A particle moving in a straight line, passes through a fixed point $O$ with a velocity of $4\text{ m/s}$. Its acceleration $t$ seconds after passing through $O$, is $(6t - 8)\text{ m s}^{-2}$. Find

1. Integrate the acceleration function to find the velocity function

$v = \int a \, dt$
$v = \int (6t - 8) \, dt$
$v = 3t^2 - 8t + c$

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

When $t = 0$, $v = 4$.
$4 = 3(0)^2 - 8(0) + c \implies c = 4$
$v = 3t^2 - 8t + 4$

3. Find the time at which the minimum velocity occurs

For minimum velocity, $\frac{dv}{dt} = a = 0$.
$6t - 8 = 0$
$t = \frac{8}{6} = \frac{4}{3}$

4. Calculate the minimum velocity

Substitute $t = \frac{4}{3}$ into the velocity equation:
$v_{\text{min}} = 3\left(\frac{4}{3}\right)^2 - 8\left(\frac{4}{3}\right) + 4$
$v_{\text{min}} = 3\left(\frac{16}{9}\right) - \frac{32}{3} + 4$
$v_{\text{min}} = \frac{16}{3} - \frac{32}{3} + \frac{12}{3} = -\frac{4}{3}$

1. Find the expression for velocity $v(t)$ by integrating acceleration.

$v = \int (6t - 8) \, dt = 3t^2 - 8t + c$
Given $v = 4$ when $t = 0$, $c = 4$.
$\therefore v = 3t^2 - 8t + 4$

2. Set velocity to zero to find times of instantaneous rest.

For instantaneous rest, $v = 0$:
$3t^2 - 8t + 4 = 0$
$(3t - 2)(t - 2) = 0$
$t = \frac{2}{3}$ or $t = 2$

3. Select the time when the particle first comes to rest.

The particle first comes to rest at the smaller positive value of $t$.
$t = \frac{2}{3}$

1. Find the velocity function $v(t)$

$v(t) = \int (6t - 8) dt = 3t^2 - 8t + c$
Given $v(0) = 4$, $c = 4$.
$v(t) = 3t^2 - 8t + 4$

2. Determine if the particle changes direction during the $2^{\text{nd}}$ second ($t=1$ to $t=2$)

$v(t) = (3t - 2)(t - 2)$
The particle is at rest at $t = \frac{2}{3}$ and $t = 2$.
Since $v(t) \le 0$ for $1 \le t \le 2$, the particle does not change direction during the $2^{\text{nd}}$ second.

3. Calculate the distance travelled in the $2^{\text{nd}}$ second

Distance $= \left| \int_{1}^{2} (3t^2 - 8t + 4) dt \right|$
$= \left| \left[ t^3 - 4t^2 + 4t \right]_{1}^{2} \right|$
$= \left| (2^3 - 4(2)^2 + 4(2)) - (1^3 - 4(1)^2 + 4(1)) \right|$
$= \left| (8 - 16 + 8) - (1 - 4 + 4) \right|$
$= \left| 0 - 1 \right| = 1\text{ m}$