Acceleration as a function of position, and velocity.

In summary, the conversation discusses the relationship between acceleration and position, and whether velocity can be expressed as a function of position without a time variable. It is possible to do so using the chain rule and conservation of energy, but the integral would be over time rather than position. The conversation also mentions a more complicated problem involving kinetic and potential energy.
  • #1
Axecutioner
32
0
If acceleration is a function of position, can you express velocity as a function of position, or would there have to be a time variable? For example, if F = kx = ma, and v0 is given, then a = (k/m)x, so what would the function for v be?

I would think that, if you know the acceleration at every point along a given path and you know initial velocity, you would also know velocity at every point as well, without having to know what time is.

My actual problem is more complicated than a spring, but it comes down to ∫ f(x) d? = ∫ a d? = v, but what would the integral be? dx or dt?

Thanks!
 
Physics news on Phys.org
  • #2
Yes, you can do this. For example you can use the chain rule

dv/dt = (dv/dx)(dx/dt)

but dx/dt = v, so

a = dv/dt = v dv/dx.
 
  • #3
Axecutioner said:
I would think that, if you know the acceleration at every point along a given path and you know initial velocity, you would also know velocity at every point as well, without having to know what time is.
That's correct, if you take what AlephZero said, and write it as 0.5*d(v2)/dx = F(x), then integrate over x, you get the concept of kinetic energy on the RHS and potential energy on the LHS, which let's you figure out v(x) without ever knowing t.
My actual problem is more complicated than a spring, but it comes down to ∫ f(x) d? = ∫ a d? = v, but what would the integral be? dx or dt?
It's over dt, to get the second integral to =v. But it's not useful, since F is a function of x not t. So get v(x) with conservation of energy, and if you want to know what t is doing, say v(x)=dx/dt so t is the integral of dx/v(x).
 

1. What is acceleration as a function of position and velocity?

Acceleration as a function of position and velocity is a mathematical relationship that describes the change in an object's acceleration as it moves through space. It takes into account both the object's position and velocity at a given time.

2. How is acceleration as a function of position and velocity calculated?

The formula for calculating acceleration as a function of position and velocity is a = dv/dt, where a is acceleration, v is velocity, and t is time. This formula can be derived from the fundamental principles of calculus.

3. What is the difference between acceleration as a function of position and velocity and acceleration as a function of time?

Acceleration as a function of position and velocity takes into account the object's changing position and velocity, while acceleration as a function of time only considers the change in velocity over time. The former provides a more detailed and accurate description of an object's acceleration.

4. How does acceleration as a function of position and velocity apply to real-world scenarios?

Acceleration as a function of position and velocity is commonly used in physics and engineering to model the motion of objects in real-world scenarios. It can help predict the behavior of objects in space, such as satellites orbiting the Earth, or in more complex systems like cars moving on a curved track.

5. Can acceleration as a function of position and velocity be negative?

Yes, acceleration as a function of position and velocity can be negative. A negative value indicates that the object is decelerating, or slowing down, as it moves through space. This could happen if the object is moving in the opposite direction of the positive direction defined by the chosen coordinate system.

Similar threads

Replies
22
Views
1K
Replies
12
Views
3K
Replies
8
Views
339
  • Mechanics
Replies
3
Views
522
Replies
1
Views
616
Replies
4
Views
935
Replies
9
Views
846
Replies
8
Views
897
  • Mechanics
Replies
11
Views
929
Back
Top