The position x of a particle at time t is given by `x=(V_(0))/(a)(1-e^(-at))`, where `V_(0)` is constant and `a gt 0`. The dimensions of `V_(0)` and a are

Options

A. `M^(-0)LT^(-1) and T^(-1)`
B. `M^(0)LT^(0) and T^(-1)`
C. `M^(0)LT^(-1) and L T^(-2)`
D. `M^(0) L T^(-1) and T`

Explanation

To find the dimensions of \( V_0 \) and \( a \) from the given equation \( x = \frac{V_0}{a} (1 - e^{-at}) \), we can follow these steps:

### Step 1: Analyze the equation
The equation given is:
\[ x = \frac{V_0}{a} (1 - e^{-at}) \]
Since \( e^{-at} \) is an exponential function, the term \( -at \) must be dimensionless. Therefore, the dimensions of \( a \) multiplied by the dimensions of \( t \) must equal the dimensions of 1.

### Step 2: Determine the dimensions of \( at \)
Let’s denote the dimensions of \( a \) as \( [a] \) and the dimensions of time \( t \) as \( [T] \). For \( -at \) to be dimensionless:
\[ [a][T] = 1 \]
This implies:
\[ [a] = [T]^{-1} \quad \text{(1)} \]
Thus, the dimensions of \( a \) are \( T^{-1} \).

### Step 3: Find the dimensions of \( V_0 \)
From the equation \( x = \frac{V_0}{a} (1 - e^{-at}) \), we can rearrange it to find \( V_0 \):
\[ V_0 = x \cdot a \]
Now, we know that:
- The dimensions of \( x \) (position) are \( [L] \) (length).
- The dimensions of \( a \) are \( [T]^{-1} \).

Substituting these into the equation gives:
\[ [V_0] = [L] \cdot [T]^{-1} \]
Thus, the dimensions of \( V_0 \) are:
\[ [V_0] = LT^{-1} \quad \text{(2)} \]

### Step 4: Summarize the results
From the steps above, we have determined:
- The dimensions of \( a \) are \( T^{-1} \).
- The dimensions of \( V_0 \) are \( LT^{-1} \).

### Final Answer
- Dimensions of \( V_0 \): \( [L][T]^{-1} \) (or \( LT^{-1} \))
- Dimensions of \( a \): \( [T]^{-1} \)