Q.A particle executes the motion described by $x(t) = x_0\left(1 - e^{-\gamma t}\right)$; $t \ge 0$, $x_0 > 0$.

The key idea here is Kinematics Vector Differentiation, where position, velocity, and acceleration are related through differentiation with respect to time.

First, we find the velocity $v(t)$ and acceleration $a(t)$ functions by differentiating $x(t)$:

$x(t) = x_0\left(1 - e^{-\gamma t}\right)$

$v(t) = \frac{dx}{dt} = \frac{d}{dt}\left[x_0(1 - e^{-\gamma t})\right] = x_0(0 - (-\gamma)e^{-\gamma t}) = x_0\gamma e^{-\gamma t}$

$a(t) = \frac{dv}{dt} = \frac{d}{dt}\left[x_0\gamma e^{-\gamma t}\right] = x_0\gamma(-\gamma)e^{-\gamma t} = -x_0\gamma^2 e^{-\gamma t}$

We assume $\gamma > 0$ for the exponential term to decay.

(a) Where the particle starts and with what velocity:

At $t=0$:

$x(0) = x_0(1 - e^0) = x_0(1 - 1) = 0$

$v(0) = x_0\gamma e^0 = x_0\gamma$

(b) Maximum and minimum values of $x(t)$, $v(t)$, $a(t)$ and their monotonicity:

• For $x(t)$:

  • $x(0) = 0$. As $t \to \infty$, $e^{-\gamma t} \to 0$, so $\lim_{t \to \infty} x(t) = x_0(1 - 0) = x_0$.
  • Since $v(t) = x_0\gamma e^{-\gamma t} > 0$ for all $t \ge 0$ (given $x_0 > 0, \gamma > 0$), $x(t)$ is strictly increasing with time.
  • Minimum value of $x(t)$ is $0$ (at $t=0$). Maximum value of $x(t)$ is $x_0$ (as $t \to \infty$).

• For $v(t)$:

  • $v(0) = x_0\gamma$. As $t \to \infty$, $e^{-\gamma t} \to 0$, so $\lim_{t \to \infty} v(t) = 0$.
  • Since $a(t) = -x_0\gamma^2 e^{-\gamma t} < 0$ for all $t \ge 0$, $v(t)$ is strictly decreasing with time.
  • Maximum value of $v(t)$ is $x_0\gamma$ (at $t=0$). Minimum value of $v(t)$ is $0$ (as $t \to \infty$).

• For $a(t)$:

  • $a(0) = -x_0\gamma^2$. As $t \to \infty$, $e^{-\gamma t} \to 0$, so $\lim_{t \to \infty} a(t) = 0$.
  • To check monotonicity, we find $\frac{da}{dt}$: …

In kinematics, the position, velocity, and acceleration of a particle are fundamentally linked through differentiation with respect to time. Velocity is the rate of change of position, and acceleration is the rate of change of velocity. This means if we have the position function $x(t)$, we can find the velocity function $v(t)$ by differentiating $x(t)$, and the acceleration function $a(t)$ by differentiating $v(t)$.

To find initial conditions, we simply evaluate these functions at $t=0$. To determine maximum and minimum values, we analyze the function's behavior as $t \to \infty$ and check for critical points (where the derivative is zero or undefined). For monotonicity (whether a function increases or decreases), we examine the sign of its derivative: if the derivative is positive, the function is increasing; if negative, it's decreasing.

Let's apply these concepts to the given problem.

The position of the particle is described by $x(t) = x_0\left(1 - e^{-\gamma t}\right)$, where $t \ge 0$ and $x_0 > 0$.

Part (a): Initial position and velocity

1. Find the initial position: The initial position is the position of the particle at $t=0$. We substitute $t=0$ into the position function $x(t)$:

So, the particle starts at the origin.

2. Find the velocity function $v(t)$:

Velocity is the first derivative of position with respect to time.

Using the chain rule for $e^{-\gamma t}$, where $\frac{d}{dt}(e^{ku}) = k e^{ku} \frac{du}{dt}$, and here $u=t$, $k=-\gamma$:

3. Find the initial velocity: The initial velocity is the velocity of the particle at $t=0$. We substitute $t=0$ into the velocity function $v(t)$:

Part (b): Maximum and minimum values of $x(t)$, $v(t)$, $a(t)$, and their monotonicity

First, let's find the acceleration function $a(t)$.

4. Find the acceleration function $a(t)$: Acceleration is the first derivative of velocity with respect to time (or the second derivative of position).

