Relative Velocity concept help

I guess this is going to be my worst thread. I'm so sick of this stupid concept "relative velocity" that i'm in no mood to present this thread in my "exemplary style" for which i have been included in the "best homework threads" section. I'm sorry for that, but its true.
I have no problems regarding relative velocity in one-dimension, thats quite simple. But this relative velocity concept in two-dimensions is really really confusing specially when it comes to solving problems. Believe me or not, i have been trying to understand this concept for the past almost one year or so. But its impossible. I have never faced such a problem in my life, and i guess i wont face it in future either. I'm so agitated that i now go around saying that even if i become one of the greatest physicists in the world, even then i will not be able to do problems from one chapter- and that is relative velocity. I understand all other concepts and dont have much trouble solving problems from other chapters myself(and ofcourse with PF help at times, thanks for that). But i am not even able to solve the simplest problems of this "relative velocity".
I'll explain my problem in detail. See, i do understand the theoretic part, somewhat. But whenever i sit down to do problems, i have varied difficulties and am unable to solve them. I have gone thru many books and solved exaples but still dont get it. each problem seems to pose more problems and they seem to be done differently from each other. I have so many ... anyway here i'll just post a few:
"Rain is falling vertically with speed 4m/s and the man is moving due east with a speed 3m/s. with what speed the rain appears to be coming to man and where should he hold the umbrella to protect himself from rains?"
QUESTIONS THAT TROUBLE MY MIND: (1) what should be the vector relation or equation? (2) Is it v(man wrt ground) + v(rain wrt ground) = v(rain wrt man) ? (all r vectors ofcourse).
Again there is this weird swimmer problem, shortest path, shortest time, all that crap. I cant even form the equation. If only i can form the equation, then only will i resolve the vectors along chosen axes and then find the answer. but i cant form the vector relation. whose velocity with respect to whom plus whose velocity wrt whom will i add? Its all too confusing, even i cant explain my problem. This is the worst topic in physics. No other concept in the world is as confusing as "relative velocity". I feel suicidal. Please please help me, i beg u.
Once again sorry for not presenting my thread well, as always. But i promise that i'll be back with my well organised threads as soon as i understand this concept. If PF cant explain it, then no one on earth can. Thanks a lot.

SGT
Relative velocity is always the difference between the two velocity vectors.
In your problem of the man in the rain, if you put your positive x axis directed to east and the positive y axis directed up, you have:
$$v_{man} = (3,0)$$
$$v_{rain} = (0,-4)$$
So the relative velocity is
$$v_{rel} = (3,0) - (0,-4) = (3,4)$$
All other problems can be solved the same way.

Kurdt
Staff Emeritus
Gold Member
I think that this subject is taught rather poorly because it is also to do with frames of reference and certainly that was never mentioned when I was taught this. I believe I see where you are having difficulty and thus I hope this explaination helps.

Taking your example of the man in the rain, what you must think of first is which frame of reference you wish to use. Since the question asks what way the man perceives the rain falling you must use the rest frame of the man. In the rest frame of the man the rain has two components of velocity. It has a vertical component directed toward the ground of 4m/s and a westerly component of 3m/s. So to find the perceived or relative velocity we just add the vectors together. Then to find the magnitude and the angle all you have to do is apply pythagorean theorems as you should know from maths classes with vectors.

i dont think one can do all rel. vel. problems with that...

SGT said:
Relative velocity is always the difference between the two velocity vectors.
In your problem of the man in the rain, if you put your positive x axis directed to east and the positive y axis directed up, you have:
$$v_{man} = (3,0)$$
$$v_{rain} = (0,-4)$$
So the relative velocity is
$$v_{rel} = (3,0) - (0,-4) = (3,4)$$
All other problems can be solved the same way.

Thanks a lot. But now, what if the rain is not vertical, that is no component is zero. Lets say: "To a man moving due north with a speed 5m/s, the rain appears to fall vertically. When the man doubles his speed, the rain appears to fall at 60 degrees. find the actual speed of rain and its direction."
This cant be solved "the same way"? lets say the actual rain makes an angle 'a' with the vertical. so its component along vertical is v cos a. But the other component will be equal to the velocity of the man. so we get v sin a = v(man with respect to ground). But what do we do about the second part. i thought i had understood , but the second part of the problem, where this stupid man suddenly wants to double his speed, then i dont know how to proceed. all relative velocity problems dont seem to be done in the same way. Anyway, thanks for helping with the first one. But see, every relative velocity problem i try to solve, i am stuck somewhere or the other. At my level of studies, i can do almost all other problems, at least understand them, but i really dont get this. Sorry for troubling you, but any help will be appreciated. thanks.

