The distance traveled by a particle from a fixed point is given by the function \(s(t) = t^3 - t^2 - t + 5\). To find the minimum distance that the particle can cover from the fixed point, we need to find the minimum value of this function.

To find the minimum value of the function, we can take its derivative and set it equal to zero to find the critical points. The derivative of \(s(t)\) is given by:

\(s'(t) = 3t^2 - 2t - 1\)

Setting this equal to zero, we get:

\(3t^2 - 2t - 1 = 0\)

Using the quadratic formula, we can solve for t:

\(t = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 3 \cdot (-1)}}{2 \cdot 3}\)

\(= \frac{2 \pm \sqrt{4 + 12}}{6}\)

\(= \frac{2 \pm \sqrt{16}}{6}\)

\(= \frac{2 \pm 4}{6}\)

So, t has two possible values: \(t = \frac{2 + 4}{6} = 1\) and \(t = \frac{2 - 4}{6} = -\frac{1}{3}\).

To determine which of these values gives the minimum value of \(s(t)\), we can use the second derivative test. The second derivative of \(s(t)\) is given by:

\(s''(t) = 6t - 2\)

Substituting \(t = 1\) into this expression, we get:

\(s''(1) = 6 \cdot 1 - 2\)

\(= 4\)

Since \(s''(1)\) is positive, this means that \(t = 1\) is a local minimum of \(s(t)\). Substituting \(t = 1\) into the expression for \(s(t)\), we get:

\(s(1) = (1)^3 - (1)^2 - (1) + 5\)

\(= 4\)

So, the minimum distance that the particle can cover from the fixed point is 4 cm.