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Here $\vec{v}$ represents velocity and $dv / dt$ gives the instantaneous acceleration. $|\vec v|$ gives the magnitude of velocity. When acceleration is zero, then change in magnitude of velocity is zero. In case of circular motion, the magnitude of velocity does not change (if tangential acceleration is zero) while the direction changes. Can circular motion be an example?

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    $\begingroup$ Answer: Yes. You're welcome. $\endgroup$ – Robert Filter Jul 29 '15 at 18:52
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    $\begingroup$ You answered your own question. Note that "When acceleration is perpendicular to the instantaneous velocity, the magnitude of the velocity does not change" is a more accurate way of stating what you said. $\endgroup$ – Floris Jul 29 '15 at 18:55
  • $\begingroup$ As stated by @Floris , if the magnitude of the velocity remains unchanged (i.e ciruclar motion) then $\frac{d |\vec{v}|}{dt} = 0$ , but since the vector changes direction $|\frac{d \vec{v}}{dt}| = mv^2/r > 0$ $\endgroup$ – Andre Maizel Jul 29 '15 at 19:01
  • $\begingroup$ Since $t$ is a scalar, $$\| \frac{{\rm d}\vec{v}}{{\rm d}t} \| = \frac{\| {\rm d}\vec{v}\|}{{\rm d}t} = \frac{{\rm d}\| \vec{v}\|}{{\rm d}t} $$ $\endgroup$ – John Alexiou Jul 29 '15 at 19:08
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    $\begingroup$ What you said is wrong @ja72 , how can the magnitude of the derivative be the same as the derivative of the magnitude if clearly this does not occur in the circular motion? This would only apply if $\vec{v}$ was a scalar. $|d\vec{v}|$ indicates the magnitude of the variation, which can be a simple change in direction. $\endgroup$ – Andre Maizel Jul 29 '15 at 19:19
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$|\frac{d\vec{v}}{dt}|$ = magnitude of rate of change of $\vec{v}$ = |$\vec{a}$|

while

$\frac{d|\vec{v}|}{dt}$= rate of change of |$\vec{v}$| = rate of change of speed.

If you consider any situation where speed of object is not changing and object is not moving in straight line , i.e. , moving in any curved path , both these conditions will be true. This is because in such a case rate of change of speed is 0 (obvious as we are not moving it causing any change in speed) and net acceleration is non-zero as velocity (as a vector) changes.

And hence it also includes uniform circular motion.

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  • $\begingroup$ I attempted to give an easier and concise answer. $\endgroup$ – BEWARB Jul 30 '15 at 16:31
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A particle moving along a curved path has its motion decomposed as

$$ \vec{v} = v \hat{e} \\ \vec{a} = \dot{v} \hat{e} + \frac{v^2}{\rho} \hat{n} $$

where $\hat{e}$ is the unit tangent vector, and $\hat{n}$ the unit normal vector pointing towards the inside of the curve. $v$ is the scalar speed and $\rho$ is the radius of curvature at that moment.

1) You have $$ \frac{{\rm d}}{{\rm d} t} | \vec{v} | = 0 $$

which means that tangential acceleration must be zero

$$ \frac{{\rm d}v}{{\rm d} t} | \vec{e} | = \dot{v} = 0 $$

2) You also have $$ \left| \frac{{\rm d} \vec{v}}{{\rm d}t} \right| = | \vec{a} | \neq 0$$

which means that

$$ |\vec{a}|= \sqrt{ \left( \dot{v} \hat{e} + \frac{v^2}{\rho} \hat{n} \right) \cdot \left( \dot{v} \hat{e} + \frac{v^2}{\rho} \hat{n} \right) } = \sqrt{\dot{v}^2 + \left( \frac{v^2}{\rho} \right)^2} \neq 0$$

For both to be true it means that $v \neq 0$ and $\dot{v}=0$ making $| \vec{a} | = \frac{v^2}{\rho}$

So a path that provides constant (non-zero) speed along any curved path with no tangential acceleration.

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  • $\begingroup$ Point taken. I will correct. $\endgroup$ – John Alexiou Jul 29 '15 at 21:00

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