一个质点从点 P 向点 O 运动。在通过点 P 之后 $t$ 秒，质点的速度可以用公式 $v = (k - 2t)^n$ 来建模，其中 $k$ 和 $n$ 是常数。 当 $t = 3.5$ 时，质点的速度为 27 m/s；当 $t = 8$ 时，质点在点 O 右侧 4.8 米处瞬时静止。

1. Differentiate the velocity function with respect to time to find acceleration

$a = \frac{dv}{dt} = n(k - 2t)^{n-1} \cdot \frac{d}{dt}(k - 2t)$
$a = -2n(k - 2t)^{n-1}$

1. Use the condition for instantaneous rest to find $k$.

The particle comes to instantaneous rest when $t = 8$, so $v = 0$.
$(k - 2(8))^n = 0$
$k - 16 = 0$
$k = 16$

2. Use the given speed at $t = 3.5$ to form an equation for $n$.

When $t = 3.5$, $v = 27$. Substituting $k = 16$:
$27 = (16 - 2(3.5))^n$
$27 = (16 - 7)^n$
$27 = 9^n$

3. Solve the exponential equation to find $n$.

$3^3 = (3^2)^n$
$3^3 = 3^{2n}$
$2n = 3$
$n = 1.5$

1. State the velocity function using the values of $k$ and $n$ from part (b)

From part (b), $k = 16$ and $n = 1.5$.
$v = (16 - 2t)^{1.5}$

2. Set up the definite integral for the total distance travelled

Since $v \ge 0$ for $0 \le t \le 8$, the particle does not change direction.
$\text{Distance} = \int_{0}^{8} (16 - 2t)^{1.5} \, dt$

3. Integrate the velocity function

$\int (16 - 2t)^{1.5} \, dt = \frac{(16 - 2t)^{2.5}}{2.5 \times (-2)} = -\frac{(16 - 2t)^{2.5}}{5}$

4. Evaluate the definite integral

$\text{Distance} = \left[ -\frac{(16 - 2t)^{2.5}}{5} \right]_{0}^{8}$
$= \left( -\frac{0}{5} \right) - \left( -\frac{16^{2.5}}{5} \right)$
$= 0 - \left( -\frac{1024}{5} \right) = 204.8$

1. Determine the expression for $v$ using the values of $k$ and $n$ from part (b) and evaluate its validity for $t > 8$.

From part (b), $k = 16$ and $n = 1.5$, so $v = (16 - 2t)^{1.5}$.
For $t > 8$, $16 - 2t < 0$.

2. Conclude whether the model can continue to describe the motion.

Since $16 - 2t$ is negative for $t > 8$, $(16 - 2t)^{1.5}$ involves the square root of a negative number, which is undefined. Therefore, the model cannot continue to describe the motion.