A particle moves in a velocity field V(x, y) = x2, x + y2 . If it is at position (x, y) = (7, 2) at time t = 3, estimate its location at tim

Question

A particle moves in a velocity field V(x, y) = x2, x + y2 . If it is at position (x, y) = (7, 2) at time t = 3, estimate its location at time t = 3.01.

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Nem 4 years 2021-08-14T09:10:52+00:00 1 Answers 32 views 0

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  1. dangkhoa
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    2021-08-14T09:12:33+00:00
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    Answer:

    New location at time 3.01 is given by: (7.49, 2.11)

    Explanation:

    Let’s start by understanding what is the particle’s velocity (in component form) in that velocity field at time 3:

    V_x=x^2=7^2=49\\V_y=x+y^2=7+2^2=11

    With such velocities in the x direction and in the y-direction respectively, we can find the displacement in x and y at a time 0.01 units later by using the formula:

    distance=v\,*\, t

    distance_x=49\,(0.01)=0.49\\distance_y=11\,(0.01)=0.11

    Therefore, adding these displacements in component form to the original particle’s position, we get:

    New position: (7 + 0.49, 2 + 0.11) = (7.49, 2.11)

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