The velocity, $v\text{ ms}^{-1}$, of a particle moving in a straight line, $t$ seconds after passing a fixed point $O$, is given by $v = \frac{3}{t+1} - \frac{t+1}{3}$. It is given that the particle comes instantaneously at rest at point $A$.

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

$v = 0$
$\frac{3}{t+1} - \frac{t+1}{3} = 0$

2. Solve the resulting equation for $t$

$\frac{3}{t+1} = \frac{t+1}{3}$
$(t+1)^2 = 9$
$t+1 = \pm 3$
$t = 2 \quad \text{or} \quad t = -4$

3. Reject the invalid time value and state the final answer

Since $t \geqslant 0$, $t = -4$ is rejected.
$t = 2$

1. Find the time $t$ when the particle is at $A$ (from part a)

At $A$, $v = 0$.
$\frac{3}{t+1} - \frac{t+1}{3} = 0$
$\frac{3}{t+1} = \frac{t+1}{3}$
$(t+1)^2 = 9$
$t+1 = 3$ (since $t \ge 0$)
$t = 2$

2. Set up the definite integral for displacement

Distance $OA = \int_0^2 v \, dt$
$s = \int_0^2 \left( \frac{3}{t+1} - \frac{t+1}{3} \right) \, dt$

3. Integrate the velocity function

$\int \left( \frac{3}{t+1} - \frac{1}{3}(t+1) \right) \, dt = 3\ln(t+1) - \frac{1}{3} \frac{(t+1)^2}{2} = 3\ln(t+1) - \frac{(t+1)^2}{6}$

4. Evaluate the definite integral

$s = \left[ 3\ln(t+1) - \frac{(t+1)^2}{6} \right]_0^2$
$= \left( 3\ln 3 - \frac{9}{6} \right) - \left( 3\ln 1 - \frac{1}{6} \right)$
$= 3\ln 3 - \frac{3}{2} - 0 + \frac{1}{6}$
$= 3\ln 3 - \frac{8}{6}$
$= 3\ln 3 - \frac{4}{3} \approx 1.9626$

1. Find the expression for displacement $s$ in terms of $t$.

$s = \int v \, dt = \int \left( \frac{3}{t+1} - \frac{t+1}{3} \right) dt$
$s = 3 \ln(t+1) - \frac{(t+1)^2}{6} + c$

2. Find the constant of integration $c$.

When $t = 0$, $s = 0$ (since it passes through $O$ at $t=0$).
$0 = 3 \ln(1) - \frac{1^2}{6} + c$
$0 = 0 - \frac{1}{6} + c \implies c = \frac{1}{6}$
So, $s = 3 \ln(t+1) - \frac{(t+1)^2}{6} + \frac{1}{6}$.

3. Evaluate displacement at the start and end of the fifth second ($t=4$ and $t=5$).

At $t = 4$:
$s(4) = 3 \ln(5) - \frac{25}{6} + \frac{1}{6} = 3 \ln 5 - 4 \approx 4.828 - 4 = 0.828 > 0$

At $t = 5$:
$s(5) = 3 \ln(6) - \frac{36}{6} + \frac{1}{6} = 3 \ln 6 - \frac{35}{6} \approx 5.375 - 5.833 = -0.458 < 0$

4. Conclude that the particle passes through $O$ during the fifth second.

Since $s(4) > 0$ and $s(5) < 0$, and $s(t)$ is continuous, there must be some instant $t$ between 4 and 5 where $s(t) = 0$. Thus, the particle is again at $O$ during the fifth second.

1. Differentiate the velocity function to find the expression for acceleration

$v = 3(t+1)^{-1} - \frac{1}{3}(t+1)$

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

2. Substitute $t = 5$ to find the acceleration

When $t = 5$,
$a = -\frac{3}{(5+1)^2} - \frac{1}{3}$
$a = -\frac{3}{36} - \frac{1}{3}$
$a = -\frac{1}{12} - \frac{4}{12}$
$a = -\frac{5}{12}$