质点 P 沿直线运动并经过定点 O，其速度 $v$ m/s 由 $v = t^2 - 5t + 4$ 给出，其中 $t$ 是通过 O 点后的时间（以秒为单位）。

1. Set velocity to zero for instantaneous rest

$v = 0$
$t^2 - 5t + 4 = 0$

2. Solve the quadratic equation for $t$

$(t - 1)(t - 4) = 0$
$t = 1 \text{ or } t = 4$

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

$s = \int v \, dt = \int (t^2 - 5t + 4) \, dt$
$s = \frac{t^3}{3} - \frac{5t^2}{2} + 4t + C$

2. Determine the constant of integration using the initial condition.

Since the particle passes through O at $t = 0$, $s = 0$ when $t = 0$.
$0 = 0 - 0 + 0 + C \implies C = 0$
$s = \frac{t^3}{3} - \frac{5t^2}{2} + 4t$

3. Set displacement to zero to find when the particle returns to O.

For the particle to return to O, $s = 0$ and $t > 0$.
$\frac{t^3}{3} - \frac{5t^2}{2} + 4t = 0$
$t \left( \frac{t^2}{3} - \frac{5t}{2} + 4 \right) = 0$

4. Solve the resulting quadratic equation for $t$.

Since $t > 0$, we solve $\frac{t^2}{3} - \frac{5t}{2} + 4 = 0$.
Multiply by 6:
$2t^2 - 15t + 24 = 0$
Using the quadratic formula:
$t = \frac{-(-15) \pm \sqrt{(-15)^2 - 4(2)(24)}}{2(2)}$
$t = \frac{15 \pm \sqrt{225 - 192}}{4}$
$t = \frac{15 \pm \sqrt{33}}{4}$
$t \approx 5.1861$ or $t \approx 2.3139$
$t = 5.19$ or $t = 2.31$ (to 3 s.f.)

1. Find the time $t$ when the particle first returns to O (from part b)

Displacement $s = \int (t^2 - 5t + 4) \, dt = \frac{t^3}{3} - \frac{5t^2}{2} + 4t$.
When $s = 0$ and $t > 0$, $2t^2 - 15t + 24 = 0$.
$t = \frac{15 \pm \sqrt{15^2 - 4(2)(24)}}{4} = \frac{15 \pm \sqrt{33}}{4}$.
The particle first returns to O at $t = \frac{15 - \sqrt{33}}{4}$.

2. Find the acceleration function and evaluate it at the found time

Acceleration $a = \frac{dv}{dt} = 2t - 5$.
When $t = \frac{15 - \sqrt{33}}{4}$,
$a = 2\left(\frac{15 - \sqrt{33}}{4}\right) - 5 = \frac{15 - \sqrt{33}}{2} - \frac{10}{2} = \frac{5 - \sqrt{33}}{2} \approx -0.372$ m/s$^2$.

1. Identify the intervals where the particle moves in different directions

From part (a), the particle is at rest at $t = 1$ and $t = 4$.
For $0 \le t < 1$, $v > 0$.
For $1 < t < 4$, $v < 0$.
For $4 < t \le 5$, $v > 0$.

2. Find the displacement function by integrating velocity

$s = \int (t^2 - 5t + 4) \, dt = \frac{t^3}{3} - \frac{5t^2}{2} + 4t + c$
Since $s = 0$ when $t = 0$, $c = 0$.
$s(t) = \frac{t^3}{3} - \frac{5t^2}{2} + 4t$

3. Calculate the total distance travelled in the first 5 seconds

Distance $D_1$ for $0 \le t \le 1$:
$s(1) = \frac{1}{3} - \frac{5}{2} + 4 = \frac{11}{6}$
$D_1 = |s(1) - s(0)| = \frac{11}{6}$

Distance $D_2$ for $1 \le t \le 4$:
$s(4) = \frac{64}{3} - \frac{80}{2} + 16 = -\frac{8}{3}$
$D_2 = |s(4) - s(1)| = |-\frac{8}{3} - \frac{11}{6}| = |-\frac{27}{6}| = \frac{9}{2}$

Distance $D_3$ for $4 \le t \le 5$:
$s(5) = \frac{125}{3} - \frac{125}{2} + 20 = -\frac{5}{6}$
$D_3 = |s(5) - s(4)| = |-\frac{5}{6} - (-\frac{8}{3})| = \frac{11}{6}$

Total distance $= \frac{11}{6} + \frac{27}{6} + \frac{11}{6} = \frac{49}{6}\text{ m}$

4. Calculate the average speed

Average speed $= \frac{\text{Total distance}}{\text{Total time}} = \frac{\frac{49}{6}}{5} = \frac{49}{30}\text{ m/s}$