Q preguntado en Science & MathematicsPhysics · hace 1 mes

At time   t = 1,  a particle is located at position   (x, y) = (5, 3).  If it moves in the velocity field  ?

At time  

t = 1, 

a particle is located at position  

(x, y) = (5, 3). 

If it moves in the velocity field  

F(x, y) =  (xy − 1, y2 − 9 )

 

find its approximate location at time  

t = 1.02

1 respuesta

Relevancia
  • hace 1 mes
    Respuesta preferida

    F(x, y) = (xy - 1, y² - 9)

    Velocity at t=1, at point (5,3), is:

    F(5, 3) = (5*3 − 1, 3² − 9) = (14, 0)

    We assume that in the short time interval Δt = 0.02 (from t=1 to t=1.02)  the velocity is approximately constant.

    New position vector = original position vector + displacement

    = original position vector + velocity*time

    = (5, 3) + (14, 0)(0.02)

    = (5, 3) + (0.28, 0)

    = (5.28, 3)

¿Aún tienes preguntas? Pregunta ahora para obtener respuestas.