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For a particle in projectile motion with a constant upward acceleration of $-9.8\,\mathrm{m/s},$ the time derivative of its speed looks like this:

enter image description here

The time derivative of velocity is acceleration, but what is the time derivative of speed, specifically $\sqrt{(x'(t))^2+(y'(t))^2}$ ?

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  • $\begingroup$ Looks like a sigmoid function. en.wikipedia.org/wiki/Sigmoid_function $\endgroup$ Commented May 24 at 0:51
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    $\begingroup$ I am surprised that your range is almost $\pm20$ rather than $\pm 9.8$ $\endgroup$
    – Henry
    Commented May 24 at 0:53
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    $\begingroup$ @Henry thanks for pointing that out thats on me I didn't put in the 2 in the the denominator $\endgroup$
    – Jiamu Yue
    Commented May 24 at 1:11
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    $\begingroup$ How did you calculate the derivative without a function? $\endgroup$
    – Gwen
    Commented May 24 at 1:11
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    $\begingroup$ I think this can answer your question $\endgroup$
    – Marco
    Commented May 24 at 1:34

4 Answers 4

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Position is $(x(t), y(t))$ and velocity is $v(t) = (\dot{x}(t), \dot{y}(t))$, acceleration is $a=\dot{v} = (\ddot{x}, \ddot{y})$, and speed is the scalar function $s(t) = |v(t)| = \sqrt{\dot{x}^2 + \dot{y}^2}$.

I am using the dot notation, $\dot{u} = u' = \frac{du}{dt}$, because it is easier to read with exponents.

So let's use the chain rule: $$ \dot{s} = \frac{1}{2\sqrt{\dot{x}^2 +\dot{y}^2}} \bigg( 2\dot{x}\ddot{x} + 2\dot{y}\ddot{y} \bigg) = \frac{\dot{x}\ddot{x} + \dot{y}\ddot{y}}{\sqrt{\dot{x}^2 +\dot{y}^2}}$$

If you are familiar with dot products, this is: $$ \dot{s} = \frac{ v\cdot a }{s} = \left(\frac{v}{|v|}\right)\cdot a $$

Notice: $v/s = v/|v|$ is the unit vector in the direction of motion. The dot product of a vector with a unit vector gives the component of the vector in the direction of the unit vector. So this means:

The derivative of speed is the component of acceleration in the direction of motion.


So let's look at your graph. The vertical axis is indeed the derivative of speed.

In projectile motion, a ball is shot upward. Its speed is initially declining as gravity acts on it, so the derivative of speed is negative. As the ball reaches its apex, its motion becomes horizontal, perpendicular to gravity, so its speed doesn't change so much; at the apex, the ball's speed is constant, and that's the instant that $\dot{s}$ passes through $0$.

Once the ball is falling downward, its speed increases: positive $\dot{s}$, and as its motion becomes more and more downward, its speed derivative asymptotically tends toward the magnitude of gravity's acceleration.

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The derivative of speed is called the tangential acceleration. It shares the same unit with the acceleration. It is equal to $$\frac{\mathbf{a}\cdot \mathbf{v}}{\|\mathbf{v}\|}$$ where $\mathbf{v}$ is the velocity and $\mathbf{a} = \dot{\mathbf{v}}$ is the acceleration

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Rewritten:

$v:=||\overrightarrow{v} ||$;

$v^2=\overrightarrow{v} \cdot \overrightarrow{v}$;

Differentiting:

$2v\dfrac{dv}{dt}=2\dfrac{d\overrightarrow{v}}{dt} \cdot \overrightarrow{v}$;

$\dfrac{dv}{dt}=\overrightarrow{a} \cdot \dfrac{\overrightarrow{v}}{v}$;

Recall:

$\dfrac{\overrightarrow{v}}{v}$ is a unit tangential vector.

Now interpret this result.

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As pointed out above, the time derivative of speed (i.e., the rate of change of speed) is the tangential component of acceleration (i.e., the component of acceleration in the direction of velocity). It can also be characterised as the additive inverse of retardation, since retardation is defined as the rate of decrease of speed (note that retardation is not the additive inverse of acceleration: the trajectory $e^{-t}$ has the exact same retardation and acceleration).

Incidentally, the circular trajectory $\cos(t)\mathbf{i}+\sin(t)\mathbf{j}$ shows that the derivative of speed does not equal the magnitude of acceleration: the derivative of speed is $0,$ whereas the acceleration has magnitude $1.$

In this case, retardation's sign is in the direction of decreasing speed while acceleration's sign refers to the assigned positive direction in (one-dimensional) space.

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  • $\begingroup$ Among the people who know what they're talking about and still use the term retardation, do they actually use the term retardation in a context like the context of this question -- that is, when describing motion where the position of the object has more than one degree of freedom? $\endgroup$
    – David K
    Commented May 26 at 18:05
  • $\begingroup$ Yes, among the usual kinematics scalars (distance from the origin, distance travelled, instantaneous speed, average speed, rate of change of speed, retardation), retardation is the most niche; it doesn't help that "deceleration" and "retardation" are frequently shoddily defined (in terms of velocity) and confusingly considered synonymous with "negative acceleration" (whatever this phrase means). $\endgroup$
    – ryang
    Commented May 27 at 4:04
  • $\begingroup$ I ask because I went looking for it and only found the shoddy definitions. Do you have good examples that aren't self-citations? $\endgroup$
    – David K
    Commented May 27 at 4:09
  • $\begingroup$ Your search results are as good as mine! I do think that unlike "deceleration" and "negative acceleration", "retardation" in itself (in the absence of shoddy definitions) is neither potentially ambiguous or confusing. Regarding the OP's query, I'd just stick to "the rate of change of speed". $\endgroup$
    – ryang
    Commented May 27 at 4:22
  • $\begingroup$ We have a word for reduction in speed but not increase - that's weird. $\endgroup$
    – badjohn
    Commented May 31 at 16:52

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