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We often calculate the velocity of a boat moving upstream as the (velocity of boat in still water$-$velocity of stream), and add them for downstream. My teacher told me that it's "net velocity" of a body, but can someone prove mathematically that net velocity is the sum of all the velocities acting on the body? I get it intuitively but I want to know the maths clearly behind arriving at such a conclusion.

Edit: some of the answerers think that I'm talking about relative velocity. I'm not talking of the relative velocity, I'm talking about the "final" velocity of the boat which a stationary observer from, say, the shore of the stream will observe the boat to travel with.

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  • $\begingroup$ Well the math is just v_net = v_boat - v_stream. So this follows from the specific definition of net velocity in this case. There is nothing more to it. The more fundamental question lies in what net velocity will tell you (e.g. which simplifications are implied). $\endgroup$
    – Alexander
    Nov 24 at 8:53
  • $\begingroup$ So is it like, a "fundamental" formula? $\endgroup$
    – Mehmer
    Nov 24 at 9:17
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    $\begingroup$ No matter what you call it, all velocities are relative. What you do here is a change from velocity relative to water to velocity relative to the shore. $\endgroup$
    – nasu
    Nov 24 at 12:20
  • $\begingroup$ @nasu you are missing what I mean, please try to understand me. I mean that suppose a boat is sailing against the flow of water, i.e., upstream. The speed will be slower compared to the velocity of the boat in still water right? This slower velocity is calculated by the formula I mentioned in my question, and I wanted to know how could we just subtract the velocities to get the "final" velocity of the boat. You can check the answer I posted on this question itself to understand what I meant. Thanks. $\endgroup$
    – Mehmer
    Nov 24 at 15:44
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    $\begingroup$ This slower velocity is the velocity relative to the shore. The velocity relative to water is the same in both cases. I did not miss anything. The "proof" is the same no matter how you call it $\endgroup$
    – nasu
    Nov 24 at 16:00
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What you are dealing with here is typically called relative velocity. Honestly, I've never heard the term net velocity before and it's doesn't seem right to talk about several velocities "acting" on an object.

Rather, an object has just one single velocity. We can talk about its velocity vector components if we will, but that's all.

Relative velocity in a non-relativistic sense (for not-too-high speed) is simply the velocity as seen from different frames and is simply the difference. You consider the speed of some other object and then subtract your own speed "away".

  • If you are floating downstream on a bathing ring with the speed of the stream, then you will pass the upstream-sailing boat with the difference between your velocities. Remember to include signs, and then it's a simple subtraction: $$v_\text{of boat relative to you}=v_\text{boat} - v_\text{you}=v_\text{boat} - v_\text{stream}.$$ This calculation holds true from any reference frame, meaning for speeds measured from any frame (still assuming not-too-high speeds).

  • If you are standing on the ground, then the boat will pass you as well with the relative speed: $$v_\text{of boat relative to you}=v_\text{boat} - v_\text{you}=v_\text{boat}.$$ The last reduction can be made when the speeds are measured from the ground frame, where your own speed is zero.

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  • $\begingroup$ I'm aware of relative velocity, thanks for the answer. I'll edit my post to make sure what I'm actually asking for but actually, I've figured out the answer myself, so I'll post that too myself $\endgroup$
    – Mehmer
    Nov 24 at 9:57
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    $\begingroup$ I have heard the term net velocity only once: in a poorly-worded example problem here on Physics Stack Exchange. (I can't remember if it was homework or an exam, or if it was published generally) $\endgroup$
    – garyp
    Nov 24 at 13:06
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    $\begingroup$ +1 for mentioning the difference in reference frames. Net velocity of anything on earth would be a very different calculation, if such a concept could exist. $\endgroup$
    – mascoj
    Nov 24 at 20:27
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A simple mathematical basis for determining relative quantities

Because all positions, velocities, and speeds are relative (i.e., there is no absolute frame that we know about), we need a structured approach for describing these quantities. We can use vectors to describe these quantities. Without getting into formal linear algebra, let's first consider positions. enter image description here This figure shows 3 points in a Cartesian geometry and the arrows connecting them represent relative position vectors. The arrow pointing toward B from A is the position of B relative to A and in notation we will call it $\vec{r}_{BA}$. The other two arrows/vectors would be $\vec{r}_{CB}$ and $\vec{r}_{AC}$.

Geometrically we see that $$\vec{r}_{BA}+\vec{r}_{CB}+\vec{r}_{AC}=0.$$

That means if we know the position of object A relative to C, and C relative to B, we can calculate the position of B relative to A: $$\vec{r}_{BA}=-\vec{r}_{CB}-\vec{r}_{AC}.$$ If we need the position of A relative to B, we simply take the negative of the vector: $$\vec{r}_{AB}=-\vec{r}_{BA}=\vec{r}_{AC}+\vec{r}_{CB}\tag{1}$$

Now we can calculate the velocity of A relative to B by taking the time derivative of both sides of the equation (1): $$\vec{v}_{AB}=\vec{v}_{AC}+\vec{v}_{CB}\tag{2}$$

We apply this to your problem by saying A is the boat, B is the observer on the shore, and C is the water:

$$\vec{v}_{\mathrm{boat,shore}}=\vec{v}_{\mathrm{boat,water}}+\vec{v}_{\mathrm{water,shore}}\tag{3}$$

For 1D motion, if the boat velocity relative to the water is $+V_b$ and the water velocity relative to the shore is $-V_w$, then the velocity of the boat relative to the shore is $V_b-V_w$.