Again, using the chain rule:

Now we analyze each function:

Analysis of $x(t)$

5. Initial value of $x(t)$: We found $x(0) = 0$.

6. Behavior of $x(t)$ as $t \to \infty$:

   We evaluate the limit of $x(t)$ as $t$ approaches infinity:

Since $\gamma > 0$, as $t \to \infty$, $e^{-\gamma t} \to 0$.

7. Monotonicity of $x(t)$: We examine the sign of its derivative, $v(t)$.

Given $x_0 > 0$ and $\gamma > 0$, and $e^{-\gamma t}$ is always positive, it follows that $v(t) > 0$ for all $t \ge 0$.
Since $v(t) > 0$, $x(t)$ is an increasing function of time.

8. Maximum and minimum values of $x(t)$:

Since $x(t)$ starts at $0$ and continuously increases towards $x_0$ as $t \to \infty$, it never actually reaches $x_0$ but approaches it.

* The minimum value of $x(t)$ is its initial value: $\min(x(t)) = x(0) = 0$.

* The maximum value of $x(t)$ is the limit it approaches: $\max(x(t)) = x_0$.

Analysis of $v(t)$

9. Initial value of $v(t)$: We found $v(0) = x_0 \gamma$.

10. Behavior of $v(t)$ as $t \to \infty$:

As $t \to \infty$, $e^{-\gamma t} \to 0$.

11. Monotonicity of $v(t)$: We examine the sign of its derivative, $a(t)$.

Given $x_0 > 0$ and $\gamma > 0$, $x_0 \gamma^2 e^{-\gamma t}$ is always positive. Therefore, $-x_0 \gamma^2 e^{-\gamma t}$ is always negative.
So, $a(t) < 0$ for all $t \ge 0$.
Since $a(t) < 0$, $v(t)$ is a decreasing function of time.

12. Maximum and minimum values of $v(t)$:

Since $v(t)$ starts at $x_0 \gamma$ and continuously decreases towards $0$ as $t \to \infty$. …

Concept: Expressing Velocity and Acceleration as Functions of Remaining Distance, Not of Time

Method: Phase-Space Elimination — Write $v$ and $a$ in Terms of $(x_0-x)$ (eliminate $t$ entirely, no limits-as-$t\to\infty$ needed)

The stored answer differentiates in the time domain, finds $v(t)$ and $a(t)$ explicitly, then separately evaluates limits as $t\to\infty$ and checks the sign of each derivative. This method instead eliminates $t$ from the start, expressing $v$ and $a$ directly in terms of how much distance remains to the asymptote $x_0$ — revealing the same monotonicity and extreme-value results as an immediate algebraic consequence, with no limits needed at all.

Steps

1. Rewrite the given position function to isolate the "remaining distance." From $x(t)=x_0(1-e^{-\gamma t})$:

This quantity — call it $r(t) = x_0-x(t)$, the remaining distance to $x_0$ — is exactly $x_0 e^{-\gamma t}$.

2. Differentiate $x(t)$ directly and substitute $r(t)$ back in:

So velocity is exactly $\gamma$ times the remaining distance — a clean $v$-versus-$x$ relation, $v = \gamma(x_0-x)$, with no $t$ or exponential appearing explicitly at all.

3. Differentiate once more, in the same way:

So acceleration is exactly $-\gamma^2$ times the remaining distance — again expressed purely in terms of $(x_0-x)$.

4. Read off monotonicity of $x(t)$ without ever solving for $t$. Since $r(t)=x_0-x(t) \ge 0$ (the particle starts at $0<x_0$ and, by Step 2, $v=\gamma r \ge 0$ whenever $r\ge0$ — so $x$ can only ever increase, never overshoot $x_0$, which is self-consistent): $x(t)$ is increasing for as long as $r(t)>0$, i.e. always (since $r=0$ is only reached in the limit, never at a finite $t$).

5. Read off monotonicity of $v$ and $a$ directly from $r(t)$'s own behaviour. $r(t)=x_0e^{-\gamma t}$ is manifestly a strictly decreasing, always-positive quantity for $t\ge0$ (a positive number times a decaying exponential). Since $v=\gamma r$ (positive constant times $r$), $v$ inherits $r$'s strict decrease exactly. Since $a=-\gamma^2 r$ (negative constant times $r$), and $|a|=\gamma^2r$ decreases as $r$ decreases, $a$ itself — being negative — increases toward $0$ as $r$ shrinks. …