# Acceleration From Rest Under Constant Power

• I
Gold Member
Impossible?!?

By my estimation the equation that describes this motion is given by:

$$Pt = \frac{1}{2}m{ \dot x}^2$$

or

$$\dot{x} = \sqrt{\frac{2P}{m}} \sqrt{t}$$

but this implies:

$$\ddot{x} = \sqrt{\frac{2P}{m}} \frac{1}{\sqrt{t}}$$

So, no matter how small we make the power, we appear to get an infinite acceleration at ##t=0##( or in the case of ##P=0, t= 0##) an indeterminate form ##\frac{0}{\infty}##.

Conclusion: Acceleration from rest under constant power is non-physical. However, with any non-zero initial velocity things seem to be ok with the following:

$$\ddot{x} = \frac{P}{m} \frac{1}{\sqrt{ \frac{2P}{m}t + v_o^2}}$$

Anything interesting here, something I'm missing?

Staff Emeritus
2021 Award
Acceleration from rest under constant power is non-physical.
Correct. However, this situation only lasts for an instant,

Gold Member
Correct. However, this situation only lasts for an instant.
Well, how far does it travel in that instant?

Homework Helper
Look at your own equations. Simply because things go to zero and we can can choose to devide by them does not make the world unphysical.

bob012345, Lord Jestocost and russ_watters
Gold Member
Look at your own equations. Simply because things go to zero and we can can choose to devide by them does not make the world unphysical.
I didn't say the world was unphysical. I'm saying the model I'm choosing for the "world" here yields an unphysical result at the initial conditions. I don't think those two things are equivalent.

Homework Helper
Your model is an approximation. In the jargon it is integrable. To be blunt your point is true but not interesting

Gold Member
Well, how far does it travel in that instant?
I see that apparently the displacement is indifferent to this:

$$x = \frac{2}{3} \sqrt{ \frac{2P}{m}} \sqrt[3]{t}$$

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Gold Member
Your model is an approximation. In the jargon it is integrable. To be blunt your point is true but not interesting
What about it is an approximation though? Its CoE.

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Mentor
What about it is an approximation though?
The premise that you can apply constant power is false at low speed, for the obvious reason you discovered.

Homework Helper
In application and in theory. No body is perfectly rigid. Even then no influence tops the speed of light. Science, by its very incarnation, deals with the testable.
As long as the theory conforms to test, a few incidental singularities is not dispositive.

Staff Emeritus
2021 Award
he assumption that you can apply constant power is false at low speed
Exactly. P = Fv. As v →0, for constant P, F →∞.

Approximations break down when approximations break down.

phinds, jack action, Lord Jestocost and 1 other person
Gold Member
The premise that you can apply constant power is false at low speed, for the obvious reason you discovered.
$$\ddot{x} = \frac{P}{m} \frac{1}{\sqrt{ \frac{2P}{m}t + v_o^2}}$$

I disagree. At low speeds applying constant power is completely fine. The singularity vanishes. ##\ddot x## may be arbitrarily large or small depending on ##P## but none the less finite when ##v_o > 0##.

It is when we have no initial speed ##v_o = 0## the any applied power ##P>0## yeilds the singularity. So it seems more like its impossible to apply power at zero velocity, not that you cant apply constant power at low speed. The sticky part is how does anything begin to move if we cant apply any power without visiting ##\infty##?

Mentor
$$\ddot{x} = \frac{P}{m} \frac{1}{\sqrt{ \frac{2P}{m}t + v_o^2}}$$

I disagree. At low speeds applying constant power is completely fine. The singularity vanishes.
So again, that's math, not reality. In reality there are reasons (some already specified) why it doesn't work that way. It is, as @hutchphd said, the difference between the theory and the application (reality).

Staff Emeritus
2021 Award
that's math, not reality. In reality
This.

The sticky part is how does anything begin to move if we cant apply any power without visiting infinity?
You have rediscovered a variant of Zeno's paradox.

erobz and hutchphd
Gold Member
So again, that's math, not reality. In reality there are reasons (some already specified) why it doesn't work that way. It is, as @hutchphd said, the difference between the theory and the application (reality).
Yes, but is it not the supposition in Physics that reality is perfect, and when we encounter these singularities in our models of it, it is due to an imperfect model. We naturally should wonder what fixes the model so the singularity yields?

Mentor
Yes, but is it not the supposition in Physics that reality is perfect, and when we encounter these singularities in our models of it, it is due to an imperfect model. We naturally should wonder what fixes the C.o.E. so the singularity yields?
What? I think you've forgotten the pont of physics.

And by the way there are actually scenarios where the singularity you noticed is not a problem. Elastic collisions for example.

Gold Member
Yes, but is it not the supposition in Physics that reality is perfect, and when we encounter these singularities in our models of it, it is due to an imperfect model. We naturally should wonder what fixes the model so the singularity yields?
The results for your "model" are not nonsensical results in case your "power supply" is able to supply an infinite high force for a moment. It's theory taken at face value.
To be realistic, replace the word “infinite acceleration” with “very high acceleration".

