A particle passes a fixed point $O$ and moves in a straight line such that, $t$ s after leaving $O$, its velocity, $v\text{ m/s}$, is given by $v = \frac{3}{t+2} - \frac{t+2}{3}$. The particle reaches its greatest distance from $O$ at $P$.

1. Set velocity to zero to find the time when the particle reaches its greatest distance

At the greatest distance from $O$, the particle is instantaneously at rest, so $v = 0$.
$\frac{3}{t+2} - \frac{t+2}{3} = 0$

2. Solve the equation for $t$

$\frac{3}{t+2} = \frac{t+2}{3}$
$(t+2)^2 = 9$
$t+2 = 3 \quad \text{or} \quad t+2 = -3$
$t = 1 \quad \text{or} \quad t = -5$
Since $t \geqslant 0$, $t = 1$.

1. Find the time when the particle changes direction (from part a).

The particle changes direction when $v = 0$.
$\frac{3}{t+2} - \frac{t+2}{3} = 0$
$(t+2)^2 = 9$
$t+2 = 3 \implies t = 1$

2. Integrate velocity to find the displacement function $s(t)$.

$s(t) = \int \left( \frac{3}{t+2} - \frac{t+2}{3} \right) dt$
$s(t) = 3 \ln(t+2) - \frac{(t+2)^2}{6} + C$

3. Find the constant of integration $C$.

Since the particle passes $O$ at $t = 0$, $s(0) = 0$.
$0 = 3 \ln(2) - \frac{2^2}{6} + C$
$C = \frac{2}{3} - 3 \ln 2$

4. Calculate the displacement at $t=1$ and $t=2$.

$s(1) = 3 \ln(3) - \frac{3^2}{6} + \frac{2}{3} - 3 \ln 2 = 3 \ln\left(\frac{3}{2}\right) - \frac{5}{6} \approx 0.38306$
$s(2) = 3 \ln(4) - \frac{4^2}{6} + \frac{2}{3} - 3 \ln 2 = 3 \ln 2 - 2 \approx 0.07944$

5. Calculate the total distance travelled.

Since the particle changes direction at $t=1$, the total distance is the distance from $t=0$ to $t=1$ plus the distance from $t=1$ to $t=2$.
$\text{Distance} = |s(1) - s(0)| + |s(2) - s(1)|$
$\text{Distance} = s(1) + (s(1) - s(2)) = 2s(1) - s(2)$
$\text{Distance} = 2(0.38306) - 0.07944 = 0.68668 \approx 0.687\text{ m}$

1. Differentiate the velocity function to find the acceleration $a$.

$v = 3(t+2)^{-1} - \frac{1}{3}(t+2)$
$a = \frac{dv}{dt} = -3(t+2)^{-2} - \frac{1}{3}$
$a = -\frac{3}{(t+2)^2} - \frac{1}{3}$

2. Evaluate the acceleration at $t = 1$ (the time at point $P$ from part a) to show it is negative.

From part (a), at point $P$, $t = 1$.
Substitute $t = 1$ into $a$:
$a = -\frac{3}{(1+2)^2} - \frac{1}{3}$
$a = -\frac{3}{9} - \frac{1}{3} = -\frac{2}{3}$
Since $a < 0$, the velocity of the particle is decreasing at point $P$.