电脑屏幕上的一个光点沿直线运动并经过定点 $A$。在经过 $A$ 之后 $t$ 秒内它所运动的距离 $d$ 米由以下公式给出：
$d = t^3 - 6t^2 + 9t \text{ for } t \ge 0.$

1. Find the expression for velocity by differentiating distance with respect to time

$v = \frac{dd}{dt} = \frac{d}{dt}(t^3 - 6t^2 + 9t)$
$v = 3t^2 - 12t + 9$

2. Calculate the speed at t = 10

When $t = 10$,
$v = 3(10)^2 - 12(10) + 9$
$v = 300 - 120 + 9 = 189$
Speed is $|v| = 189 \text{ m/s}$.

3. Find the expression for acceleration by differentiating velocity with respect to time

$a = \frac{dv}{dt} = \frac{d}{dt}(3t^2 - 12t + 9)$
$a = 6t - 12$

4. Calculate the acceleration at t = 10

When $t = 10$,
$a = 6(10) - 12$
$a = 60 - 12 = 48 \text{ m/s}^2$

1. Find the expression for velocity by differentiating distance with respect to time

$v = \frac{dd}{dt} = 3t^2 - 12t + 9$

2. Set velocity to zero to find when the dot is instantaneously at rest and solve for $t$

$3t^2 - 12t + 9 = 0$
$t^2 - 4t + 3 = 0$
$(t - 1)(t - 3) = 0$
$t = 1 \text{ or } t = 3$

1. Identify the times when the dot changes direction (from part ii)

From part (ii), the dot is instantaneously at rest when $v = 3t^2 - 12t + 9 = 0$, which gives $t = 1$ and $t = 3$. The dot changes direction at these times.

2. Calculate the displacement of the dot at $t = 0, 1, 3, 4$

Substitute $t = 0, 1, 3, 4$ into $d = t^3 - 6t^2 + 9t$:
$d(0) = 0$
$d(1) = 1^3 - 6(1)^2 + 9(1) = 4$
$d(3) = 3^3 - 6(3)^2 + 9(3) = 0$
$d(4) = 4^3 - 6(4)^2 + 9(4) = 4$

3. Calculate the total distance travelled by summing the distances between turning points

Distance from $t=0$ to $t=1$: $|4 - 0| = 4\text{ m}$
Distance from $t=1$ to $t=3$: $|0 - 4| = 4\text{ m}$
Distance from $t=3$ to $t=4$: $|4 - 0| = 4\text{ m}$
Total distance $= 4 + 4 + 4 = 12\text{ m}$