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Homework Helper
Where, in your theory, are you not allowed to have infinities? Its your theory and infinite acceleration is up to you. When you apply it to actual reality you will see that there are ameliorating factors and you make a note.

Lord Jestocost
Gold Member
...
So, no matter how small we make the power, we appear to get an infinite acceleration at ##t=0##.

Anything interesting here, something I'm missing?
Perhaps the lack of agreement or consistency comes from the concept that we can make power.
We can only transfer mechanical energy at certain rate, via application of a finite amount of force or torque.

Consider, for example, how slowly a space rocket starts moving upwards at the launch pad.

russ_watters and erobz
Gold Member
I disagree. At low speeds applying constant power is completely fine.
But not at zero exactly.

If you apply a force to a non-moving object, the power is zero (##P=Fv##). So the initial acceleration is known based on ##F=ma##. Once the object starts moving, then you have power. You can then alter your force input such that the power stays constant as the velocity grows.

Even if you had a wheel rotating freely with a certain power input and you lowered it to the ground, as soon as you touched the ground, the power would be zero with respect to the ground. If we assume no slipping, the wheel will decelerate to zero and a certain initial force (in this case determined by the mechanism of friction) would be applied that would set the initial forward acceleration of the wheel. And as soon as it instantly began to move forward, the force would adapt itself to the velocity to respect the constant power input.

Your equation is just not valid at ##t=0##, assuming ##v=0##, because you will always get:
$$\ddot{x} = \sqrt{\frac{2(0)}{m}} \frac{1}{\sqrt{(0)}}$$
You must use another equation which is simply ##a=\frac{F}{m}##.

Lnewqban
Gold Member
Perhaps the lack of agreement or consistency comes from the concept that we can make power.
We can only transfer mechanical energy at certain rate, via application of a finite amount of force or torque.

Consider, for example, how slowly a space rocket starts moving upwards at the launch pad.
I agree, the problem must come from the idea we can “supply” power. We can supply energy
Where, in your theory, are you not allowed to have infinities? Its your theory and infinite acceleration is up to you. When you apply it to actual reality you will see that there are ameliorating factors and you make a note.
Maybe infinite acceleration is fine. I don’t know. The displacement (which is what can be directly measured) is indifferent to the initial infinity…so I don’t know. Like you said. It’s integrable.

russ_watters and Lnewqban
Gold Member
What? I think you've forgotten the point of physics.
I'm not an expert, its arguable I never knew the point. Care to expand on it, because I don't know what the point is if it not to examine what you think you know?

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Mentor
I'm not an expert, its arguable I never knew the point. Care to expand on it, because I don't know what the point is if it not to examine what you think you know?
The point of physics is to accurately understand/describe physical reality.
Maybe infinite acceleration is fine. I don’t know.
Infinite acceleration is not possible in the real world, however in some cases it is ok to assume it happens/ignore it for the sake of modeling.

Mentor
Perhaps the lack of agreement or consistency comes from the concept that we can make power.
We can only transfer mechanical energy at certain rate, via application of a finite amount of force or torque.
Those two sentences appear to contradict each other, to me.

Homework Helper
I'm not an expert, its arguable I never knew the point. Care to expand on it, because I don't know what the point is if it not to examine what you think you know?
But do not "throw out the baby with the bath water". I agree that an unexamined theory is a bad idea, but unless the theory is completely universal there will be approximations required, because connections to the other degrees of freedom in the world will invoke suppositions. The concept of infinite supplied power is not prohibited by your theory nor would one expect it to be. Other parts of the Physical world conspire to make it not available.
I am again reminded of Feynman's Law of everything. $$U=0$$ The devil is always in the detail

erobz
Gold Member
But do not "throw out the baby with the bath water". I agree that an unexamined theory is a bad idea, but unless the theory is completely universal there will be approximations required, because connections to the other degrees of freedom in the world will invoke suppositions. The concept of infinite supplied power is not prohibited by your theory nor would one expect it to be. Other parts of the Physical world conspire to make it not available.
I am again reminded of Feynman's Law of everything. $$U=0$$ The devil is always in the detail
Just a clarification. Not infinite power, but infinite force - constant power. That is what I find interesting though, any supplied power ##P>0## causes this seemingly paradoxical result. What are the interventions in the physical world that would prevent it from actually happening? The displacement and velocity are indifferent to it, so is that in itself sufficient to say the infinite force isn't a big deal here? In other cases, infinite forces certainly raise eyebrows. I find it interesting that this happens in something as benign as Conservation of Energy, and constant power application. You say it's a bore. That difference must be between what you have experienced, and I have not.

