A particle is moving in simple harmonic motion along the -x axis. Its velocity, v, at x is given by $v^2 = 24 - 8x - 2x^2$. (i) Find all values of $x$ for which the particle is at rest. (ii) Find an expression for the acceleration of the particle, in terms of $x$. (iii) Find the maximum speed of the particle. (b) (i) Express $2 ext{cos}θ + 2 ext{cos}igg(θ + rac{π}{3}igg)$ in the form $R ext{cos}igg(θ + αigg)$, where $R > 0$ and $0 < α < rac{π}{2}$. (ii) Hence, or otherwise, solve $2 ext{cos}θ + 2 ext{cos}igg(θ + rac{π}{3}igg) = 3$, for $0 < θ < 2π$. (c) The diagram shows the parabola $x^2 = 4ay$. The point $P(2ap, ap^2)$, where $p ≠ 0$, is on the parabola. The tangent to the parabola at $P$, $y = px - aq^2$, meets the y-axis at $L$. The point $M$ is on the directrix, such that $PM$ is perpendicular to the directrix. Show that $SLMP$ is a rhombus.

Photo AI

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

Question 4

A particle is moving in simple harmonic motion along the -x axis. Its velocity, v, at x is given by $v^2 = 24 - 8x - 2x^2$. (i) Find all values of $x$ for which th... show full transcript

Worked Solution & Example Answer:A particle is moving in simple harmonic motion along the -x axis - HSC - SSCE Mathematics Extension 1 - Question 4 - 2010 - Paper 1

Step 1

Find all values of $x$ for which the particle is at rest.

96%

114 rated

Answer

To find the values of $x$ for which the particle is at rest, we need to solve for when its velocity is zero:

$v^2 = 24 - 8x - 2x^2 = 0$

Rearranging gives:

$2x^2 + 8x - 24 = 0$

Dividing by 2:

$x^2 + 4x - 12 = 0$

Using the quadratic formula, $x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}$ where $a = 1$ , $b = 4$ , and $c = -12$ :

$x = \frac{{-4 \pm \sqrt{{4^2 - 4(1)(-12)}}}}{2(1)} = \frac{{-4 \pm \sqrt{{16 + 48}}}}{2} = \frac{{-4 \pm \sqrt{64}}}{2} = \frac{{-4 \pm 8}}{2}$

This gives us the two solutions:

$x = 2 ext{ and } x = -6$ .

Step 2

Find an expression for the acceleration of the particle, in terms of $x$.

99%

104 rated

Answer

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:

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

Step 3

Find the maximum speed of the particle.

96%

101 rated

Answer

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:

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

Step 4

Express $2 ext{cos}θ + 2 ext{cos}igg(θ + rac{π}{3}igg)$ in the form $R ext{cos}igg(θ + αigg)$.

98%

120 rated

Answer

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 $0 < α < \frac{π}{2}$ .

Step 5

Hence, or otherwise, solve $2 ext{cos}θ + 2 ext{cos}igg(θ + rac{π}{3}igg) = 3$.

97%

117 rated

Answer

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 $0 < θ < 2π$ .

Step 6

Show that $SLMP$ is a rhombus.

97%

121 rated

Answer

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.

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

Worked Solution & Example Answer:A particle is moving in simple harmonic motion along the -x axis - HSC - SSCE Mathematics Extension 1 - Question 4 - 2010 - Paper 1

Find all values of $x$ for which the particle is at rest.

Find an expression for the acceleration of the particle, in terms of $x$.

Find the maximum speed of the particle.

Express $2 ext{cos}θ + 2 ext{cos}igg(θ + rac{π}{3}igg)$ in the form $R ext{cos}igg(θ + αigg)$.

Hence, or otherwise, solve $2 ext{cos}θ + 2 ext{cos}igg(θ + rac{π}{3}igg) = 3$.

Show that $SLMP$ is a rhombus.

Join the SSCE students using SimpleStudy...

Other SSCE Mathematics Extension 1 topics to explore

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

Worked Solution & Example Answer:A particle is moving in simple harmonic motion along the -x axis - HSC - SSCE Mathematics Extension 1 - Question 4 - 2010 - Paper 1

Find all values of $x$ for which the particle is at rest.

Find an expression for the acceleration of the particle, in terms of $x$.

Find the maximum speed of the particle.

Express $2 ext{cos}θ + 2 ext{cos}igg(θ + rac{π}{3}igg)$ in the form $R ext{cos}igg(θ + αigg)$.

Hence, or otherwise, solve $2 ext{cos}θ + 2 ext{cos}igg(θ + rac{π}{3}igg) = 3$.

Show that $SLMP$ is a rhombus.

Join the SSCE students using SimpleStudy...

Other SSCE Mathematics Extension 1 topics to explore

Express $2 ext{cos}θ + 2 ext{cos}igg(θ + rac{π}{3}igg)$ in the form $R ext{cos}igg(θ + αigg)$.

Hence, or otherwise, solve $2 ext{cos}θ + 2 ext{cos}igg(θ + rac{π}{3}igg) = 3$.