Acceleration changing over time

In summary, the speaker is having trouble calculating movement on a dynamic system with changing acceleration. They want to know the end distance and velocity of the object, and will need to use calculus to solve the problem. They mention using a transfer function and linearization, and also mention that the acceleration is a function of time and distance. They are aware that they will need to integrate the acceleration twice to get the distance, and mention a possible method of solution involving integration by parts.
  • #1
nicklasaau
6
0
Hey, i am having trouble doing the calculation on a dynamic system.

My acceleration is not constant during, and because of that my speed of the particle is changing from the original, and i want to know how long my particle has moved in a short periode of time Δt and what the velocity is. with the math i have learned so far, i end up with only taking into account the end velocity or the beginning velocity, instead of calculating it is changing


Know is
Beginning condition: (distance, speed)
Travel: (acceleration at a specific time, that changes)
I want to know end distance and end velocity.

At the end i need to rewrite it to a transfer function, so i have to do linearization if that can help in the calculation :)
 
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  • #2
A general solution will require calculus, but it's possible you won't need this.

velocity: v(t) = ∫a(t)dt + v0
displacement: x(t) = ∫(v(t))dt + x0 = ∫(∫a(t)dt + v0)dt + x0

v0, x0 are the velocity and displacement at t=0 (initial conditions).

You will need to integrate the acceleration a(t) twice to get a distance, and the method of solution will depend on the form of a(t). Do you know anything about the form of a(t)?
 
  • #3
The object is a levitating beam, and because of that the acceleration is a function of time and distance



a(t,x) = (μ * N^2 * Ag)/4 * [I*sin(ωt + ρ)]^2 / x^2, or something like that
 
  • #4
MikeyW said:
A general solution will require calculus, but it's possible you won't need this.

velocity: v(t) = ∫a(t)dt + v0
displacement: x(t) = ∫(v(t))dt + x0 = ∫(∫a(t)dt + v0)dt + x0

v0, x0 are the velocity and displacement at t=0 (initial conditions).

You will need to integrate the acceleration a(t) twice to get a distance, and the method of solution will depend on the form of a(t). Do you know anything about the form of a(t)?

That double integral involving a(t) can be reexpressed as a single integral by integrating by parts to obtain:

[tex]\int_0^t(t-\lambda)a(\lambda) d\lambda[/tex]

where λ is a dummy variable of integration.
 
  • #5


I understand the importance of accurately calculating the motion of a dynamic system. In this case, it seems like you are struggling with calculating the acceleration, velocity, and distance of a particle over a specific time period. It is important to note that acceleration is the rate of change of velocity, meaning it can vary over time. In order to accurately determine the end distance and velocity of your particle, you will need to take into account the changing acceleration.

One way to approach this problem is to break down the time period into smaller intervals, and calculate the acceleration, velocity, and distance at each interval. This will give you a more accurate representation of the particle's motion. You can then use these values to calculate the overall end distance and velocity.

Additionally, linearization can be a helpful tool in simplifying complex systems. However, it is important to use it carefully and only when appropriate. It involves approximating a nonlinear system with a linear one, which may not always give the most accurate results. I suggest consulting with a mathematician or using advanced mathematical techniques to accurately calculate the motion of your dynamic system.

In conclusion, accurately determining the motion of a dynamic system can be challenging, but it is important to take into account the changing acceleration in order to get a complete understanding of the system's behavior. I hope this response helps guide you in your calculations and leads to a successful outcome.
 

1. How is acceleration defined?

Acceleration is defined as the rate of change of an object's velocity over time. It is a vector quantity, meaning it has both magnitude (speed) and direction.

2. Can acceleration change over time?

Yes, acceleration can change over time. This is known as a changing or variable acceleration. It occurs when an object's velocity is changing at a non-uniform rate, meaning its speed and/or direction is changing over time.

3. What causes acceleration to change over time?

Acceleration can change over time due to a variety of factors, including the application of a force, such as gravity or friction, or changes in an object's mass or velocity.

4. How is changing acceleration represented mathematically?

Changing acceleration is represented mathematically using calculus, specifically through the derivative of velocity with respect to time. This is known as the object's acceleration function, which can vary over time depending on the object's motion.

5. How does changing acceleration affect an object's motion?

Changing acceleration can affect an object's motion in various ways. For example, if an object is experiencing a positive acceleration (speeding up) over time, its velocity will increase at a non-uniform rate. Conversely, if an object is experiencing a negative acceleration (slowing down), its velocity will decrease at a non-uniform rate.

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