The cubic polynomial $P(x) = x^3 + r x^2 + s x + t$, where $r$, $s$ and $t$ are real numbers, has three real zeros, 1, $eta$ and $-eta$ - HSC - SSCE Mathematics Extension 1 - Question 4 - 2006 - Paper 1

Using Vieta's formulas again, the sum of the products of the roots taken two at a time gives us:

s + t = 1 imes eta + 1 imes (-eta) + eta imes (-eta) = -eta^2.

Also, the product of the roots is:

s = (1)(eta)(-eta) = -eta^2.

Thus, the values of $s$ and $t$ can be calculated based on the roots.

The particle returns to its initial position every 5 seconds, leading to a periodic motion. Therefore, the equation for the position $x(t)$ can be modeled as:

$x(t) = 18 imes ext{cos}\left(\frac{2\,\pi}{5} t\right).$

In simple harmonic motion, the particle spends equal time in each section of its path. To find the time taken to move from a rest position (crest) to halfway between this crest and the equilibrium position, we note that:

$T = \frac{1}{4} T_{total} = \frac{1}{4} \times 5 seconds = 1.25 seconds.$

Given the velocity as a function of time:

$v = \frac{dx}{dt} = \frac{d}{dt}(18t^3 + 27t^2 + 9t) = 54t^2 + 54t + 9.$

We equate this to:

$9t^2(1 + x)^2,$

to prove the relationship holds, given an appropriate substitution for $x$.

To evaluate the integral:

$\int \frac{1}{x(1+x)} dx = \int \left( \frac{1}{x} - \frac{1}{1+x} \right) dx.$

Upon solving and using the relationship derived, we find:

$= -3t.$

Substituting the known values and solving: $\log\left( \frac{1}{x} \right) = 3t + c.$ By using initial conditions to find $c$, we can derive the explicit function for $x(t)$.