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**Hint:**The distance is defined as the length of the path between the initial position and the final position of a moving object in the given interval of time. It is a scalar quantity. The change in the position of an object is called displacement. Displacement is a vector quantity and has both magnitude and direction.

**Complete step by step solution:**

Given equation \[ x = 40 + 12t - t^3\].....................(1)

Let say that the particle is at a distance x from O at a time, t=0

Thus substituting the value of t=0 in the given equation, \[x = 40 + 12t - {t^3}\], we get,

\[x = 40 + 12\left( 0 \right) - {\left( 0 \right)^3} = 40m\]

It is given that the particle will come to rest which means the velocity of the particle becomes zero after traveling to a certain displacement and let us consider this time as t.

Differentiating the given equation 1 to time, we get,

$\Rightarrow v = 12 - 3{t^2}$.................(2)

When the particle comes to rest at time=t, velocity=0

Substituting the value of v=0 in equation 2, we get,

$\Rightarrow 12 - 3{t^2} = 0$

$ \Rightarrow 3{t^2} = 12$

$ \Rightarrow {t^2} = 4$

$ \Rightarrow t = 2s$

Substituting the value of t in equation 1, we get

\[\Rightarrow {x} = 40 + 12\left( 2 \right) - {\left( 2 \right)^3}\]

\[ \Rightarrow {x} = 56m\]

Thus, it is clear that the particle had started the journey when it was at a distance of 40m from point O and came to rest at 56m from point O.

Thus, the particle had traveled a distance of $56m - 40m = 16m$.

Thus, the particle had traveled a distance of 16m before coming to rest.

**Hence the correct option is D.**

**Note:**1. The velocity is defined as the rate of change of the position of the object to time. Velocity is also a vector quantity and has both direction and magnitude.

2. The distance and the displacement have the same SI unit, meter.

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