Zero velocity, Zero acceleration

[PFuser1232]

In one dimension, the acceleration of a particle can be expressed as:
$$a = \frac{dv}{dt} = \frac{dx}{dt} \frac{dv}{dx} = v \frac{dv}{dx}$$
Does this equation imply that if $v = 0$ then $a = 0$?

EDIT: I think I figured out the source of my confusion. This equation is a differential equation and I cannot just plug in values for $v$, am I right?

 

[Shinaolord]

[strike]I don't think your 3rd step is correct.
$a(t)=\frac{d}{dt}v(t)=\frac{d^2}{dt^2}x(t)=\frac{d}{dt}[\frac{d}{dt} x(t)]$.[strike] disregard this comment. I made a mistake originally.

 

[PFuser1232]

I don't think your 3rd step is correct.
$a(t)=\frac{d}{dt}v(t)=\frac{d^2}{dt^2}x(t)$.

It is correct. I applied the chain rule.

 

[PWiz]

$a=\int v$ $dt$, so there will always be the constant of integration. V may or may not be 0 when a=0.(This is entirely determined by the value of v at t=0)
EDIT: Its $v=\int a$ $dt$, major mess up on my end, sorry.

 

[Shinaolord]

$a=\int v$ $dt$, so there will always be the constant of integration. V may or may not be 0 when a=0.(This is entirely determined by the value of v at t=0)

Ah, that's what I must've been seeing wrong.
I knew it was something. But shouldn't your a be x, as your integrating v(t)?

 

[PWiz]

I don't think your 3rd step is correct.
$a(t)=\frac{d}{dt}v(t)=\frac{d^2}{dt^2}x(t)=\frac{d}{dt}[\frac{d}{dt} x(t)]$.

Please re-check your calculations.

 

[Shinaolord]

I'm in the process of doing just that.You are correct. I realized what my mistake was. My apologies.

 

[nasu]

$a=\int v$ $dt$, so there will always be the constant of integration. V may or may not be 0 when a=0.(This is entirely determined by the value of v at t=0)

This is wrong if by "a" you mean acceleration. The integral of v is the displacement and not the acceleration.

 

[PWiz]

Crap, sorry I interchanged the terms.
It should be $v=\int a dt$. (How very embarrassing for me. Please excuse it as a major slip of the finger :P)

 

[nasu]

In one dimension, the acceleration of a particle can be expressed as:
$$a = \frac{dv}{dt} = \frac{dx}{dt} \frac{dv}{dx} = v \frac{dv}{dx}$$
Does this equation imply that if $v = 0$ then $a = 0$?

EDIT: I think I figured out the source of my confusion. This equation is a differential equation and I cannot just plug in values for $v$, am I right?

It implies that only if the derivative is finite at v=0. I mean the dv/dx.
For a simple case, like a body thrown straight up for example, the velocity is zero at the top of the trajectory but the derivative dv/dx is proportional to 1/v so it will blow up at v=0. But the product vdv/dx will have a finite, non-zero value.

So you may have v=0 and a=0 at the same time, as discussed in the other thread. But is not a necessary condition. v=0 does not imply a=0.

And you did not make any mistake in your derivation of the formula.

 

[Shinaolord]

It implies that only if the derivative is finite at v=0. I mean the dv/dx.
For a simple case, like a body thrown straight up for example, the velocity is zero at the top of the trajectory but the derivative dv/dx is proportional to 1/v so it will blow up at v=0. But the product vdv/dx will have a finite, non-zero value.

So you may have v=0 and a=0 at the same time, as discussed in the other thread. But is not a necessary condition. v=0 does not imply a=0.

And you did not make any mistake in your derivation of the formula.

I would like to take this time to point out to The OP, in case he missed my edit, that I made a mistake in my calculations the first time about, and your derivation was correct. I apologize for possibly misleading you.

 

[PeroK]

In one dimension, the acceleration of a particle can be expressed as:
$$a = \frac{dv}{dt} = \frac{dx}{dt} \frac{dv}{dx} = v \frac{dv}{dx}$$
Does this equation imply that if $v = 0$ then $a = 0$?

