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The instantaneous speed of a point along a circular path is given by $v=\omega r$, where $$\omega = \frac{\Delta \theta}{\Delta t},$$ $s=\Delta \theta r$, and $v=s/t$.

However, isn’t the displacement measured as in the following figure?

enter image description here

The relationship between normal speed $v$ and angular speed $\omega$ is related via arc length $s$. The displacement $d$ above is a straight line, not the arc length. So why is $v=s/t$ when it should be $v=d/t$? Would this not invalidate the equation $v=\omega r$?

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  • $\begingroup$ I would suggest you revisit the definitions of average and instantaneous velocity. $\endgroup$
    – Declan
    Commented Aug 30, 2015 at 3:42
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    $\begingroup$ the distance traveled is $s$ and not $d$. $\endgroup$
    – John Alexiou
    Commented Aug 30, 2015 at 5:06
  • $\begingroup$ instantaneous velocity is taken in a time interval so small that displacement and distance can approximately be taken equal to each other. Hope this clears the confusion. $\endgroup$
    – Saad
    Commented Aug 30, 2015 at 7:08

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The displacement d above is a straight line, not the arc length.

The point particle travels along the arc length, not along the straight line.

So why is v=s/t when it should be v=d/t?

If should not be d/t. It should be s/t. The value s is the distance travelled and t is the time.

Would this not invalidate the equation v=ωr?

It would if it were correct, but you are wrong to use d/t. So the equation is valid.

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