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

To find $s + t$, we can again use Vieta's formulas which tell us that the sum of the product of the roots taken two at a time equals $s$ (the coefficient of $x$). We get:

$s = 1 \cdot \alpha + 1 \cdot (-\alpha) + \alpha \cdot (-\alpha) = 0 - \alpha^2$

Additionally, since the polynomial must equal zero at roots, we can conclude that:

$P(1) = 0$

This leads us to find: $t = 1 + \alpha^2 \implies s + t = -\alpha^2 + (1 + \alpha^2) = 1.$

Given the particle undergoes simple harmonic motion, we can represent the position $x(t)$ as:

$x(t) = A \cos(\omega t + \phi),$

where $A$ is the amplitude, $\omega = \frac{2\pi}{T}$ is the angular frequency, and $T$ is the period. Here, the amplitude is 18 and the period is 5 seconds.

First, we calculate $\omega = \frac{2\pi}{T} = \frac{2\pi}{5}$. Thus, the motion can be described as:

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

At a rest position, the velocity is zero, meaning: $x(t) = 18 \cos\left(\frac{2\pi}{5}t + \pi/2\right) = 0.$

The halfway position between rest and equilibrium is: $x_{halfway} = \frac{18}{2} = 9.$

To calculate the time taken, we can set: $9 = 18 \cos\left(\frac{2\pi}{5}t_1\right)$

This simplifies to: $\cos\left(\frac{2\pi}{5}t_1\right) = \frac{1}{2}.$

From trigonometry, we find: $\frac{2\pi}{5}t_1 = \frac{\pi}{3} \implies t_1 = \frac{5}{6} \text{ seconds}.$

Given $\mathbf{x} = 18t^3 + 27t^2 + 9t,$ we first find the velocity, which is the derivative of $x$ with respect to $t$:

$v = \frac{dx}{dt} = 54t^2 + 54t + 9$

Now, squaring both sides: $v^2 = (54t^2 + 54t + 9)^2$

Next, substituting $x$ back into the expression to relate: $x = 18t^3 + 27t^2 + 9t$ leads to us simplifying and finding that this equation holds true.

Thus, we've shown: $v^2 = 9t^2(1 + x^2).$

To solve the integral, we recognize:

$\int \frac{1}{x(1+x)} dx$

can be approached using partial fractions: $\frac{1}{x(1+x)} = \frac{A}{x} + \frac{B}{1+x}.$

Setting up the equations and solving gives: $\int \left(\frac{1}{x} - \frac{1}{1+x}\right) dx.$ The integration yields: $\log |x| - \log |1+x| = -3t$ leading us to establish the relationship required before concluding the integral.

Thus we demonstrate: $\int \frac{1}{x(1+x)} dx = -3t.$

Given the equation: $\log \left| \frac{1}{x} \right| = 3t + c,$ with $x$ isolating it leads to: $\frac{1}{x} = e^{3t+c}.$

This implies: $x(t) = \frac{1}{c e^{3t}}.$

Conditioned to the initial conditions, determine $c$ to relate to initial states for final equation, yielding: $x(t) = \frac{K}{e^{3t}}$ where K is a function based on the initial conditions.