一个质点在直线上运动，使得在离开固定点 O 后 $t$ 秒时，其速度 $v$ m/s 由 $v = 16\bigl(2e^{-2t} - e^{-t}\bigr)$ 给出。求

1. Set velocity to zero to find the time when the particle is at its greatest distance

For greatest distance, the particle must be instantaneously at rest, so $v = 0$.
$16(2e^{-2t} - e^{-t}) = 0$

2. Solve the exponential equation for $t$

$2e^{-2t} = e^{-t}$
Divide both sides by $e^{-2t}$:
$2 = e^{t}$
$t = \ln 2$

1. Integrate the velocity function to find the displacement function $s(t)$.

$s = \int 16(2e^{-2t} - e^{-t}) \, dt$
$s = 16\left(\frac{2e^{-2t}}{-2} - \frac{e^{-t}}{-1}\right) + C$
$s = 16(-e^{-2t} + e^{-t}) + C$

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

Since the particle leaves $O$ at $t = 0$, $s = 0$ when $t = 0$.
$0 = 16(-e^{0} + e^{0}) + C$
$0 = 16(-1 + 1) + C$
$C = 0$
So, $s = 16(e^{-t} - e^{-2t})$.

3. Substitute the time of greatest distance (from part a) to find the maximum displacement.

From part (a), the greatest distance occurs when $v = 0$, which gives $e^{-t} = \frac{1}{2}$ (or $t = \ln 2$).
Substitute $e^{-t} = \frac{1}{2}$ into the displacement equation:
$s = 16\left(\frac{1}{2} - \left(\frac{1}{2}\right)^2\right)$
$s = 16\left(\frac{1}{2} - \frac{1}{4}\right)$
$s = 16\left(\frac{1}{4}\right) = 4$

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

$v = 32e^{-2t} - 16e^{-t}$

$\frac{dv}{dt} = 32(-2e^{-2t}) - 16(-e^{-t})$
$\frac{dv}{dt} = -64e^{-2t} + 16e^{-t}$

2. Set the derivative to zero to find the time at which velocity is stationary.

For minimum velocity, $\frac{dv}{dt} = 0$:
$-64e^{-2t} + 16e^{-t} = 0$
$16e^{-t}(1 - 4e^{-t}) = 0$

Since $e^{-t} > 0$ for all $t$, we have:
$1 - 4e^{-t} = 0$
$e^{-t} = \frac{1}{4}$

3. Verify that this stationary point corresponds to a minimum.

$\frac{d^2v}{dt^2} = -64(-2e^{-2t}) + 16(-e^{-t}) = 128e^{-2t} - 16e^{-t}$

Substitute $e^{-t} = \frac{1}{4}$:
$\frac{d^2v}{dt^2} = 128\left(\frac{1}{4}\right)^2 - 16\left(\frac{1}{4}\right) = 128\left(\frac{1}{16}\right) - 4 = 8 - 4 = 4$

Since $\frac{d^2v}{dt^2} > 0$, the velocity is a minimum.

4. Calculate the minimum velocity.

Substitute $e^{-t} = \frac{1}{4}$ into $v$:
$v = 16\left[2\left(\frac{1}{4}\right)^2 - \frac{1}{4}\right]$
$v = 16\left[2\left(\frac{1}{16}\right) - \frac{1}{4}\right]$
$v = 16\left(\frac{1}{8} - \frac{2}{8}\right)$
$v = 16\left(-\frac{1}{8}\right) = -2$