# A Note on Speed

A quick note on speed.

The idea of differentiating speed to determine where it is increasing or decreasing is perfectly reasonable.

$\displaystyle \text{Speed}=s(t)=\left| {v\left( t \right)} \right|=\sqrt{{{{{\left( {v(t)} \right)}}^{2}}}}$. Then,

$\displaystyle {s}'\left( t \right)=\frac{{2v\left( t \right){v}'\left( t \right)}}{{2\sqrt{{{{{\left( {v(t)} \right)}}^{2}}}}}}=\frac{{v\left( t \right)a\left( t \right)}}{{\sqrt{{{{{\left( {v(t)} \right)}}^{2}}}}}}$

Since the denominator is positive, $\displaystyle {s}'\left( t \right)>0$ and speed is increasing when $\displaystyle v\left( t \right)$ and $\displaystyle a\left( t \right)$ have the same sign, and $\displaystyle {s}'\left( t \right)<0$ and speed is decreasing when they have different signs.

As a practical matter, this is the “long way.” It requires you to calculate the sign of the velocity and acceleration and some other stuff. So, the traditional way, without the other stuff, is faster. On the other hand, it carries over nicely to higher dimensions where the velocity and acceleration vectors do not have signs, per se.

(This occurred to me in the shower this morning; I don’t think I ever realized it before – TMI.)

## 2 thoughts on “A Note on Speed”

1. A minor typo: you have “s(t)<0" but meant to say s-prime: "s'(t) < 0" ?
I do like the way that sqrt(v^2) extends nicely to higher dimensions!

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• Thanks for catching that. Fixed it!

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