# Speed as a Function of Time

Tags: function, speed, time
 P: 99 The first time derivative of velocity is acceleration. Can we then conclude that the first time derivative of speed is the magnitude of acceleration? In the following example I will consider a one dimensional case, for the sake of argument. Suppose the velocity v of a particle as a function of time t is given by v(t) = t^2. The acceleration, a, as a function of t is therefore given by a(t) = 2t. And so the magnitude of acceleration (the absolute value, since we are dealing with a one dimensional case as I have previously stated) is a "piece-wise defined function of t", namely 2|t|. That's observation A. Now, let's go back to velocity. Since |v(t)| = v(t), we can then conclude that the derivative of speed as a function of time is given by d/dt(|v(t)|) = 2t; which, technically speaking, is not the same as |a(t)|. So am I right in my conclusion that differentiating speed does not yield the magnitude of acceleration?
P: 3,917
 Quote by MohammedRady97 The first time derivative of velocity is acceleration. Can we then conclude that the first time derivative of speed is the magnitude of acceleration?
No. See uniform circular motion.

 Quote by MohammedRady97 In the following example I will consider a one dimensional case, for the sake of argument.
That's a special case then. Not true in general.
P: 99
 Quote by A.T. No. See uniform circular motion. That's a special case then. Not true in general.
But the example I gave helped make the point that differentiating speed w.r.t. time does not yield the magnitude of acceleration. Or do you mean that I am wrong in this conclusion?

 P: 99 Speed as a Function of Time To sum it all up: Mathematically, d|f(x)|/dx $\neq$ |df(x)/dx|. Ergo the first time derivative of speed is not the magnitude of acceleration, am I right?
P: 3,917
 Quote by MohammedRady97 Or do you mean that I am wrong in this conclusion?
No, you are right.
 P: 314 dv/dt in general does not equal the magnitude of acceleration. generally v is a vector with several components and the derivative is also a vector with several components. I don't readily know if d|v|/dt = |dv/dt| is true or false, but I would assume not. There might be some special cases where it is true.
 P: 19 Aren't we certain that d|v|/dt =/= |dv/dt|? |v(t)| = v(t) and the derivative of v(t) is a(t) Then the derivative of |v(t)| would also be a(t) a(t) is not |a(t)| Therefore, the derivative of |v(t)| is not |a(t)|. So we know they aren't the same. Correct me if I have any mistakes. That seems to be the problem, I think MohammadRady97 is spot on. The magnitude of the acceleration is |df(x)/dx| but you, OP, tried to do d|f(x)|/dx, which as MohammadRady said (and I've shown explicitly above) are not the same thing.
P: 99
 Quote by Wittyname6 Aren't we certain that d|v|/dt =/= |dv/dt|? |v(t)| = v(t) and the derivative of v(t) is a(t) Then the derivative of |v(t)| would also be a(t) a(t) is not |a(t)| Therefore, the derivative of |v(t)| is not |a(t)|. So we know they aren't the same. Correct me if I have any mistakes. That seems to be the problem, I think MohammadRady97 is spot on. The magnitude of the acceleration is |df(x)/dx| but you, OP, tried to do d|f(x)|/dx, which as MohammadRady said (and I've shown explicitly above) are not the same thing.
Exactly. The time derivative of speed is, in general, by no means related to the magnitude of acceleration. It just so happens that in some cases they're both equal. (A mathematical coincidence, if you will.)
P: 314
I got bored. I think this is overkill. You're all welcome.

 Quote by Wittyname6 |v(t)| = v(t) and the derivative of v(t) is a(t) Then the derivative of |v(t)| would also be a(t) a(t) is not |a(t)| Therefore, the derivative of |v(t)| is not |a(t)|.
This logic doesn't sit well with me. Let's just do the math

Say $\vec{v}(t)=f(t)\hat{\textbf{i}}+g(t)\hat{\textbf{j}}$

$|v(t)|=\sqrt{[f(t)]^2+[g(t)]^2}$

then

$\vec{a}(t) = \frac{dv}{dt}(t)=\frac{df}{dt}\hat{\textbf{i}}+\frac{dg}{dt}\hat{ \textbf{ j}}$

where

$|a(t)|=|\frac{dv}{dt}(t)|=\sqrt{\left(\frac{df}{dt}\right)^2+\left( \frac{dg}{dt} \right)^2}$

and

$\frac{d|v|}{dt}(t)=\frac{1}{2}\left( [f(t)]^2+[g(t)]^2\right)^{-1/2}\left[2f(t)\frac{df}{dt}+2g(t)\frac{dg}{dt}\right]$

If you try and solve $|a(t)| = \frac{d|v|}{dt}$

you find that this is only true if

$\left(f(t)\frac{dg}{dt}\right)^2 + \left(g(t)\frac{df}{dt}\right)^2 = f(t)g(t)(\frac{df}{dt})(\frac{dg}{dt})$

or

$\frac{f(t)}{g(t)}\frac{dg/dt}{df/dt}+\frac{g(t)}{f(t)}\frac{df/dt}{dg/dt} = 1$

Therefore, the time derivative of |v(t)| is not |a(t)| unless that condition is met. Through a little guessing, the only solutions I've found are when either f or g = 0, or f and g are constants.

I hope this clears everything up.
P: 99
 Quote by elegysix I got bored. I think this is overkill. You're all welcome. This logic doesn't sit well with me. Let's just do the math Say $\vec{v}(t)=f(t)\hat{\textbf{i}}+g(t)\hat{\textbf{j}}$ $|v(t)|=\sqrt{[f(t)]^2+[g(t)]^2}$ then $\vec{a}(t) = \frac{dv}{dt}(t)=\frac{df}{dt}\hat{\textbf{i}}+\frac{dg}{dt}\hat{ \textbf{ j}}$ where $|a(t)|=|\frac{dv}{dt}(t)|=\sqrt{\left(\frac{df}{dt}\right)^2+\left( \frac{dg}{dt} \right)^2}$ and $\frac{d|v|}{dt}(t)=\frac{1}{2}\left( [f(t)]^2+[g(t)]^2\right)^{-1/2}\left[2f(t)\frac{df}{dt}+2g(t)\frac{dg}{dt}\right]$ If you try and solve $|a(t)| = \frac{d|v|}{dt}$ you find that this is only true if $\left(f(t)\frac{dg}{dt}\right)^2 + \left(g(t)\frac{df}{dt}\right)^2 = f(t)g(t)(\frac{df}{dt})(\frac{dg}{dt})$ or $\frac{f(t)}{g(t)}\frac{dg/dt}{df/dt}+\frac{g(t)}{f(t)}\frac{df/dt}{dg/dt} = 1$ Therefore, the time derivative of |v(t)| is not |a(t)| unless that condition is met. Through a little guessing, the only solutions I've found are when either f or g = 0, or f and g are constants. I hope this clears everything up.
Yeah. But just to be clear, aren't f and g the respective x- and y-components of velocity?
 P: 314 yes, f(t) and g(t) are the x and y components of velocity. I was lazy some and used f and g, but they are the same thing. Recall that the derivative of speed you were asking about is $\frac{d}{dt}\left(|v(t)|\right)$, and the magnitude of acceleration is |dv/dt|, so that's why the answer is in components of the velocity.

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