The displacement x of a particle varies with time according to the relation `x=(a)/(b)(1-e^(-bt))`. Then select the false alternative.

Options

A. ` At t=1 //b` the displacement of the particle is nearly (2/3) (a/b)
B. The velocity and acceleration of the particle at t = 0 are respectively
C. The particle cannot reach a point at a distance x' from its starting position if x' gt a/b
D. The particle will come back to its starting point as` t to infinity `

Explanation

To solve the problem, we need to analyze the given displacement equation and evaluate the provided options to identify the false statement.

### Given:
The displacement \( x \) of a particle varies with time according to the relation:
\[ x = \frac{a}{b} (1 - e^{-bt}) \]

### Step 1: Evaluate Option A
**Option A:** At \( t = \frac{1}{b} \), displacement is nearly equal to \( \frac{2}{3} \frac{a}{b} \).

**Solution:**
Substituting \( t = \frac{1}{b} \) into the displacement equation:
\[ x = \frac{a}{b} \left(1 - e^{-b \cdot \frac{1}{b}}\right) = \frac{a}{b} \left(1 - e^{-1}\right) \]
Using the approximation \( e^{-1} \approx 0.3679 \):
\[ x \approx \frac{a}{b} \left(1 - 0.3679\right) \approx \frac{a}{b} \cdot 0.6321 \approx \frac{2}{3} \frac{a}{b} \]
Thus, Option A is **true**.

### Step 2: Evaluate Option B
**Option B:** The velocity and acceleration of the particle at \( t = 0 \) are respectively \( a \) and \( -ab \).

**Solution:**
1. **Velocity**:
\[ v = \frac{dx}{dt} = \frac{a}{b} \cdot b e^{-bt} = a e^{-bt} \]
At \( t = 0 \):
\[ v = a e^{0} = a \]

2. **Acceleration**:
\[ a = \frac{dv}{dt} = -ab e^{-bt} \]
At \( t = 0 \):
\[ a = -ab e^{0} = -ab \]

Thus, Option B is **true**.

### Step 3: Evaluate Option C
**Option C:** The particle cannot reach a point at a distance \( x' \) from the starting point when \( x = \frac{a}{b} \).

**Solution:**
As \( t \to \infty \):
\[ x = \frac{a}{b} (1 - e^{-bt}) \to \frac{a}{b} (1 - 0) = \frac{a}{b} \]
Thus, the particle can reach \( x = \frac{a}{b} \). Therefore, Option C is **false**.

### Step 4: Evaluate Option D
**Option D:** The particle will not come back to its starting point when \( t \to \infty \).

**Solution:**
As \( t \to \infty \), the displacement approaches:
\[ x \to \frac{a}{b} \]
The particle does not return to the starting point (which is \( x = 0 \)). Thus, Option D is **true**.

### Conclusion
The false alternative is **Option C**: The particle cannot reach a point at a distance \( x' \) from the starting point when \( x = \frac{a}{b} \).