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I am having trouble solving relative velocity problems because I can't seem to understand the sign that goes along with the velocity of different objects. $$ v_{ac} =v_{ab} + v_{bc}. $$

How do the signs work? when do I chose $v_{ab}$, $ v_{ac} $ and $v_{bc}$ as a negative? When an object is moving right do I consider it's velocity as negative?

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    $\begingroup$ Have you learned about vectors yet? $\endgroup$ – G. Smith May 27 at 1:05
  • $\begingroup$ Put a vector notation over each term of the equation. $\endgroup$ – Unique May 27 at 2:44
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Usually, we consider movement to the right as positive, but you can choose whichever you want, as long as you are consistent with it, and explain it.

Remember that the first thing to do whenever you solve a Mechanics problem is to define your set of coordinates. So if you define your X-Y axis as increasing from left to right and from down to up, then movement to the right and up will be positive.

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  • $\begingroup$ So if take right as positive in my coordinate system then velocities associated with objects moving to the right will be positive and objects moving to the left will have negative velocity? So the sign for velocity basically just depends on the coordinate system I chose not some other factor? $\endgroup$ – Hema_D May 27 at 5:35
  • $\begingroup$ @Hema_D Yes. Define the positive coordinate directions and relate all your vectors to them. $\endgroup$ – Bill N May 27 at 12:08
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Your equation $v_{\rm ac} =v_{\rm ab} + v_{\rm bc}$ is actually an equation relating to vector addition, $\vec v_{\rm ac} =\vec v_{\rm ab} + \vec v_{\rm bc}$

In one dimension this equation could be written as $v_{\rm ac}\,\hat x =v_{\rm ab} \,\hat x + v_{\rm bc}\,\hat x$ where $v_{\rm ac}, \,v_{\rm ab}$ and $v_{\rm bc}$ are components of the velocity in the $\hat x$ direction.
You can think of the $\hat x$ direction as the direction that you have chosen to be positive.

Going back to your equation $v_{\rm ac} =v_{\rm ab} + v_{\rm bc}$ and noting that these are components in a positive direction you have chosen it is not difficult to assign signs.


Suppose that you chose West as your positive direction.

$\vec v_{\rm ac}$ is $30 \,\rm m\,s^{-1}\, East$ and $\vec v_{\rm bc}$ is $10\,\rm m\,s^{-1}\, West $.
$30\,\rm m\,s^{-1}\, East$ is the same as $-30\,\rm m\,s^{-1}\, West$.

$v_{\rm ac} =v_{\rm ab} + v_{\rm bc}\Rightarrow -30 = v_{\rm ab} + 10 \Rightarrow v_{\rm ab} = -40 $

So the velocity of $a$ relative to $b$ is $-40\,\rm m\,s^{-1}\, West$ is the same as $40\,\rm m\,s^{-1}\, East$


On the other hand Suppose that you chose East as your positive direction.

$\vec v_{\rm ac}$ is $30 \,\rm m\,s^{-1}\, East$ and $\vec v_{\rm bc}$ is $10\,\rm m\,s^{-1}\, West $.
$10\,\rm m\,s^{-1}\, West$ is the same as $-10\,\rm m\,s^{-1}\, East$.

$v_{\rm ac} =v_{\rm ab} + v_{\rm bc}\Rightarrow 30 = v_{\rm ab} + (-10) \Rightarrow v_{\rm ab} = 40 $

So the velocity of $a$ relative to $b$ is still $40\,\rm m\,s^{-1}\, East$


If it was a two or three dimensional problem then all you would need to do is to define two (or three) unit vectors (positive directions) and deal with each of the components in the the two (or three) directions as before.

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The doppler shift formula is the main example where all physics textbooks disagree about the positive or negative sign of relative velocities. Some variants are (link):

$f = \left( \frac{c + v_r}{c + v_{s}} \right) f_0 $, $f = \left( \frac{c - v_r}{c - v_{s}} \right) f_0 $, $f = \left( \frac{c + v_r}{c - v_{s}} \right) f_0 $, $f = \left( \frac{c \pm v_{\rm r}}{c \mp v_{\rm s}}\right) f_0 $, $f = \left( \frac{c-\vec{v}_{r}\cdot \vec{e}_{sr}}{c-\vec{v}_{s}\cdot \vec{e}_{sr}} \right) f_0 $

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  • $\begingroup$ And where is the relative velocity in these formulas? The sign conventions refer to the source and detector moving towards each other or moving away. $ c-v_r $ (for example)is not a relative velocity, is it? $\endgroup$ – nasu May 27 at 13:04
  • $\begingroup$ $v_{r}$ and $v_{s}$ are the relative velocities. In the first example $v_{r}$ is positive for a receiver moving towards the source, in the second example it is positive for a receiver moving away from the source, etc, as described in the link $\endgroup$ – jkien May 27 at 13:24
  • $\begingroup$ Relative to what? They are velocities in respect to the medium (air). Not in respect to a moving object (unless is a windy day). $\endgroup$ – nasu May 27 at 13:28
  • $\begingroup$ The first order approximation is in all cases Δf/f0 = Δv/c, where Δv is the relative velocity of source and receiver. In the first 4 variants Δv is (vr-vs), -(vr-vs), (vr+vs), ±(vr+vs). So in two variants Δv is (vr-vs), in two other variants Δv is (vr+vs). The latter is due to opposite positive directions for vr and vs. $\endgroup$ – jkien May 27 at 14:49
  • $\begingroup$ It's true that the series expansion of the formulas contain the relative velocity. But this expansion is nowhere in your answer. I did not say that your answer contains an error. Just that is not relevant to the question. As it is now, there is no relative velocity in any of these formulas. Here various signs are due to various sign conventions for approaching or moving away. This does not mean that there is an ambiguity in the definition of relatie velocity. $\endgroup$ – nasu May 28 at 19:45

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