EDIT: I think I figured out the source of my confusion. This equation is a differential equation and I cannot just plug in values for $v$, am I right?

You had them all fooled for a bit with this one. To be a little more precise: your equation holds when all the functions involved are differentiable. The problem is that dv/dx may not be differentiable at certain points. That's when the use of the chain rule breaks down. The statement of the chain rule would say "... and if v(x) is differentiable at x ..".

 

[PFuser1232]

Here's the problem:

Suppose a particle $P$ has an acceleration $a$. As shown above:

$$a = v \frac{dv}{dx}$$

Now, both $v$ and $\frac{dv}{dx}$ are functions of $x$. I will call them $f(x)$ and $g(x)$ respectively.

$$a = f(x) g(x)$$

Where $g = f'$.

Suppose one solution to the equation $f(x) = 0$ is $x = b$, and I wish to find the acceleration of $P$ at this particular value of $x$.

$$a = f(b) g(b) = 0 ⋅ g(b) = 0$$

The reason I find this result bizarre is that I can think of many situations where a body is at instantaneous rest, but still possesses a nonzero acceleration component due to a nonzero net force at that point, so I know I must have gone wrong somewhere. Where did I go wrong?

 

[PeroK]

Where did I go wrong?

I already explained that above. You assumed that v is a differentiable function of x.

"The problem is that dv/dx may not be differentiable at certain points. That's when your use of the chain rule breaks down."

 

[PFuser1232]

I already explained that above. You assumed that v is a differentiable function of x.

"The problem is that dv/dx may not be differentiable at certain points. That's when your use of the chain rule breaks down."

So if $v$ is a differentiable function of $x$, then $v = 0$ always implies $a = 0$?

 

[PeroK]

So if $v$ is a differentiable function of $x$, then $v = 0$ always implies $a = 0$?

Yes, but:

$\frac{dv}{dx} = \frac{\frac{dv}{dt}}{\frac{dx}{dt}} = \frac{a(t)}{v(t)}$

So, if v = 0, then dv/dx has a tendency to be undefined at that point.

 

[PFuser1232]

Yes, but:

$\frac{dv}{dx} = \frac{\frac{dv}{dt}}{\frac{dx}{dt}} = \frac{a(t)}{v(t)}$

So, if v = 0, then dv/dx has a tendency to be undefined at that point.

This is a $\frac{0}{0}$ situation, which means we cannot compute the derivative using this equation alone; we need to be given $v$ as a function of $x$, right?

 

[PeroK]

This is a $\frac{0}{0}$ situation, which means we cannot compute the derivative using this equation alone; we need to be given $v$ as a function of $x$, right?

You're missing the point that if v = 0, then dv/dx is undefined.

Try to find an example of a function v(t) where v(t) = 0 at some point and dv/dx exists (finite) at that point.

 

[Amin2014]

In one dimension, the acceleration of a particle can be expressed as:
$$a = \frac{dv}{dt} = \frac{dx}{dt} \frac{dv}{dx} = v \frac{dv}{dx}$$
Does this equation imply that if $v = 0$ then $a = 0$?

No it does not. If v = 0, and v STAYS equal to zero, then yes, a=0. Said in more detail, if v=0 at time t, and v is still zero a moment later, then a=0. As you've said yourself, this is a differential equation, and v, a and t all represent variables. You can't just plug in numbers like that.

To give you another example, consider two objects that have equal velocities for some instant in time. Are they accelerating in the same manner? In other words, if v₁=v₂, is a₁ = a₂?

Answer: not unless v₁ is always equal to v₂. In other words, if the velocities are variables over some time, then yes the accelerations must be equal, but if they only represent numbers (or points in time), then then the accelerations are not necessarily equal.

 

[Amin2014]

If x₁=x₂, does v₁=v₂? In other words, if two objects are next to each other at the same position next to each other, does this mean they have the same velocities?
What if they are ALWAYS next to each other?