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The concept of velocity depends on the observer. If you are on the river shore and you see someone on a boat rowing upstream with the tempo that would result in, for example 3 m/s on a still lake, and if the river flows with the velocity of 2 m/s, then it's logical that you will perceive that the boat will move with the velocity of 1 m/s. However, if you're in the boat, the boat seems stationary to you, and the observer on the shore is the one that you perceive is moving (the whole shoreline is moving from that perspective actually). So to iterate, this net velocity you are interested in depends on the observer.

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I'm answering my own question as I think I came up right with this:

When a boat is travelling up a stream; say velocity of the boat is $v_b$(in still water), the velocity of the stream is $-v_s$. Now let's along still water, in a time $t$ the displacement of boat $d_b= v_bt$. After the boat has travelled the distance, the stream moves backwards, and since the boat is on the stream, it moves back the same distance the stream travels, so, the displacement of the boat now is the displacement of the stream, which is $d_s=-v_st$. So the final displacement of the boat from initial position is $$d_b+d_s$$$$=v_bt - v_st$$$$=t(v_b-v_s)$$ and so the speed of the average velocity boat is $(v_b-v_s)$.

We observe this as the final velocity as the time interval between the boat covering distance $d_b$ and then moving back by $d_s$ is $0$, i.e., the "moving forward" and "coming back" occurs simultaneously and what we observe is the final displacement divided by time taken.

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    $\begingroup$ This might work along a single Cartesian direction but wouldn't in 2 or 3 dimensions. And you "final velocity" and "final displacement" are still relative to some point you consider fixed (but might not be). In physics, we call it relative velocity and relative position. $\endgroup$
    – Bill N
    Nov 24 at 13:35
  • $\begingroup$ Alright, thanks. And yes, I do know what's relative velocity, and here, the final velocity of the boat is the velocity with respect to an observer on the shore $\endgroup$
    – Mehmer
    Nov 24 at 15:41
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    $\begingroup$ Mehmer, no mathematical proof can "proove" what's actually happening in nature, because there are always unknowns. All you can do is come up with a mathematical model (description), and experimentally check how well it matches reality, and under what circumstances. And if you had a very precise experimental setup, and sensitive enough equipment, you'd find out that this law is only approximately true (but for everyday speeds and circumstances, the discrepancy is minuscule) - which is what special relativity is all about. $\endgroup$ Nov 24 at 18:38
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I agree with you that this is not clear. It rests on the idea that the law of physics applicable here (Newtonian mechanics, fluid mechanics) are Galilean invariant. Which they are. So you just transfer to the rest frame of the stream, have your boat speed there and then transfer back.

One can see that this assumption is really made, since it does not follow for relativistic situations.

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Suppose we have a series of buoys floating in the water, each one meter downstream from the last (we'll pretend that the current is stable enough that they stay the same relative position from each other). Suppose at 0 sec the boat is at buoy0, and both are assigned the x coordinate 0m. At 1 sec the boat is at buoy5. So relative to the water, the boat is travelling 5m/s. At time 0 sec, buoy5 was at x = 5m. If the river current is 1m/s, then at time 1 sec, buoy5 is at x = 6m. So the boat travelled 6m in 1s, for a velocity of 6m/s, which is equal to the current velocity plus the velocity of the boat relative to the water.

To some extent, "relative velocity" is, by definition, total velocity minus the velocity of the thing it's relative to, so it follows that total velocity is relative velocity plus the velocity of the thing it's relative to. One way of thinking about a position vector is that it's the displacement vector between the origin and the point being considered. That is, given a point $A$, $\vec A= A-O$ (in mathematical terms, a vector space can be defined from an affine space by picking an origin point). Note that $A$ and $O$ are actual points, while $\vec A$ is a position vector. If we want to find a velocity, it doesn't matter what point we choose as the origin, as long as it's consistent, because velocity is calculated in terms of difference in position vectors. $\vec v = (\vec p_f - \vec p_i)/(t_f-t_i)$ and $\vec p_f - \vec p_i = (P_f-O)-(P_i-O)=P_f-P_i$, so $O$ cancels out. If, however, we are measuring the position relative to the current, then the origin point is moving with the current, so we have two different $O$, so we have $(P_f-O_f)-(P_i-O_i)=(P_f-P_i)-(O_f-O_i)$

Calculating the velocity relative to the current is different from calculating the velocity in the current's frame of reference. With the speeds you'll encounter with rivers, they end up being very close to the same thing, but at relativistic speeds they are different.

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There is no mathematical "proof" i think why the net velocity has to be the vector sum of the individual velocities. It might well not be. For everyday situations, we observe that this is generally what happens in the real world. However, when the speed gets much higher (closer to the speed of light), then vector addition of velocities turn out to be false. So this is really a physics theory problem, and cannot be proved by mathemetics.

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A change of reference frame to a moving coordinate system is known as a boost. That boosts should compose as you say (vector addition) is a postulate of non-relativistic classical mechanics. In technical language, non-relativistic classical mechanics assumes that physics is invariant under the action of the Galilean group.

In other words, you can't hope to prove this relationship because it's a fundamental assumption. As physicists later learned, it's an assumption that doesn't even agree with observation. It turns out that physics is actually (locally) invariant under the action of the Poincare group. The situation is much more complicated here. Instead of simple vector addition, we can find that two boosts compose to give us a boost plus a rotation.

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