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, $1$, $\alpha$ and $-\alpha$ - HSC - SSCE Mathematics Extension 1 - Question 4 - 2006 - Paper 1

Next, we find the values of $s$ and $t$. From Vieta's formulas, the sum of the products of the roots taken two at a time is given by $s$. Thus, we have:

$1 \cdot \alpha + 1 \cdot (-\alpha) + \alpha \cdot (-\alpha) = s$

This simplifies to:

$0 - \alpha^2 = s \Rightarrow s = -\alpha^2.$

Now, for the product of roots, we have:

$1 \cdot \alpha \cdot (-\alpha) = -t \Rightarrow t = \alpha.$

Thus, we find:

$s + t = -\alpha^2 + \alpha.$

The position of the particle undergoing simple harmonic motion can be modeled by the equation:

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

Where:

• $A$ is the amplitude of motion, which is 18.
• $\omega$ is the angular frequency, given by: $\omega = \frac{2\pi}{T} = \frac{2\pi}{5},$ where $T$ is the period.
• Since the particle starts from the extreme positive position, we can set the phase constant $\phi = 0$. Thus:

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

To answer this, we need to determine when the particle reaches halfway between the maximum amplitude and the equilibrium point. Since the amplitude is 18, the equilibrium position is 0. The halfway point is:

$x_{half} = \frac{18 + 0}{2} = 9.$

The velocity at the rest position is zero, indicating that the particle is at $x = 18$. To find the time taken to reach $9$, we set the position equation equal to 9:

$18 \cos\left(\frac{2\pi}{5} t\right) = 9 \Rightarrow \cos\left(\frac{2\pi}{5} t\right) = \frac{1}{2}.$

The time for this cosine function occurs at:

$\frac{2\pi}{5} t = \frac{\pi}{3} \Rightarrow t = \frac{5}{6}.$

Thus, it takes approximately $\frac{5}{6}$ seconds.

To find the expression for $v^2$, we first derive the equation for velocity given by:

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

Now we will compute $v^2$:

$v^2 = (54t^2 + 54t + 9)^2 = 9t^2(1 + t^2).$

To show this, we need to integrate the function:

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

Using partial fraction decomposition, we express:

$\frac{1}{x(1+x)} = \frac{A}{x} + \frac{B}{1+x}.$

Solving gives $A = 1$, $B = -1$. Thus:

$\int \frac{1}{x(1+x)} dx = \int \frac{1}{x} dx - \int \frac{1}{1+x} dx = \ln|x| - \ln|1+x| + C.$

Applying the initial conditions and adjusting yields:

$= -3t.$

This relationship allows us to express:

$\log_e{\left| \frac{1}{x} \right|} = 3t + c.$

To find $x$ in terms of $t$, we exponentiate both sides yielding:

$\frac{1}{x} = e^{3t+c} \implies x = \frac{1}{e^{3t+c}}.$

Replacing $c$ appropriately will yield the final expression for $x$ in terms of $t$.