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26 December, 04:42

The velocity components u and v in a two-dimensional flow field are given by: u = 4yt ft./s, v = 4xt ft./s, where t is time. What is the time rate of change of the velocity vector V (i. e., the acceleration vector) for a fluid particle at x = 1 ft. and y = 1 ft. at time t = 1 second?

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  1. 26 December, 04:59
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    vec (a) = 16 i + 16 j

    mag (a) = 22.63 ft/s^2

    Explanation:

    Given,

    - The two components of velocity are given for fluid flow:

    u = 4*y ft/s

    v = 4*x ft/s

    Find:

    What is the time rate of change of the velocity vector V (i. e., the acceleration vector) for a fluid particle at x = 1 ft. and y = 1 ft. at time t = 1 second?

    Solution:

    - The rate of change of velocity is given to be acceleration. We will take derivative of each components of velocity with respect to time t:

    a_x = du / dt

    a_x = 4*dy/dt

    a_y = dv/dt

    a_y = 4*dx/dt

    - The expressions dx/dt is the velocity component u and dy/dt is the velocity component v:

    a_x = 4 * (4*y) = 16y

    a_y = 4 * (4*x) = 16x

    - The acceleration vector can be expressed by:

    vec (a) = 16y i + 16x j

    - Evaluate vector (a) at x = 1 and y = 1:

    vec (a) = 16*1 i + 16*1 j = 16 i + 16 j

    - The magnitude of acceleration is given by:

    mag (a) = sqrt (a^2_x + a^2_y)

    mag (a) = sqrt (16^2 + 16^2)

    mag (a) = 22.63 ft/s^2
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