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Gold Member
2021 Award
I haven't follow the entire thread in detail, but I think it's simplest to just solve the equation of motion than some debates with many words. We always have Newton's equation of motion
$$m \ddot{x}=F.$$
Here we don't know ##F## but have given that the power is constant during the motion, i.e.,
$$P=F v=F \dot{x}=\text{const} \; \Rightarrow \; F=\frac{P}{\dot{x}}.$$
Plugging this into the equation of motion yields
$$m \dot{x} \ddot{x}=P=\text{const}.$$
This you can write as
$$\frac{m}{2} \mathrm{d}_t (\dot{x}^2)=P \; \Rightarrow \; \dot{x}^2=\frac{2 P}{m} t+C.$$
The initial condition says ##v(0)=0##, i.e., ##C=0##. That means
$$\dot{x}=\sqrt{2 P t/m}$$
Integrating this once more you get
$$x(t)=\frac{2}{3} \sqrt{\frac{2 P}{m}} t^{3/2},$$
where I assumed that the position at ##t=0## is ##x(0)=0##.

erobz
Gold Member
I haven't follow the entire thread in detail, but I think it's simplest to just solve the equation of motion than some debates with many words. We always have Newton's equation of motion
$$m \ddot{x}=F.$$
Here we don't know ##F## but have given that the power is constant during the motion, i.e.,
$$P=F v=F \dot{x}=\text{const} \; \Rightarrow \; F=\frac{P}{\dot{x}}.$$
Plugging this into the equation of motion yields
$$m \dot{x} \ddot{x}=P=\text{const}.$$
This you can write as
$$\frac{m}{2} \mathrm{d}_t (\dot{x}^2)=P \; \Rightarrow \; \dot{x}^2=\frac{2 P}{m} t+C.$$
The initial condition says ##v(0)=0##, i.e., ##C=0##. That means
$$\dot{x}=\sqrt{2 P t/m}$$
Integrating this once more you get
$$x(t)=\frac{2}{3} \sqrt{\frac{2 P}{m}} t^{3/2},$$
where I assumed that the position at ##t=0## is ##x(0)=0##.
Thats not the problem I'm having. The problem is ## \lim_{t \to 0} \ddot x \rightarrow \infty## for all ##P > 0##.

vanhees71
Gold Member
2021 Award
Yes, but acceleration with ##P=\text{const}## is a pretty artificial situation to begin with ;-)).

bob012345 and Lord Jestocost
Gold Member
Yes, but acceleration with ##P=\text{const}## is a pretty artificial situation to begin with ;-)).
Yeah, why exactly though? You can't even say ##P=0## at ##t=0## to start the motion. If ##P=0## there is no motion. The very moment ##P>0## we have a singularity. It's seeming to be a twist on Zenos's Paradox.

vanhees71
Mentor
That is what I find interesting though, any supplied power ##P>0## causes this seemingly paradoxical result. What are the interventions in the physical world that would prevent it from actually happening?
Depends on the device. A car has a transmission with a limited range of gear ratios, a clutch or torque converter that slips and tires with limited grip. If you rev the engine and pop the clutch, you'll spin the tires, stall the engine or break the transmission or drive shaft. The first/most common limiter is tire friction.

Lnewqban and jack action
Staff Emeritus
2021 Award
I can't believe this is still going on.

1. Building a device where this approximation is valid is impossible. Actual devices do not have constant power at v = 0. Here is the very first graph Google founmd:

As engine speed goes to zero, so does power. This is unavoidable.

2. Constant power implies infinite force. And infinite force does not give you infinite acceleration. It breaks your device. Smacking your car's engine with a sledgehammer does not make it go any faster.

Dale, Lnewqban, russ_watters and 1 other person
Homework Helper
That is what I find interesting though, any supplied power P>0 causes this seemingly paradoxical result.
I do not understand your particular vexation here. Suppose I collide two billiard balls head-on at t=0. I require energy be conserved. At t=0 the velocity is zero, yet an instant later all is copacetic. Paradox? Of course not because the balls cannot instantaneously reverse. Why? because other physics intervenes in that limit. The (physical) limit is a useful approximate fiction.

Gold Member
Thats not the problem I'm having. The problem is ## \lim_{t \to 0} \ddot x \rightarrow \infty## for all ##P > 0##.
Perhaps we should move away from transduction capable machines in this discussion.
What if instead we analyze the first instants of the simple free fall of a body, or the highest point reached by a vertical projectile trajectory?

If we consider a simple lever, or gear’s pair, yes, the input force should tend to infinite as the arm distace to the fulcrum or axis of rotation tends to zero.
But at that point, do we still have a lever, or rather a support point unable to transfer our “constant power”?

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jack action and vanhees71
Gold Member
2. Constant power implies infinite force. And infinite force does not give you infinite acceleration. It breaks your device. Smacking your car's engine with a sledgehammer does not make it go any faster.
This is more of hitting the engine with a pea breaks your car... ##P## can be arbitrarily small. Which is why I found it troubling...

I can't believe this is still going on.

1. Building a device where this approximation is valid is impossible.
I'm fine with this...I implied as much in the OP. It was the implied paradox created by the constant power assumption that I was interested in.

Simply letting ##P## be the next simplest model ##P = kt## resolves the issue. ##\ddot x (0) = \rm{const.}##

Sorry for bothering everyone, I was genuinely surprised by this...not trolling.

Lnewqban