一个质点在直线上运动，其速度 $v\text{ m/s}$ 由 $v = 9\sin\left(\frac{t}{2}\right)$ 给出，其中 $t$ 是通过固定点 $O$ 之后的秒数。

1. Set the velocity equation to zero to find when the particle is at instantaneous rest

$v = 9\sin\left(\frac{t}{2}\right) = 0$

2. Solve the trigonometric equation for the first positive value of t

$\sin\left(\frac{t}{2}\right) = 0$
$\frac{t}{2} = 0, \pi, 2\pi, \dots$
Since it is after leaving $O$ ($t > 0$), the first instantaneous rest occurs when $\frac{t}{2} = \pi$.

3. Calculate the final value of t

$t = 2\pi$

1. Identify the condition for maximum velocity

The maximum value of $v = 9\sin\left(\frac{t}{2}\right)$ occurs when $\sin\left(\frac{t}{2}\right) = 1$.

2. Solve for the first positive value of $t$

$\frac{t}{2} = \frac{\pi}{2}$
$t = \pi$

1. Identify the time when the particle changes direction

From part (a), the particle is at instantaneous rest when $v = 0$.
$9\sin\left(\frac{t}{2}\right) = 0$
$\frac{t}{2} = \pi \implies t = 2\pi \approx 6.283\text{ s}$
Since $2\pi < 10$, the particle changes direction at $t = 2\pi$.

2. Set up the integral for total distance

Total distance $= \int_0^{2\pi} v \, dt + \left| \int_{2\pi}^{10} v \, dt \right|$

3. Integrate the velocity function

$\int 9\sin\left(\frac{t}{2}\right) \, dt = \frac{9}{1/2}\left(-\cos\left(\frac{t}{2}\right)\right) = -18\cos\left(\frac{t}{2}\right)$

4. Evaluate the definite integrals to find the total distance

Distance from $t=0$ to $t=2\pi$:
$[-18\cos(t/2)]_0^{2\pi} = -18\cos(\pi) - (-18\cos(0)) = 18 - (-18) = 36$

Distance from $t=2\pi$ to $t=10$:
$\left| [-18\cos(t/2)]_{2\pi}^{10} \right| = \left| -18\cos(5) - (-18\cos(\pi)) \right|$
$= \left| -18\cos(5) - 18 \right| = 18\cos(5) + 18$

Total distance $= 36 + 18\cos(5) + 18 = 54 + 18\cos(5)$
$= 54 + 5.1059 \approx 59.1059$
Total distance $\approx 59.1\text{ m}$ (to 3 s.f.)

1. Integrate the velocity function to find the displacement $s$.

$s = \int v \, dt = \int 9\sin\left(\frac{t}{2}\right) \, dt$
$s = -18\cos\left(\frac{t}{2}\right) + C$

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

Since the particle is at $O$ when $t = 0$, $s = 0$ when $t = 0$.
$0 = -18\cos(0) + C$
$0 = -18(1) + C \implies C = 18$
$s = 18 - 18\cos\left(\frac{t}{2}\right)$

3. Set displacement to zero and solve for the next time $t$.

The particle passes through $O$ again when $s = 0$ and $t > 0$.
$18 - 18\cos\left(\frac{t}{2}\right) = 0$
$\cos\left(\frac{t}{2}\right) = 1$
$\frac{t}{2} = 2\pi$
$t = 4\pi \approx 12.6 \text{ s}$