SGT
app said:
Thanks a lot. But now, what if the rain is not vertical, that is no component is zero. Lets say: "To a man moving due north with a speed 5m/s, the rain appears to fall vertically. When the man doubles his speed, the rain appears to fall at 60 degrees. find the actual speed of rain and its direction."
This cant be solved "the same way"? lets say the actual rain makes an angle 'a' with the vertical. so its component along vertical is v cos a. But the other component will be equal to the velocity of the man. so we get v sin a = v(man with respect to ground). But what do we do about the second part. i thought i had understood , but the second part of the problem, where this stupid man suddenly wants to double his speed, then i dont know how to proceed. all relative velocity problems dont seem to be done in the same way. Anyway, thanks for helping with the first one. But see, every relative velocity problem i try to solve, i am stuck somewhere or the other. At my level of studies, i can do almost all other problems, at least understand them, but i really dont get this. Sorry for troubling you, but any help will be appreciated. thanks.

It is solved the same way. assuming the positive x axis points north the velocity of the rain is
$$v_{rain} = (-V cos a, V sin a) = (-V cos a, 5)$$
The relative velocity is $$v_{rel} = v_{man} - v_{rain}$$
So, when the man doubles his velocity you have
$$v_{rel} = (10,0) - (-V cos a, 5) = (10+V cos a, -5)$$
You know that the angle the rain forms is 60 degrees, so you can calculate V.

SGT said:
It is solved the same way. assuming the positive x axis points north the velocity of the rain is
$$v_{rain} = (-V cos a, V sin a) = (-V cos a, 5)$$
The relative velocity is $$v_{rel} = v_{man} - v_{rain}$$
So, when the man doubles his velocity you have
$$v_{rel} = (10,0) - (-V cos a, 5) = (10+V cos a, -5)$$
You know that the angle the rain forms is 60 degrees, so you can calculate V.

Thanks, I feel i am starting to understand, but could u please tell me why it is $$v_{rel} = v_{man} - v_{rain}$$ and not $$v_{rel} = v_{man} + v_{rain}$$
or why not $$v_{rel} = v_{rain} - v_{man}$$ ?
Thanks a lot (Thats all i can give u).But of course i will always remember you as the "GREAT MAN WHO MADE ME UNDERSTAND THE WIERDEST CHAPTER "RELATIVE VELOCITY".

SGT
app said:
Thanks, I feel i am starting to understand, but could u please tell me why it is $$v_{rel} = v_{man} - v_{rain}$$ and not $$v_{rel} = v_{man} + v_{rain}$$
or why not $$v_{rel} = v_{rain} - v_{man}$$ ?
Thanks a lot (Thats all i can give u).But of course i will always remember you as the "GREAT MAN WHO MADE ME UNDERSTAND THE WIERDEST CHAPTER "RELATIVE VELOCITY".
As Kurdt pointed, it is a question of reference frame. We can consider 3 reference frames for your problem:
1. Earth's reference frame. In this frame the man's velocity is 5m/s north and the rains velocity is Vsin a = 5m/s north and -V cos a up.
2. Man's reference frame. In this frame man's velocity is zero and the relative rain's velocity is $$v_{rain} - v_{man}$$ = -V cos a up and 0m/s north
3. Rain's reference frame. In this frame the rain is immobile and the relative man's velocity is $$v_{man} - v_{rain}$$ = V cos a up and 0 m/s north
Any of these are valid. I was using the rain's reference frame in my example, but the more intuitive would be to use man's reference. Please, make the necessary adjustments in my previous answer.
$$v_{rel} = v_{rain} - v_{man}$$

SGT said:
As Kurdt pointed, it is a question of reference frame. We can consider 3 reference frames for your problem:
1. Earth's reference frame. In this frame the man's velocity is 5m/s north and the rains velocity is Vsin a = 5m/s north and -V cos a up.
2. Man's reference frame. In this frame man's velocity is zero and the relative rain's velocity is $$v_{rain} - v_{man}$$ = -V cos a up and 0m/s north
3. Rain's reference frame. In this frame the rain is immobile and the relative man's velocity is $$v_{man} - v_{rain}$$ = V cos a up and 0 m/s north
Any of these are valid. I was using the rain's reference frame in my example, but the more intuitive would be to use man's reference. Please, make the necessary adjustments in my previous answer.
$$v_{rel} = v_{rain} - v_{man}$$

WHY 0 m/s? why have u added 0m/s north in each of earth's ref. frame and rain's ref. frame?

SGT
app said:
WHY 0 m/s? why have u added 0m/s north in each of earth's ref. frame and rain's ref. frame?
No, it is 0m/s north in man's and rain's reference frames, because rain is falling vertically on the man, so there is no horizontal component in relative velocity.