Is the Derivative of Speed the Magnitude of Acceleration?

In summary, the first time derivative of velocity is acceleration. However, the first time derivative of speed is not necessarily the magnitude of acceleration, as shown by the example of a one-dimensional case. The derivative of speed is a piece-wise defined function of time, while the magnitude of acceleration is simply a function of time. Therefore, in general, differentiating speed does not yield the magnitude of acceleration.
  • #1
PFuser1232
479
20
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?
 
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  • #2
MohammedRady97 said:
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.

MohammedRady97 said:
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.
 
  • #3
A.T. said:
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?
 
  • #4
To sum it all up:
Mathematically, d|f(x)|/dx [itex]\neq[/itex] |df(x)/dx|. Ergo the first time derivative of speed is not the magnitude of acceleration, am I right?
 
  • #5
MohammedRady97 said:
Or do you mean that I am wrong in this conclusion?
No, you are right.
 
  • #6
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.
 
  • #7
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.
 
  • #8
Wittyname6 said:
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.)
 
  • #9
I got bored. I think this is overkill. You're all welcome.

Wittyname6 said:
|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 [itex]\vec{v}(t)=f(t)\hat{\textbf{i}}+g(t)\hat{\textbf{j}}[/itex]

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

then

[itex]\vec{a}(t) = \frac{dv}{dt}(t)=\frac{df}{dt}\hat{\textbf{i}}+\frac{dg}{dt}\hat{ \textbf{ j}}[/itex]

where

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

and

[itex]\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][/itex]


If you try and solve [itex]|a(t)| = \frac{d|v|}{dt}[/itex]

you find that this is only true if

[itex]\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})[/itex]

or

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


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.
 
  • #10
elegysix said:
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 [itex]\vec{v}(t)=f(t)\hat{\textbf{i}}+g(t)\hat{\textbf{j}}[/itex]

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

then

[itex]\vec{a}(t) = \frac{dv}{dt}(t)=\frac{df}{dt}\hat{\textbf{i}}+\frac{dg}{dt}\hat{ \textbf{ j}}[/itex]

where

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

and

[itex]\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][/itex]


If you try and solve [itex]|a(t)| = \frac{d|v|}{dt}[/itex]

you find that this is only true if

[itex]\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})[/itex]

or

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


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?
 
  • #11
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 [itex]\frac{d}{dt}\left(|v(t)|\right)[/itex], and the magnitude of acceleration is |dv/dt|, so that's why the answer is in components of the velocity.
 
Last edited:

1. What is speed as a function of time?

Speed as a function of time is a mathematical representation of the relationship between an object's speed and the amount of time it takes to travel a certain distance. It shows how the speed of an object changes over time.

2. How is speed as a function of time calculated?

Speed as a function of time is calculated by dividing the distance traveled by the time it took to travel that distance. The formula is: speed = distance / time. This can be represented graphically as a slope on a distance-time graph.

3. What does a straight line on a speed vs. time graph indicate?

A straight line on a speed vs. time graph indicates that the speed of the object is constant. This means that the object is traveling at the same speed over a period of time, without any change in its velocity.

4. How does acceleration affect speed as a function of time?

Acceleration affects speed as a function of time by causing a change in the object's velocity. If the acceleration is positive, the speed of the object will increase over time. If the acceleration is negative, the speed of the object will decrease over time.

5. How can speed as a function of time be used in real life?

Speed as a function of time can be used in real life to calculate the average speed of a moving object, such as a car or a runner. It can also be used to determine the acceleration or deceleration of an object, which is important in fields such as physics and engineering. Additionally, it can help predict the time it will take an object to reach a certain distance or destination.

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