The position of a particle at time `t` is given by the relation `x(t) = ( v_(0) /( alpha)) ( 1 - c^(-at))`, where `v_(0)` is a constant and `alpha gt 0`. Find the dimensions of `v_(0) and alpha`.

Options

A. `[L^(-1)T^(-1)]``,[ T^(-2)]`
B. `[L^(2)T^(-1)]``,[ T^(-1)]`
C. `[LT^(-1)]``,[ T^(-1)]`
D. `[LT^(-2)]``,[ T^(1)]`

Explanation

To find the dimensions of \( v_0 \) and \( \alpha \) in the equation \( x(t) = \frac{v_0}{\alpha} (1 - c^{-at}) \), we will analyze the equation step by step.

### Step 1: Understand the equation
The equation describes the position \( x(t) \) of a particle at time \( t \). The term \( \frac{v_0}{\alpha} \) must have dimensions that allow it to be multiplied by a dimensionless term (since \( 1 - c^{-at} \) is dimensionless).

### Step 2: Analyze the term \( c^{-at} \)
The term \( c^{-at} \) must be dimensionless. This means that the exponent \( -at \) must also be dimensionless.

### Step 3: Determine the dimensions of \( a \)
Since \( t \) has the dimension of time \([T]\), for \( -at \) to be dimensionless, \( a \) must have dimensions of \( [T^{-1}] \).

### Step 4: Analyze the term \( \alpha t \)
From the previous analysis, we know \( at \) is dimensionless. Thus, \( \alpha t \) must also be dimensionless. Since \( t \) has the dimension of time \([T]\), \( \alpha \) must have dimensions of \( [T^{-1}] \).

### Step 5: Analyze the term \( \frac{v_0}{\alpha} \)
Now, we need to analyze the term \( \frac{v_0}{\alpha} \). Since \( x(t) \) represents position, it has dimensions of length \([L]\). Therefore, we can write:

\[ \frac{v_0}{\alpha} \text{ has dimensions of } [L] \]

### Step 6: Substitute the dimensions of \( \alpha \)
We know that \( \alpha \) has dimensions of \( [T^{-1}] \). Therefore, we can express the dimensions of \( v_0 \):

\[ \frac{v_0}{[T^{-1}]} = [L] \]

### Step 7: Solve for the dimensions of \( v_0 \)
To isolate \( v_0 \), we multiply both sides by \( [T^{-1}] \):

\[ v_0 = [L] \cdot [T^{-1}] = [L T^{-1}] \]

### Summary of Results
- The dimensions of \( v_0 \) are \( [L T^{-1}] \).
- The dimensions of \( \alpha \) are \( [T^{-1}] \).

### Final Answer
- Dimensions of \( v_0 \): \( [L T^{-1}] \)
- Dimensions of \( \alpha \): \( [T^{-1}] \)