# Minimum speed (~V3N0M~'s question at Yahoo! Answers)

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#### Fernando Revilla

##### Well-known member
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Hello ~V3N0M~,

We have $r'(t)=(-6t,-2,2t+4)$ so, $$|r'(t)|=\sqrt{(-6t)^2+(-2)^2+(2t+4)^2}=\sqrt{40t^2+16t+20}$$ Then, $$\frac{d}{dt}\left(|r'(t)|\right)=\frac{80t+16}{2\sqrt{40t^2+16t+20}}=\frac{8(5t+1)}{\sqrt{40t^2+16t+20}}=0\Leftrightarrow t=-\frac{1}{5}$$ If $t<-1/5$ then $\frac{d}{dt}\left(|r'(t)|\right)<0$ and if $t>-1/5$ then $\frac{d}{dt}\left(|r'(t)|\right)>0$. This implies that the speed has an strict absolute minimum at $t=-1/5$.

#### topsquark

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Just my two cents.

Fernando's solution is, of course, correct. But I've never come across a Mechanics problem where the time was negative. This has me wondering whether the correct answer to this problem is when t = 0 s.

-Dan

#### Fernando Revilla

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It is natural consider $t\in (-\infty,+\infty)$, think (for example) about some interpretations of the orbits in the phase space as $t\to +\infty$ and as $t\to -\infty$.

#### Prove It

##### Well-known member
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It is natural consider $t\in (-\infty,+\infty)$, think (for example) about some interpretations of the orbits in the phase space as $t\to +\infty$ and as $t\to -\infty$.
You seem to have missed topsquark's point entirely, which is that the question asks "At what TIME is the speed at its minimum?" which implies that we are using t to represent time. Topsquark has said it does not make sense to include negative values for t in this case as there is no such thing as negative time.

#### Fernando Revilla

##### Well-known member
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You seem to have missed topsquark's point entirely, which is that the question asks "At what TIME is the speed at its minimum?" which implies that we are using t to represent time. Topsquark has said it does not make sense to include negative values for t in this case as there is no such thing as negative time.
I am astonished. I can't explain you such a trivial thing.

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#### Prove It

##### Well-known member
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I am astonished.
In a good way I hope #### Fernando Revilla

##### Well-known member
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In a good way I hope Sorry, you have no idea of this subjet. Browse on the net and you'll find a non denumerable cardinal of references. A good begining: Dynamic Systems.

#### topsquark

##### Well-known member
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Both Fernando and I have good points. I had presumed that the OP's question was of a lower level of Physics and thus my response is probably correct. However if the question is of a higher level then Fernando's response is more likely to be correct. And upon reflection, this question probably came from a Calculus class as opposed to Physics. In this case I'd lean toward Fernando's answer.

-Dan

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#### Ackbach

##### Indicium Physicus
Staff member
Just my two cents.

Fernando's solution is, of course, correct. But I've never come across a Mechanics problem where the time was negative. This has me wondering whether the correct answer to this problem is when t = 0 s.

-Dan
You seem to have missed topsquark's point entirely, which is that the question asks "At what TIME is the speed at its minimum?" which implies that we are using t to represent time. Topsquark has said it does not make sense to include negative values for t in this case as there is no such thing as negative time.
I have seen negative time in mechanics problems. In fact, I'll wager you have as well, although you might not have recognized it.

Suppose a boy throws a ball in the air at an initial $10$ m/s from a height of $1$ m. When will the ball hit the ground, if you ignore air resistance?

Ok, standard problem in kinematics. Set up
$$y=y_{0}+v_{0y}t+\frac{a}{2}t^{2}=1+10t-\frac{9.8}{2}t^{2},$$
and set this equal to zero. Mathematically, you get two solutions: $t=2.14, -0.096$, both in seconds. So, we throw out the negative solution and say the answer is $t=2.14$ seconds. Done.

Or are we? Is there any physical significance to the negative time solution? I claim there is. It's the time at which you would have had to throw the ball from the ground height in order to reach the height of $1$ m at $t=0$ with a velocity of $+10$ m/s.

What's happening here? What's happening is that the time at which you "start the clock" is completely arbitrary. Now, once you've started it, you shouldn't generally restart it, unless you keep careful track of how you're doing so.

Time can be negative in exactly the same way that displacement can be negative. For displacement, it depends on where you put the origin of your meter stick. For time, it depends on when you start the clock.

Here's an alternative problem with the same physics: at $t=-2.136$ seconds, a boy throws a ball in the air at $10$ m/s. Ignoring air resistance, when will the ball hit the ground? The usual kinematic equation won't work here, because it will assume we know the velocity at time $t=0$. So, start from the acceleration and integrate, using initial conditions.
\begin{align*}
a&=-g \\
v&= -gt+C \\
10 &=-9.8(-2.136)+C \\
C&=-10.933 \\
v&= -gt-10.933 \\
y&= -\frac{g}{2} t^{2}-10.933t+C \\
1&= -4.9 (-2.136)^{2}-10.933(-2.136)+C \\
C&=0.0033 \quad \text{This should be zero, but isn't because of rounding.} \\
0&=-\frac{g}{2} t^{2}-10.933t+0.0033 \\
t&=3.018 \times 10^{-4}, -2.23 \; \text{seconds}.
\end{align*}
The first solution is essentially zero, as you'd expect.

In fact, time reversal is a standard topic in high-energy particle physics. It's the T in CPT symmetry.

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#### topsquark

##### Well-known member
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...
Or are we? Is there any physical significance to the negative time solution? I claim there is.
...
True, and I've thought about that when I came up with my class notes. On the one hand when teaching Intro Physics (or AP Physics) I find that talking about the negative time aspect usually just confuses them. On the other hand it's fun to bring it up in Advanced Mechanics and let them try to sort it out on their own.

What can I say? I'm evil. And yes, the use of an anti-Hermitian operator (time reversal) is ubiquitous in QM. But I don't think our OP is quite at that level. -Dan