一质点从点 O 出发并在一条直线上运动，其速度 $v$ m/s 由 $v = 5 - e^{3-2t}$ 给出，其中 $t$ 是离开 O 后的时间（以秒为单位）。

1. Set velocity to zero for instantaneous rest

For the particle to be instantaneously at rest, $v = 0$.
$5 - e^{3-2t} = 0$
$e^{3-2t} = 5$

2. Solve the exponential equation using natural logarithms

Take the natural logarithm ($\ln$) of both sides:
$3 - 2t = \ln 5$
$2t = 3 - \ln 5$

3. Calculate the final value of $t$

$t = \frac{3 - \ln 5}{2}$
$t \approx 0.69528$
$t = 0.695 \text{ (to 3 s.f.)}$

1. Identify the time when the particle is at rest (from part a) to split the distance calculation.

From part (a), $v = 0$ when $5 - e^{3-2t} = 0$.
$e^{3-2t} = 5 \implies 3 - 2t = \ln 5 \implies t = \frac{3 - \ln 5}{2} \approx 0.69534$ s.
Since $0 < 0.69534 < 1$, the particle changes direction within the first second.

2. Find the displacement function $s(t)$ by integrating the velocity.

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

3. Evaluate the displacement at $t = 0$, $t = \frac{3 - \ln 5}{2}$, and $t = 1$.

Let $t_1 = \frac{3 - \ln 5}{2}$.
$s(0) = 5(0) + \frac{1}{2}e^3 = \frac{1}{2}e^3 \approx 10.0428$
$s(t_1) = 5\left(\frac{3 - \ln 5}{2}\right) + \frac{1}{2}e^{3-2t_1} = 7.5 - 2.5\ln 5 + \frac{1}{2}(5) = 10 - 2.5\ln 5 \approx 5.9764$
$s(1) = 5(1) + \frac{1}{2}e^{3-2(1)} = 5 + \frac{1}{2}e \approx 6.3591$

4. Calculate the total distance travelled.

Distance $= |s(t_1) - s(0)| + |s(1) - s(t_1)|$
$= |5.9764 - 10.0428| + |6.3591 - 5.9764|$
$= 4.0664 + 0.3827$
$= 4.4491 \approx 4.45$ m (to 3 s.f.)

1. Identify the range of the exponential term

For all real values of $t$, the exponential function is always positive, so $e^{3-2t} > 0$.

2. Deduce the upper bound for the velocity

Since we are subtracting a positive value from 5, $v = 5 - e^{3-2t} < 5$. Thus, the velocity cannot exceed 5 m/s.