一质点沿直线运动，其初始速度为 21 cm/s。在 $t$ 秒后，其加速度 $a$ cm/s$^2$ 由 $a = 2t - 10$ 给出。当 $t = 0$ 时，它相对于固定点 $O$ 的位移 $s$ cm 为 4 cm。 求该质点在头 7 秒内运动的路程。

1. Integrate acceleration to find the velocity function $v(t)$.

$v = \int a \, dt$
$v = \int (2t - 10) \, dt$
$v = t^2 - 10t + C$

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

Given initial velocity is $21$ cm/s, when $t = 0$, $v = 21$.
$21 = (0)^2 - 10(0) + C$
$C = 21$
$v = t^2 - 10t + 21$

3. Find the times when the particle is instantaneously at rest.

Set $v = 0$:
$t^2 - 10t + 21 = 0$
$(t - 3)(t - 7) = 0$
$t = 3 \text{ or } t = 7$

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

$s = \int v \, dt$
$s = \int (t^2 - 10t + 21) \, dt$
$s = \frac{1}{3}t^3 - 5t^2 + 21t + D$

5. Use the initial condition to find the constant of integration for displacement.

When $t = 0$, $s = 4$.
$4 = \frac{1}{3}(0)^3 - 5(0)^2 + 21(0) + D$
$D = 4$
$s = \frac{1}{3}t^3 - 5t^2 + 21t + 4$

6. Calculate the displacement at the key times $t=0$, $t=3$, and $t=7$.

At $t = 0$, $s = 4$
At $t = 3$, $s = \frac{1}{3}(3)^3 - 5(3)^2 + 21(3) + 4 = 9 - 45 + 63 + 4 = 31$
At $t = 7$, $s = \frac{1}{3}(7)^3 - 5(7)^2 + 21(7) + 4 = \frac{343}{3} - 245 + 147 + 4 = \frac{343}{3} - 94 = \frac{61}{3} = 20\frac{1}{3}$

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

Distance from $t=0$ to $t=3$: $|31 - 4| = 27$
Distance from $t=3$ to $t=7$: $|20\frac{1}{3} - 31| = |-10\frac{2}{3}| = 10\frac{2}{3}$
Total distance $= 27 + 10\frac{2}{3} = 37\frac{2}{3}$ cm