Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 1 - Question 12 - 2014 - Paper 1

At rest implies that the velocity is zero. The velocity is given by the derivative of the displacement:

$v = \frac{dx}{dt} = 2 \cdot 3\cos{3t} = 6\cos{3t}.$

Setting this to zero gives:

$6\cos{3t} = 0 \implies \cos{3t} = 0 \implies 3t = \frac{(2n + 1)\pi}{2}$

For $n=0$, we have:

$t = \frac{\pi}{6}.$

Now, we evaluate the acceleration, which is the second derivative of displacement:

$a = \frac{d^2x}{dt^2} = -18\sin{3t}.$

Substituting $t = \frac{\pi}{6}$ gives:

$$a = -18\sin{\left(3 \cdot \frac{\pi}{6}\right)} = -18 \cdot 1 = -18 ext{ m/s}^2.$

To find the volume of the solid obtained by rotating the area bounded by $y = ext{cos } 4x$, the x-axis, and the lines $x = 0$ and $x = \frac{\pi}{4}$ around the x-axis, we use the formula:

$V = \pi \int_{0}^{\frac{\pi}{4}} (y^2) \, dx = \pi \int_{0}^{\frac{\pi}{4}} (\text{cos}^2(4x)) \, dx.$

Utilizing the double angle formula $ext{cos}^2 \theta = \frac{1 + \text{cos}(2\theta)}{2}$, we get:

$V = \pi \int_{0}^{\frac{\pi}{4}} \frac{1 + \text{cos}(8x)}{2} \, dx.$

Computing this integral gives:

$$= \frac{\pi}{2} \left[ x + \frac{\text{sin}(8x)}{8} \right]_{0}^{\frac{\pi}{4}} = \frac{\pi}{2} \left[ \frac{\pi}{4} + 0 - (0 + 0) \right] = \frac{\pi^2}{8}.$

The acceleration of the particle is given by

$\frac{d^2x}{dt^2} = 2 - \frac{x}{2}.$

We can express the velocity as

$v = \frac{dx}{dt}.$

To find $v^2$ in terms of $x$, we can relate acceleration and velocity by using

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

Substituting the acceleration expression gives:

$v\frac{dv}{dx} = 2 - \frac{x}{2}.$

This can be rearranged and integrated to yield:

$\int v \, dv = \int (2 - \frac{x}{2}) \, dx.$

Evaluating both sides leads to the relationship between $v^2$ and $x$ and applying the boundary conditions will yield the final expression.