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According to me the modulus of $\frac {\text{d}v}{\text{d}t}$ gives the magnitude of acceleration without accounting for direction but I am stumbled on how to interpret $\left(\frac{\text{d}}{\text{d}t}\right)|velocity|$

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As you said, $|d{\bf v}/dt|$ is the magnitude of acceleration.

$d|{\bf v}|/dt$ is the rate of change of speed.

In general they are not the same.

For example, in uniform circular motion

$$\bigg|\frac{d{\bf v}}{dt}\bigg|=\frac{|{\bf v}|^2}{R}$$ while $$\frac{d|{\bf v}|}{dt}=0$$

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If we are dealing with motion in one dimension, then the velocity function is a scalar (i.e. $v: \mathbb{R} \to \mathbb{R}$), and it makes sense to talk about "positive velocity" and "negative velocity" (and likewise for acceleration). I'm going to assume that in what follows.

First, it may be helpful to know that $\frac{d}{dt} |t| = sgn(t)$, where $sgn(t)$ is the signum function: $sgn(t) = 1$ for $t>0$, $sgn(t) = -1$ for $t<0$, and $sgn(0)$ is undefined. You can also write this as $\frac{d}{dt} |t| = \frac{|t|}{t}$ for $ t\ne 0$.

By the chain rule, then, $$\frac{d}{dt} |v(t)| = sgn(v(t)) \space a(t)$$

That is, $\frac{d}{dt} |v(t)| $ tells you the acceleration times the sign of the velocity.

There are four cases to consider:

  • If an object has positive velocity and positive acceleration, $\frac{d}{dt} |v(t)| $ is positive (and is equal to the acceleration).

  • If an object has positive velocity but negative acceleration, $\frac{d}{dt} |v(t)| $ is negative (and is equal to the acceleration, but not equal to the magnitude of the acceleration, which is of course positive).

  • If an object has negative velocity but positive acceleration, $\frac{d}{dt} |v(t)| $ is negative (and opposite in sign to both the acceleration and the magnitude of the acceleration, which would be positive and equal to each other).

  • If an object has negative velocity and negative acceleration, $\frac{d}{dt} |v(t)| $ is positive (and is equal to the magnitude of the acceleration).

Of the four cases above, in only the first and last is $\frac{d}{dt} |v(t)| $ equal to $|\frac{d}{dt} v(t) | $.

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EDIT: This answer is too wrong to correct. Striking it out.

If $v:\mathbb R\to \mathbb R$, then numerically, they are the same where they both exist. That's a consequence of the fact that $(-v)' = -(v')$.

But where $v$ changes sign, $|v|$ may not be differentiable. This is because the absolute value function $x\mapsto |x|$ isn't differentiable at $x=0$.

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    $\begingroup$ They are not the same. For example, in uniform circular motion, $|d{\bf v}/dt| = v^2/R$, while $dv/dt=0$. $\endgroup$ – velut luna Apr 21 '16 at 15:52
  • $\begingroup$ @Mathaholic: I think you intended to say $d|v|/dt = 0$ $\endgroup$ – Henry Apr 21 '16 at 15:56
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    $\begingroup$ My $v=|{\bf v}|$ is the speed. $\endgroup$ – velut luna Apr 21 '16 at 15:57
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    $\begingroup$ They are not the same numerically, even where they are defined. If $v$ is positive and $a$ is negative, then the derivative of $|v|$ is negative, whereas the magnitude of $a$ is positive. $\endgroup$ – mweiss Apr 21 '16 at 16:43
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    $\begingroup$ @mweiss: Yeah, okay, you're right. I think my answer is too off the mark to fix now. Since I missed the original gist of the question anyway, I'm striking this out. But I'm leaving it here, which is always my philosophy for a wrong answer. Otherwise, nobody can learn from it. +1 for you. $\endgroup$ – MPW Apr 21 '16 at 20:31

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