一个质点在直线上运动，其离开固定点 $O$ 之后 $t$ 秒的速度 $v\text{ ms}^{-1}$ 由 $v = 10\left(3 - \mathrm{e}^{-\frac{1}{2}t}\right)$ 给出。

1. Find the value of $e^{-\frac{1}{2}t}$ when $v = 23$.

$23 = 10\left(3 - \mathrm{e}^{-\frac{1}{2}t}\right)$
$2.3 = 3 - \mathrm{e}^{-\frac{1}{2}t}$
$\mathrm{e}^{-\frac{1}{2}t} = 0.7$

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

$v = 30 - 10\mathrm{e}^{-\frac{1}{2}t}$
$a = \frac{\mathrm{d}v}{\mathrm{d}t} = -10\left(-\frac{1}{2}\right)\mathrm{e}^{-\frac{1}{2}t}$
$a = 5\mathrm{e}^{-\frac{1}{2}t}$

3. Calculate the acceleration when $v = 23$.

$a = 5(0.7)$
$a = 3.5$

1. Set up the integral for displacement

The displacement $s$ is the integral of velocity $v$ with respect to time $t$.
$s = \int_{0}^{5} 10\left(3 - \mathrm{e}^{-\frac{1}{2}t}\right) \, dt$

2. Integrate the velocity function

$s = \left[ 30t - 10\left(\frac{\mathrm{e}^{-\frac{1}{2}t}}{-\frac{1}{2}}\right) \right]_{0}^{5}$
$s = \left[ 30t + 20\mathrm{e}^{-\frac{1}{2}t} \right]_{0}^{5}$

3. Evaluate the definite integral and round to the nearest metre

$s = \left( 30(5) + 20\mathrm{e}^{-\frac{5}{2}} \right) - \left( 30(0) + 20\mathrm{e}^{0} \right)$
$s = 150 + 20\mathrm{e}^{-2.5} - 20$
$s = 130 + 20\mathrm{e}^{-2.5}$
$s \approx 131.6417$
To the nearest metre, $s = 132\text{ m}$.

1. Determine the limit of the exponential term as t approaches infinity

As $t \to \infty$, $\mathrm{e}^{-\frac{1}{2}t} \to 0$.

2. Calculate the limiting value of v

$v \to 10(3 - 0) = 30$