A particle moves in a straight line, so that $t$ seconds after passing through a fixed point $O$, its acceleration, $a\text{ ms}^{-2}$, is given by $a = -12e^{2t}$. The initial velocity is $9\text{ ms}^{-1}$.

1. Integrate the acceleration function to find the velocity function

$v = \int a \, dt$
$v = \int -12e^{2t} \, dt$
$v = -6e^{2t} + C$

2. Use the initial condition to find the constant of integration

Given initial velocity is $9\text{ ms}^{-1}$, so when $t = 0$, $v = 9$.
$9 = -6e^{2(0)} + C$
$9 = -6(1) + C$
$C = 15$
$v = 15 - 6e^{2t}$

3. Set velocity to zero to find the time when the particle is instantaneously at rest

When the particle is instantaneously at rest, $v = 0$.
$15 - 6e^{2t} = 0$
$6e^{2t} = 15$
$e^{2t} = \frac{15}{6}$

4. Solve for $t$ to show the required result

$2t = \ln\left(\frac{15}{6}\right)$
$t = \frac{1}{2}\ln\left(\frac{15}{6}\right)$ (shown)

1. Find the expression for velocity $v$ by integrating acceleration $a$.

$v = \int -12e^{2t} \, dt = -6e^{2t} + C$
Given $v = 9$ when $t = 0$:
$9 = -6e^{0} + C \implies C = 15$
$v = 15 - 6e^{2t}$

2. Find the expression for displacement $s$ by integrating velocity $v$.

$s = \int (15 - 6e^{2t}) \, dt = 15t - 3e^{2t} + D$
Since the particle starts at $O$, $s = 0$ when $t = 0$:
$0 = 15(0) - 3e^{0} + D \implies D = 3$
$s = 15t - 3e^{2t} + 3$

3. Calculate the maximum displacement from $O$.

The particle is instantaneously at rest at $t = \frac{1}{2}\ln\frac{15}{6} = \frac{1}{2}\ln 2.5$ (from part i).
Maximum displacement, $s_{max} = 15\left(\frac{1}{2}\ln 2.5\right) - 3e^{2\left(\frac{1}{2}\ln 2.5\right)} + 3$
$s_{max} = 7.5\ln 2.5 - 3(2.5) + 3$
$s_{max} = 7.5\ln 2.5 - 4.5$

4. Determine the total distance travelled when the particle returns to $O$.

The particle moves from $O$ to $s_{max}$ and then reverses direction to return to $O$.
Total distance $= 2 \times s_{max}$
$= 2(7.5\ln 2.5 - 4.5)$
$= 15\ln 2.5 - 9 \approx 4.74 \text{ m}$