$A$ particle executes the motion described by $x(t) = x_0 (1 - e^{-\gamma t})$ for $t \geqslant 0$ and $x_0 > 0$.
$(a)$ Where does the particle start and with what velocity?
$(b)$ Find the maximum and minimum values of $x(t)$,$v(t)$,and $a(t)$. Show that $x(t)$ increases with time,$v(t)$ decreases with time,and $a(t)$ increases with time.

(N/A) Given,$x(t) = x_0 (1 - e^{-\gamma t})$.
Velocity $v(t) = \frac{dx}{dt} = x_0 \gamma e^{-\gamma t}$.
Acceleration $a(t) = \frac{dv}{dt} = -x_0 \gamma^2 e^{-\gamma t}$.
$(a)$ At $t = 0$,$x(0) = x_0(1 - e^0) = 0$. The particle starts at the origin.
Velocity at $t = 0$ is $v(0) = x_0 \gamma e^0 = x_0 \gamma$.
$(b)$
$x(t)$: As $t \to \infty$,$x(t) \to x_0$ (maximum). At $t = 0$,$x(t) = 0$ (minimum). Since $\frac{dx}{dt} > 0$,$x(t)$ increases with time.
$v(t)$: At $t = 0$,$v(0) = x_0 \gamma$ (maximum). As $t \to \infty$,$v(t) \to 0$ (minimum). Since $\frac{dv}{dt} < 0$,$v(t)$ decreases with time.
$a(t)$: At $t = 0$,$a(0) = -x_0 \gamma^2$ (minimum). As $t \to \infty$,$a(t) \to 0$ (maximum). Since $\frac{da}{dt} = x_0 \gamma^3 e^{-\gamma t} > 0$,$a(t)$ increases with time.