First, let's find the acceleration of the particle. The equation describing the motion of the particle down the inclined plane is

`d = v_0t + at^2/2` , where d = 3 m is the distance, the initial velocity is

`v_0 = 0` since the particle is released from rest, t = 3 s is the time during which the particle moved the distance d, and a is acceleration, which is directed down the inclined plane.

From here acceleration can be found:

`3 = a3^2/2`

`a = 6/9 = 2/3 = 0.67 m/s^2`

Now we can calculate the coefficient of the kinetic friction between the surface of the inclined plane and the particle. The forces acting on the particle are the gravity, normal force from the surface, and the friction. The net vector sum of these forces provides the acceleration, according to Newton's second law:

`mvecg + vecN + vecF_(fr) = mveca`

Let's direct the x-axis down the inclined surface and the y-axis up, perpendicular to the surface (please see the attached image for the illustration.)

Then, the x-component of the terms in the equation will be

`mgcos(60) - F = ma`

(The angle between the force of gravity and the x-axis is 90-30 = 60 degrees, from geometric considerations.)

The y-component of the equation will be

`-mgcos(30) + N = 0`

Finally, the relationship between the normal force N and the friction force F is

`F=muN` , where `mu` is the coefficient of kinetic friction. Combining the three equation, we get

`mgcos(60) - mumgcos(30) = ma`

The mass m cancels because it is a common factor in all the terms of the equation. Then, `mu` can be expressed in terms of g and a:

`mu=(gcos(60)-a)/(gcos(30)) = (gsqrt(3)/2 -a)/(g*1/2)`

`mu=(gsqrt(3)-2a)/(g) = 1.6`

**The acceleration of the particle is 0.67 m/s^2 and the coefficient of kinetic friction is 1.6.**

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