一质点从 $A$ 点由静止开始运动，并在 $B$ 点停止。它从 $A$ 运动到 $B$ 的速度（单位：m/s）由方程 $v = 10t - \frac{1}{2}t^2$ 给出，其中 $t$ 是从 $A$ 开始计时的秒数。
证明该质点的速度不低于 5 m/s 的持续时间为 $6\sqrt{10}$ 秒。

1. Set up the inequality for the speed being 5 m/s or more

$10t - \frac{1}{2}t^2 \ge 5$

2. Rearrange the inequality into standard form

$20t - t^2 \ge 10$
$t^2 - 20t + 10 \le 0$

3. Find the roots of the corresponding quadratic equation

Solving $t^2 - 20t + 10 = 0$ using the quadratic formula:
$t = \frac{-(-20) \pm \sqrt{(-20)^2 - 4(1)(10)}}{2(1)}$
$t = \frac{20 \pm \sqrt{400 - 40}}{2}$
$t = \frac{20 \pm \sqrt{360}}{2}$
$t = \frac{20 \pm 6\sqrt{10}}{2}$
$t = 10 \pm 3\sqrt{10}$

4. Determine the time interval and calculate the duration

The inequality $t^2 - 20t + 10 \le 0$ is satisfied for:
$10 - 3\sqrt{10} \le t \le 10 + 3\sqrt{10}$
The duration for which the speed is 5 m/s or more is the difference between these two times:
$\text{Duration} = (10 + 3\sqrt{10}) - (10 - 3\sqrt{10})$
$\text{Duration} = 6\sqrt{10} \text{ s (shown)}$