Let $s=$distance (a variable) we define instantaneous speed = magnitude $\left[\frac{ds}{dt}\right]$. However instantaneous speed is also defined as magnitude of instantaneous velocity i.e. instantaneous speed=mod $v$ [$v$=velocity]. How can we prove mathematically that magnitude $\left[\frac{ds}{dt}\right]$= magnitude $v$.
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$\begingroup$ I have rewritten the terms as used in standards/Please help $\endgroup$– pik selvanCommented Jun 11, 2018 at 17:51
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1$\begingroup$ "instantaneous speed=mod(dv/dt)[v=velocity] ": that's wrong - where did you see that? And what is "mod"? $\endgroup$– NickDCommented Jun 11, 2018 at 17:55
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$\begingroup$ @Nick It looks like "mod" is supposed to mean magnitude perhaps? $\endgroup$– BioPhysicistCommented Jun 11, 2018 at 17:57
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$\begingroup$ mod= modulus=magnitude of velocity $\endgroup$– pik selvanCommented Jun 11, 2018 at 17:58
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2$\begingroup$ Instantaneous speed is not ds/dt is your issue. Speed is always positive. ds/dt can be negative. Therefore, the correct relation for instantaneous speed is the magnitude of ds/dt. $\endgroup$– BioPhysicistCommented Jun 11, 2018 at 17:59
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2 Answers
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What you are trying to prove is just a definition.
$\frac{ds}{dt}$ is the velocity $v$.
$\frac{ds}{dt} = v$.
Therefore the magnitude of $\frac{ds}{dt}$ is the same thing as the magnitude of the velocity $v$
answered Jun 11, 2018 at 18:07
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I tried and worked as under for small displacement dr ( dr is a vector,where ever written in this answer) |dr|=|ds| , where ds is small distance so v=dr/dt, v is velocity vector and |v|=|dr|/dt =|ds|/dt = |ds/dt| =instantaneous speed so instantaneous speed is defined as |ds|/dt as well as |v| i.e. the magnitude of instantaneous velocity