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I am working on a problem that states:

"A swimmer capable of swimming at a speed $c$ in still water is swimming in a stream in which the current is $u$ (which we assume to be less than $c$). Suppose the swimmer swims upstream a distance $L$ and then returns downstream to the starting point. Find the time necessary to make the round trip, and compare it with the time to swim across the stream a distance $L$ and return "

I was able to get to the point where I got the value of the velocity of the swimmer relative to the stream to be $v = -c$ Hence relative to the observer, the velocity of the swimmer would be $v = -c + u $

In the solutions manual they state

"As expected, the velocity relative to the ground has magnitude smaller than $c$ ; it is also negative, since the swimmer is swimming in the negative $x$ direction, so $|v| = c − u$.) "

I understand why the magnitude of $v$ would be $c - u$, but does this mean everytime I see $u - c = v$, I should change it to $c - u$ because mathematically, $|v| = \sqrt{(u-c)^2} = c - u$?

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  • $\begingroup$ What do you mean by classical relativity? And does c have anything to do with the speed of light? If not, and not relativistic speeds just add and subtract velocities. If yes, the whole thing is nonsense because for a speed c of light, it would not interact with the water at all, and it would always be c. Which it is not in water. $\endgroup$
    – Bob Bee
    Jan 30, 2017 at 6:25
  • $\begingroup$ Just with relation to your last line. $\sqrt{(u-c)^2}=u-c$ $\endgroup$
    – Rumplestillskin
    Jan 30, 2017 at 7:59
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    $\begingroup$ @BobBee I also wondered that for a second. We tend to think of relativity as only post-Einstein, but we usually forget there was (and is) a Galileean relativity for classical mechanics. $\endgroup$
    – FGSUZ
    Nov 11, 2017 at 22:58
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    $\begingroup$ @BobBee Classical relativity is Galilean relativity. It's clear from the context of the title and the idea of problem that c is not the speed of light. $\endgroup$
    – Bill N
    Nov 12, 2017 at 0:48
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    $\begingroup$ @Continuum Always establish a coordinate system with positive and negative directions when you're working kinematics and dynamics problems. Use vectors with directions described consistent with the coordinates. That saves a LOT of headaches. $\endgroup$
    – Bill N
    Nov 12, 2017 at 0:50

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Depends on what's being asked of you. Whenever you are asked for the speed, you should provide $|v|$. If asked for the velocity, you should provide $v$.

When solving the problem, you should always use $ v$ and not $|v|$. Velocity is a vector quantity and sign MATTERS. The excerpt is saying that when the velocity is negative the object is moving with a speed $|v|$ in the negative $x$ direction.

Whether or not $|v| = c-u$ or $|v| = u-c$ depends on whether $u < c\space $ or $\space c < u$.

In this instance $u< c$ but that will not always be the case.

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