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Magnitude of instantaneous velocity = $|dr/dt|$ but this is not always equal to $ds/dt$ where ds is the infinitesimally small change in distance in the interval $dt$

Which one is speed?

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  • $\begingroup$ Why would they not be equal? For both you are only travelling an infinitesimal amount. The difference between $\Delta r$ and $\Delta s$ goes away as $\Delta$ becomes smaller. In this case it's $ds$ and $dr$ which is the limit where they become equal. $\endgroup$ – JMac Feb 9 '17 at 17:09
  • $\begingroup$ Can you post that as an answer so I can accept it? Also, is this always the case? (Instantaneous speed being equal to the modulus of dr/dt and also equal to ds/dt?) $\endgroup$ – xasthor Feb 9 '17 at 17:14
  • $\begingroup$ Sure, I just thought there might have been something from the question that I misinterpreted. It's also good practice to wait before accepting answers in case a better one comes along; but I'll leave that to your judgement. $\endgroup$ – JMac Feb 9 '17 at 17:15
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For both you are only travelling an infinitesimal amount.

The difference between $\Delta r$ and $\Delta s$ goes away as $\Delta$ becomes smaller. In this case it's $ds$ and $dr$ which is the limit where they become equal.

You can look at your drawing and imagine as you decrease r and s, the space between the line $dr$ and the arc of the circle will keep getting smaller until at an infinitesimally small value they are the same.

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More precisely,

\begin{align*} s &=\int |\mathbf{v}| \, dt \\ \frac{ds}{dt} &= |\mathbf{v}| \\ &= \left| \frac{d\mathbf{x}}{dt} \right| \\ ds &= |d\mathbf{x}| \\ &= |\mathbf{v}| \, dt \end{align*}

For linear motion without "U-turn", we can mix the usage of $s$ and $x$.

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