A particle P moves along a horizontal straight line so that t seconds after motion has begun, its displacement, s m, from a fixed point O, is given by $s = t^3 + 5t^2 - 2t + 4.$

1. Differentiate the displacement function with respect to time to find the velocity

$v = \frac{ds}{dt} = \frac{d}{dt}(t^3 + 5t^2 - 2t + 4)$
$v = 3t^2 + 10t - 2$

1. Integrate acceleration to find the general expression for velocity

$v = \int a \, dt$
$v = \int (6t + 4) \, dt$
$v = 3t^2 + 4t + C$

2. Use the given condition to find the constant of integration for velocity

When $t = 1$, $v = 4$:
$4 = 3(1)^2 + 4(1) + C$
$4 = 7 + C$
$C = -3$
So, $v = 3t^2 + 4t - 3$.

3. Integrate velocity to find the general expression for displacement

$s = \int v \, dt$
$s = \int (3t^2 + 4t - 3) \, dt$
$s = t^3 + 2t^2 - 3t + D$

4. Use the initial displacement to find the constant of integration for displacement

Initial displacement is 6 m, so when $t = 0$, $s = 6$:
$6 = (0)^3 + 2(0)^2 - 3(0) + D$
$D = 6$
So, $s = t^3 + 2t^2 - 3t + 6$.

1. Equate the displacements of P and Q to find the time of collision

From earlier parts, $s_P = t^3 + 5t^2 - 2t + 4$ and $s_Q = t^3 + 2t^2 - 3t + 6$.
At collision, $s_P = s_Q$:
$t^3 + 5t^2 - 2t + 4 = t^3 + 2t^2 - 3t + 6$
$3t^2 + t - 2 = 0$

2. Solve the quadratic equation for $t$

$(3t - 2)(t + 1) = 0$
$t = \frac{2}{3}$ or $t = -1$
Since $t \ge 0$, $t = \frac{2}{3}$.

3. Evaluate the velocities of P and Q at $t = \frac{2}{3}$

From earlier parts, $v_P = 3t^2 + 10t - 2$ and $v_Q = 3t^2 + 4t - 3$.
When $t = \frac{2}{3}$:
$v_P = 3\left(\frac{2}{3}\right)^2 + 10\left(\frac{2}{3}\right) - 2 = \frac{4}{3} + \frac{20}{3} - 2 = 6 \text{ m/s}$
$v_Q = 3\left(\frac{2}{3}\right)^2 + 4\left(\frac{2}{3}\right) - 3 = \frac{4}{3} + \frac{8}{3} - 3 = 1 \text{ m/s}$

4. Determine if they are travelling in the same direction

Since both $v_P = 6 > 0$ and $v_Q = 1 > 0$, they are both travelling in the positive direction.
Therefore, they are travelling in the same direction.