一个质点在直线上运动，其速度 $v$ m/s 由 $v = t^2 - 4t + 3$ 给出，其中 $t$ 是通过固定点 $O$ 后的时间（以秒为单位）。

1. Find the expression for acceleration by differentiating the velocity function

$a = \frac{dv}{dt}$
$a = \frac{d}{dt}(t^2 - 4t + 3)$
$a = 2t - 4$

2. Substitute $t = 1$ to find the acceleration

When $t = 1$,
$a = 2(1) - 4$
$a = -2$

1. Set velocity to zero for instantaneous rest

$v = 0$
$t^2 - 4t + 3 = 0$

2. Solve the quadratic equation for $t$

$(t - 1)(t - 3) = 0$
$t = 1 \text{ or } t = 3$

1. Find the expression for displacement by integrating velocity

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

2. Determine the constant of integration

Since the particle passes a fixed point $O$ at $t = 0$, $s = 0$ when $t = 0$.
$0 = \frac{0^3}{3} - 2(0)^2 + 3(0) + c \implies c = 0$
$s = \frac{t^3}{3} - 2t^2 + 3t$

3. Calculate the displacements at the times when the particle is at rest

From part (b), the particle is at rest when $t = 1$ and $t = 3$.
When $t = 1$:
$s = \frac{1^3}{3} - 2(1)^2 + 3(1) = \frac{1}{3} - 2 + 3 = \frac{4}{3} \text{ m}$
When $t = 3$:
$s = \frac{3^3}{3} - 2(3)^2 + 3(3) = 9 - 18 + 9 = 0 \text{ m}$