Velocity with respect to acceleration

In summary, the conversation discusses the possibility of differentiating a function with respect to acceleration, expressed in terms of velocity. The derivative of velocity with respect to acceleration is the fraction of acceleration and its rate of change. The questioner is trying to determine the relationship between energy and acceleration, and has arrived at an expression involving the chain rule. They are unsure of how to proceed and ask for guidance.
  • #1
Zman
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Is it possible to differentiate a function with respect to acceleration where the function is expressed in terms of velocity?

[tex]\frac{dy}{da} = \frac{d}{da}{\frac{1}{\sqrt{1 - v^2/c^2}}}[/tex]
 
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  • #2
Well, if say the acceleration/time-relationship is invertible, so that time may be regarded as a function of the acceleration, we would have:
[tex]\frac{dv}{da}=\frac{dv}{dt}\frac{dt}{da}=a\frac{dt}{da}=\frac{a}{\frac{da}{dt}}[/tex]

Thus, the derivative of velocity wrt. to acceleration is the fraction between the acceleration itself and its rate of change.
 
  • #3
Thanks for your reply.
I tried to keep my question simple but I think that that was a mistake. My maths is extremely rusty and I definitely feel uncomfortable with it.

The situation that I am dealing with is the relationship between the energy of a body and its acceleration.

I want to determine the relationship dE/da (E is energy, a is acceleration)
I have arrived at the expression;
[tex]\frac{dE}{da} = \frac{d}{da}{mc^2\frac{1}{\sqrt{1 - v^2/c^2}}}[/tex]

and I am not sure how to proceed from this point.
 
  • #4
Zman said:
Thanks for your reply.
I tried to keep my question simple but I think that that was a mistake. My maths is extremely rusty and I definitely feel uncomfortable with it.

The situation that I am dealing with is the relationship between the energy of a body and its acceleration.

I want to determine the relationship dE/da (E is energy, a is acceleration)
I have arrived at the expression;
[tex]\frac{dE}{da} = \frac{d}{da}{mc^2\frac{1}{\sqrt{1 - v^2/c^2}}}[/tex]

and I am not sure how to proceed from this point.

[tex]mc^2(1- v^2/c^2)^{-1/2}[/tex]
Now use the chain rule.
 
  • #5
The velocity v is the only variable in the equation. Surely I need to express v in terms of acceleration ‘a’ before I can differentiate the expression using the chain rule?

If I was resolving dE/dv, I believe that I could go ahead and differentiate the expression using the chain rule but I am trying to resolve dE/da.
 
  • #6
dE/da = dE/dv*dv/da

aka the chain rule. Go for it
 

1. What is velocity with respect to acceleration?

Velocity with respect to acceleration refers to the change in speed or direction of an object over time. It is a measure of how quickly an object's velocity is changing.

2. How is velocity with respect to acceleration calculated?

Velocity with respect to acceleration can be calculated by dividing the change in velocity by the change in time. This gives the average acceleration of the object over a certain period.

3. What is the difference between velocity and acceleration?

Velocity is a measure of an object's speed and direction of motion, while acceleration is a measure of the change in velocity over time. In other words, velocity is the rate of change of position, while acceleration is the rate of change of velocity.

4. What are some real-life examples of velocity with respect to acceleration?

A car accelerating on a highway, a ball being thrown in the air, and a person riding a roller coaster are all examples of velocity with respect to acceleration. In each of these scenarios, the object's velocity is changing due to the effects of acceleration.

5. How does velocity with respect to acceleration relate to Newton's laws of motion?

According to Newton's Second Law of Motion, an object's acceleration is directly proportional to the net force acting on it and inversely proportional to its mass. This means that an object with a greater mass will require more force to accelerate it to the same velocity as an object with a smaller mass.

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