一粒子沿直线运动，在通过一固定点 $O$ 时的速度为 3 m/s。粒子在通过 $O$ 点 $t$ 秒后的加速度 $a$ m/s$^2$ 由 $a = -e^{-0.3t}$ 给出。该粒子在点 $P$ 处瞬时静止。

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

$v = \int a \, dt = \int -e^{-0.3t} \, dt$
$v = -\frac{e^{-0.3t}}{-0.3} + c$
$v = \frac{10}{3}e^{-0.3t} + c$

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

When $t = 0$, $v = 3$.
$3 = \frac{10}{3}e^{0} + c$
$c = 3 - \frac{10}{3} = -\frac{1}{3}$
So, $v = \frac{10}{3}e^{-0.3t} - \frac{1}{3}$

3. Set velocity to zero to find the time when the particle is at instantaneous rest.

At instantaneous rest, $v = 0$.
$\frac{10}{3}e^{-0.3t} - \frac{1}{3} = 0$
$\frac{10}{3}e^{-0.3t} = \frac{1}{3}$
$10e^{-0.3t} = 1$
$e^{-0.3t} = \frac{1}{10}$

4. Solve for $t$ using logarithms.

Take natural logarithm on both sides:
$-0.3t = \ln\left(\frac{1}{10}\right)$
$-0.3t = -\ln 10$
$t = \frac{\ln 10}{0.3} = \frac{10}{3} \ln 10$ (Shown)

1. Find the expression for velocity $v$ by integrating acceleration $a$.

$v = \int a \, dt = \int -e^{-0.3t} \, dt$
$v = \frac{-1}{-0.3}e^{-0.3t} + C = \frac{10}{3}e^{-0.3t} + C$
Given $v = 3$ when $t = 0$:
$3 = \frac{10}{3}e^{0} + C \implies C = 3 - \frac{10}{3} = -\frac{1}{3}$
$v = \frac{10}{3}e^{-0.3t} - \frac{1}{3}$

2. Set up the definite integral for displacement from $O$ to $P$.

Since the particle moves in one direction until it comes to rest at $P$, the distance $OP$ is the displacement at $t = \frac{10}{3} \ln 10$.
$s = \int_0^{\frac{10}{3} \ln 10} \left( \frac{10}{3}e^{-0.3t} - \frac{1}{3} \right) \, dt$

3. Evaluate the definite integral to find the distance.

$s = \left[ \frac{10/3}{-0.3}e^{-0.3t} - \frac{1}{3}t \right]_0^{\frac{10}{3} \ln 10}$
$s = \left[ -\frac{100}{9}e^{-0.3t} - \frac{1}{3}t \right]_0^{\frac{10}{3} \ln 10}$
At $t = \frac{10}{3} \ln 10$, $-0.3t = -\ln 10$, so $e^{-0.3t} = e^{-\ln 10} = \frac{1}{10}$.
Upper limit: $-\frac{100}{9}\left(\frac{1}{10}\right) - \frac{1}{3}\left(\frac{10}{3} \ln 10\right) = -\frac{10}{9} - \frac{10}{9} \ln 10$
Lower limit ($t=0$): $-\frac{100}{9}(1) - 0 = -\frac{100}{9}$
$s = \left( -\frac{10}{9} - \frac{10}{9} \ln 10 \right) - \left( -\frac{100}{9} \right)$
$s = \frac{90}{9} - \frac{10}{9} \ln 10 = 10 - \frac{10}{9} \ln 10 \approx 7.44$ m

1. Find the expression for velocity $v$ in terms of $t$.

$v = \int -e^{-0.3t} \, dt = \frac{10}{3}e^{-0.3t} + C$
When $t = 0$, $v = 3$:
$3 = \frac{10}{3}e^0 + C \implies C = -\frac{1}{3}$
$v = \frac{10}{3}e^{-0.3t} - \frac{1}{3}$

2. Find the expression for displacement $s$ in terms of $t$.

$s = \int \left(\frac{10}{3}e^{-0.3t} - \frac{1}{3}\right) \, dt = -\frac{100}{9}e^{-0.3t} - \frac{1}{3}t + D$
When $t = 0$, $s = 0$:
$0 = -\frac{100}{9}e^0 - 0 + D \implies D = \frac{100}{9}$
$s = \frac{100}{9} - \frac{100}{9}e^{-0.3t} - \frac{1}{3}t$

3. Evaluate the displacement at $t = 33$ and $t = 34$.

At $t = 33$:
$s(33) = \frac{100}{9} - \frac{100}{9}e^{-9.9} - \frac{33}{3} \approx 0.1105 > 0$

At $t = 34$:
$s(34) = \frac{100}{9} - \frac{100}{9}e^{-10.2} - \frac{34}{3} \approx -0.2226 < 0$

4. Conclude using the sign change.

Since $s(33) > 0$ and $s(34) < 0$, and $s(t)$ is continuous, there must be a time $t$ between 33 and 34 where $s(t) = 0$. Therefore, the particle is at $O$ during the thirty-fourth second.