Displacement after Time T in a Velocity Field

In summary, the speaker is trying to find the displacement of a particle after T seconds in a static velocity field. They consider a one dimensional case and use a separable differential equation to find the final time, T. However, to find the ending position, they would need to take the inverse of the function given by the equation, which may not be possible for an arbitrary velocity function. In a two dimensional problem, this process would need to be repeated for each component of the position vector.
  • #1
Teg Veece
8
0
I have a velocity field that is static in time. At every location, x, there is a corresponding velocity vector.

I'm trying to work out the displacement of a particle after T seconds if I drop it into the velocity field at time t=0 and at location x_0.

I was thinking something along the lines of ∫v(x_0)dt between the limits of t=0 and t=T but it doesn't seem right because the integrand doesn't seem to depend on t at all (in fact v(x_0) is a constant so it can't be right). Maybe there's a double integral involved...

Anyone know what a suitable form for this equation would be?
 
Physics news on Phys.org
  • #2
For simplicity, you could consider a one dimensional case:

[itex]\frac{dx}{dt}=v(x)[/itex]

This is a separable differential equation, so:

[itex]\int_{x_0}^{x_f}\frac{dx}{v(x)}=\int_0^T dt[/itex]

But it gets you a solution for the final time, T.

[itex]T = \int_{x_0}^{x_f}\frac{dx}{v(x)} [/itex]

If for a fixed x0, you can suppose that the total time is a function of the ending position.

[itex]T = f(x_f) [/itex], where this function is given by the right hand side of the equation above.

So, to get the ending position for a given total time, you would need to take the inverse of that function

[itex]x_f = f^{-1}(T)[/itex]

In order to get a closed form for this, you need to plug in the velocity field. I don't think there's a way to get a general expression for an arbitrary velocity function.

Now, for a two dimensional problem you need to do this for each component of the position vector.
 

Related to Displacement after Time T in a Velocity Field

1. How is displacement after time T calculated in a velocity field?

Displacement after time T in a velocity field is calculated by multiplying the velocity at a certain point by the amount of time that has passed. This gives the change in position or displacement from the starting point.

2. What is the difference between displacement and velocity?

Displacement is a vector quantity that describes the change in position of an object from a starting point, while velocity is a vector quantity that describes the rate of change of an object's displacement over time.

3. Can displacement after time T be negative?

Yes, displacement after time T can be negative. This indicates that the object has moved in the opposite direction of its starting point.

4. What is the unit of measurement for displacement after time T?

The unit of measurement for displacement after time T depends on the unit of measurement for velocity and time. It is typically measured in meters (m) or feet (ft).

5. How does the velocity field affect displacement after time T?

The velocity field determines the direction and magnitude of the velocity at different points. This directly affects the displacement after time T, as the object will move in the direction and at the rate determined by the velocity field.

Similar threads

Replies
49
Views
2K
  • Mechanics
Replies
30
Views
1K
Replies
3
Views
292
Replies
22
Views
2K
Replies
4
Views
1K
Replies
23
Views
1K
  • Special and General Relativity
Replies
24
Views
692
Replies
2
Views
1K
Replies
13
Views
2K
Replies
20
Views
1K
Back
Top