreover, your diagram, even if not to scale, should indicate that ##v_{pw}## is much larger than ## v_{pf}##.

Alternatively, you could draw ##\vec v_{pw}##, and then add ##\vec v_{fp}## to it (the velocity of the ferry relative to the person), to give you ##\vec v_{fw}## as the resultant vector.

Note that in your diagram you cannot have ##\vec v_{pf}## starting from the origin. The origin is the initial point in the water. ##\vec v_{pf}## must start from where the ferry is. It's ##\vec v_{pw}## and ##\vec v_{fw}## that start from the origin.

 

[LR5]

This is not right. The vector ##\vec v_{pw}## is the resultant vector. It is the sum of ##\vec v_{fw} + \vec v_{pf}##. Moreover, your diagram, even if not to scale, should indicate that ##v_{pw}## is much larger than ## v_{pf}##.

Alternatively, you could draw ##\vec v_{pw}##, and then add ##\vec v_{fp}## to it (the velocity of the ferry relative to the person), to give you ##\vec v_{fw}## as the resultant vector.

Note that in your diagram you cannot have ##\vec v_{pf}## starting from the origin. The origin is the initial point in the water. ##\vec v_{pf}## must start from where the ferry is. It's ##\vec v_{pw}## and ##\vec v_{fw}## that start from the origin.

Ok,that makes more sense. Thing is, when I change the formula from Vpw = Vpf + Vfw to
Vfw = Vpw - Vpf, I'm not sure if I'm still supposed to switch the pw to wp (and therefore the direction) and I'm pretty sure the x component for Vpw would still be negative as well. So,lots of negatives...

 

[PeroK]

View attachment 256545
Ok,that makes more sense. Thing is, when I change the formula from Vpw = Vpf + Vfw to
Vfw = Vpw - Vpf, I'm not sure if I'm still supposed to switch the pw to wp (and therefore the direction) and I'm pretty sure the x component for Vpw would still be negative as well. So,lots of negatives...

That's still not right.

The method is simple. The water is the fixed reference frame here:

Either:

A) You draw the boat moving in the water, then add the person moving in the boat (vector addition, tail to tip). This gives the person moving relative to the water (resultant vector).

or

B) You draw the person moving relative to the water, then add the boat moving relative to the person. This gives the boat moving relative to the water (resultant vector).

But, you can't add the person moving relative to the boat to the person moving relative to the water. That makes no sense.

I'd figure this out first with vector addition, before thinking about subtracting vectors. Note that subtracting a vector is really just adding the opposite vector.

 

[LR5]

That's still not right.

The method is simple. The water is the fixed reference frame here:

Either:

A) You draw the boat moving in the water, then add the person moving in the boat (vector addition, tail to tip). This gives the person moving relative to the water (resultant vector).

or

B) You draw the person moving relative to the water, then add the boat moving relative to the person. This gives the boat moving relative to the water (resultant vector).

But, you can't add the person moving relative to the boat to the person moving relative to the water. That makes no sense.

I'd figure this out first with vector addition, before thinking about subtracting vectors. Note that subtracting a vector is really just adding the opposite vector.

?

 

[PeroK]

View attachment 256547
?

In general, yes, but those arrows don't correspond to the vectors in this particular case. If the vector on the right is pointing north, then ##\vec v_{pw}## looks ENE to me.

 

[PeroK]

PS Maybe something you are struggling with here is the concept of an "unknown" vector. It's true that you can't draw immediately an unknown vector. So, you have to put in the vectors you know first.

So, first I would draw a diagram with just ##\vec v_{pw}## and ##\vec v_{pf}## on it. You can then deduce ##\vec v_{fw}## from what's missing.

Or, I've hinted at this twice now. Taking ##\vec v_{fp} = - \vec v_{pf}## and then you can draw your tip to tail vector addition directly.

 

[LR5]

I think I'm focusing on the person walking North relative to the ferry and I keep wanting to place that vector pointing north when it is in fact not north relative to the water.

 

[PeroK]

 

[PeroK]

I think I'm focusing on the person walking North relative to the ferry and I keep wanting to place that vector pointing north when it is in fact not north relative to the water.

North is north! The north pole is so far away that a few metres here or there doesn't make any difference. Even if you were looking at a lighthouse a few miles away, the direction "to the lighthouse" would be fine as a reference direction for north for a short time.

If you look at my diagram. Imagine two people who start next to each other. One remains there relative to the ferry (they follow the ferry's path: ##\vec v_{fw}##). The other moves along the path of ##\vec v_{pw}##. Now, imagine how that person moves relative to the first. They appear to walk north: ##\vec v_{pf}##.

That's what the diagram is showing.

 

[LR5]

Thank you. I think I have it now.

 

[phinds]

Thank you. I think I have it now.

Then you should post your solution, with the corrected diagram.

 

[LR5]

Vpw = Vpf + Vfw
Vfw = Vpw - Vpf
Vfw = Vpw + Vfp
Vpw = (-4.5cos30)x + (4.5sin30)y
Vfp = 0x + (-1.5sin30)y
Vfw = (-3.90)x + (2.25 - 0.75)y
= 4.18m/s at (10.9 +30) = 40.9 degrees west of north

 

[PeroK]

Vpw = Vpf + Vfw
Vfw = Vpw - Vpf
Vfw = Vpw + Vfp
Vpw = (-4.5cos30)x + (4.5sin30)y
Vfp = 0x + (-1.5sin30)y
Vfw = (-3.90)x + (2.25 - 0.75)y
= 4.18m/s at (10.9 +30) = 40.9 degrees west of north
View attachment 256555

Sadly I think everything is wrong there!

 

[LR5]

Lol, even the diagram? Please help...

 

[PeroK]

Lol, even the diagram? Please help...

Well, the diagram was one I gave you! So, that's okay. You have four lines of calculations:

Sines and cosine mixed up
Sine inappropriately applied
##v_{fw}## is wrong
Also, the final bearing doesn't match your ##v_{fw}##. You have a large (negative) x-coordinate and a smaller (positive) y coordinate, so that's more west than north.

 

[LR5]

I sort of thought I had sine & cos mixed up..
Vpw = (-4.5sin30)x +(4.5cos30)y
Vfp = 0x + (-1.5)y
Vfw = -2.25x + 2.4y
= 3.28m/s and 43.17 + 30 = 73.17 west of north

 

[PeroK]

I sort of thought I had sine & cos mixed up..
Vpw = (-4.5sin30)x +(4.5cos30)y
Vfp = 0x + (-1.5)y
Vfw = -2.25x + 2.4y
= 3.28m/s and 43.17 + 30 = 73.17 west of north

Why add 30°? ##\vec v_{fw} = (-2.25, 2.4)m/s##. That's your resultant vector.

 

[LR5]

Thanks again. I don't know why this one had me so puzzled, though.