A particle is moving in simple harmonic motion along the -x axis - HSC - SSCE Mathematics Extension 1 - Question 4 - 2010 - Paper 1

Acceleration is the derivative of velocity. Starting with:

$v = \sqrt{24 - 8x - 2x^2}$

We differentiate $v$ with respect to $t$, using the chain rule:

$a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = \frac{dv}{dx} v$

Calculating $\frac{dv}{dx}$ gives:

$\frac{dv}{dx} = \frac{1}{2\sqrt{24 - 8x - 2x^2}} \cdot (-8 - 4x)$

Substituting $v$ back into the equation, we have:

$a = \left(\frac{(-8 - 4x)}{2\sqrt{24 - 8x - 2x^2}}\right) \cdot \sqrt{24 - 8x - 2x^2}$

Thus, the expression for acceleration simplifies to:

$a = \frac{-8 - 4x}{2} = -4 - 2x.$

To find the maximum speed of the particle, we explore the maximum value of $v$:

The maximum occurs where the velocity function becomes zero:

$v^2 = 24 - 8x - 2x^2.$

To maximize $v$, we need the vertex of the equation.

The maximum point is at:

$x = -\frac{b}{2a} = -\frac{{-8}}{2(-2)} = 2.$

Substituting $x=2$ into the equation for $v^2$:

$v^2 = 24 - 8(2) - 2(2^2) = 24 - 16 - 8 = 0.$

Thus, the maximum velocity is:

$v = \sqrt{0} = 0.$

Using the cosine addition formula:

ext{cos}(θ + rac{π}{3}) = ext{cos}θ ext{cos}rac{π}{3} - ext{sin}θ ext{sin}rac{π}{3}

Gives:

2 ext{cos}θ + 2igg( ext{cos}θ \cdot \frac{1}{2} - ext{sin}θ \cdot \frac{\sqrt{3}}{2}\bigg)

This simplifies to:

R \text{cos}igg(θ + α\bigg) where R = 2 ext{cos}igg(θ + α\bigg).

Further, we find $R$ and $α$ to satisfy conditions where $R > 0$ and $0 < α < \frac{π}{2}$.

Setting our expression equal to 3, we have:

$R \text{cos}(θ + α) = 3.$

Substituting $R$ derived from the earlier expression, we solve for $θ$:

$θ + α =$, and solve for values in the range $0 < θ < 2π$.

Using the properties of the tangents and coordinates in conjunction with the graphics of the parabola:

We show all sides of $SL$, $LM$, $MP$, and $PS$ are equal in length, therefore confirming:

$|SL| = |LM| = |MP| = |PS|$

Thus, confirming that $SLMP$ is indeed a rhombus.