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

Options

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

Explanation

To find the dimensions of \( v_0 \) and \( \text{prop} \) in the equation \( x(t) = \frac{v_0}{\text{prop}} (1 - e^{-ut}) \), we will analyze the components step by step.

### Step 1: Understanding the equation
The position \( x(t) \) is given as a function of time \( t \). The term \( (1 - e^{-ut}) \) is dimensionless because the exponential function must be dimensionless. Therefore, the argument of the exponential, \( -ut \), must also be dimensionless.

### Step 2: Analyzing the term \( ut \)
Since \( ut \) is dimensionless, we can write:
\[ [u][t] = 1 \]
Where:
- \( [u] \) is the dimension of \( u \)
- \( [t] \) is the dimension of time, which is \( T \)

Thus, we can express the dimension of \( u \) as:
\[ [u] = \frac{1}{[t]} = T^{-1} \]

### Step 3: Analyzing the term \( x(t) \)
The position \( x(t) \) has dimensions of length, which we denote as:
\[ [x] = L \]

### Step 4: Relating \( v_0 \) and \( \text{prop} \)
From the equation \( x(t) = \frac{v_0}{\text{prop}} (1 - e^{-ut}) \), we can isolate \( \frac{v_0}{\text{prop}} \):
\[ [x] = \frac{[v_0]}{[\text{prop}]} \cdot 1 \]
This implies:
\[ [x] = \frac{[v_0]}{[\text{prop}]} \]

### Step 5: Finding the dimensions of \( v_0 \)
Assuming \( v_0 \) is a velocity, we have:
\[ [v_0] = L T^{-1} \]

### Step 6: Finding the dimensions of \( \text{prop} \)
Substituting the dimensions into the equation:
\[ L = \frac{L T^{-1}}{[\text{prop}]} \]
Rearranging gives:
\[ [\text{prop}] = \frac{L T^{-1}}{L} = T^{-1} \]

### Conclusion
Thus, we have:
- The dimension of \( v_0 \) is \( L T^{-1} \) (velocity).
- The dimension of \( \text{prop} \) is \( T^{-1} \).

### Final Answer:
- Dimensions of \( v_0 \): \( L T^{-1} \)
- Dimensions of \( \text{prop} \): \( T^{-1} \)