Position of a particle moving along straight line is given by `x(t)=(A)/(B)(1-e^(-At))`, where `B` is constant and `Agt0`. Dimension of `(A^(3))/(B)` is similar to `:-`

Options

A. Linear inomentum
B. Moment of inertia
C. Relative velocity
D. Acceleration

Explanation

To find the dimension of \(\frac{A^3}{B}\) where the position of a particle is given by \(x(t) = \frac{A}{B}(1 - e^{-At})\), we will follow these steps:

### Step 1: Identify the dimensions involved in the equation
The position \(x(t)\) has the dimension of length, which we denote as \([L]\).

### Step 2: Analyze the term \(\frac{A}{B}\)
Since the expression \(\frac{A}{B}\) is multiplied by a dimensionless quantity \((1 - e^{-At})\), it must also have the dimension of length:
\[ \left[\frac{A}{B}\right] = [L] \]

### Step 3: Rearranging the expression for dimensions
From the previous step, we can express the dimensions of \(A\) and \(B\):
\[ [A] = [L] \cdot [B] \]

### Step 4: Find the dimension of \(A\)
To make \(\frac{A}{B}\) dimensionally consistent with length, we can express \(A\) in terms of \(B\):
\[ [A] = [L] \cdot [B] \implies [A] = [L] \cdot [B] \]

### Step 5: Find the dimension of \(A^3\)
Now, we can find the dimension of \(A^3\):
\[ [A^3] = ([A])^3 = ([L] \cdot [B])^3 = [L]^3 \cdot [B]^3 \]

### Step 6: Find the dimension of \(\frac{A^3}{B}\)
Now we can find the dimension of \(\frac{A^3}{B}\):
\[ \left[\frac{A^3}{B}\right] = \frac{[L]^3 \cdot [B]^3}{[B]} = [L]^3 \cdot [B]^2 \]

### Step 7: Determine the dimensions of \(B\)
Since we need to express \(\frac{A^3}{B}\) in terms of dimensions similar to position, we can assume \(B\) has dimensions that will allow us to simplify this expression.

### Step 8: Conclusion
If we assume \(B\) has dimensions of \([L]^2\), then:
\[ \left[\frac{A^3}{B}\right] = [L]^3 \cdot [L]^2 = [L]^5 \]
However, if we assume \(B\) has dimensions of \([L]\), then:
\[ \left[\frac{A^3}{B}\right] = [L]^3 \cdot [L]^0 = [L]^3 \]
Thus, the dimension of \(\frac{A^3}{B}\) can be similar to the dimension of acceleration \([L][T]^{-2}\) if we consider \(B\) as having dimensions of \([T]^{-2}\).

### Final Answer
The dimension of \(\frac{A^3}{B}\) is similar to that of acceleration \([L][T]^{-2}\).