The velocity v of a particle at time t is given by `v=at+b/(t+c)`, where a, b and c are constants. The dimensions of a, b, c are respectively :-

Options

A. `LT^(-2)`, L and T
B. `L^(2)`, T and `LT^(2)`
C. `LT^(2)`, LT and L
D. L, LT and `T^(2)`

Explanation

To solve the problem, we need to determine the dimensions of the constants \( a \), \( b \), and \( c \) in the given velocity equation \( v = at + \frac{b}{t + c} \).

### Step-by-Step Solution:

1. **Identify the dimensions of velocity \( v \):**
- The dimension of velocity \( v \) is \([L T^{-1}]\), where \( L \) represents length and \( T \) represents time.

2. **Determine the dimensions of \( a \):**
- The term \( at \) must have the same dimensions as velocity \( v \).
- Given \( v = at \), we have:
\[ [v] = [at] \]
\[ [L T^{-1}] = [a][T] \]
- Solving for the dimension of \( a \):
\[ [a] = \frac{[L T^{-1}]}{[T]} = [L T^{-2}] \]

3. **Determine the dimensions of \( c \):**
- The term \( t + c \) implies that \( c \) must have the same dimensions as \( t \).
- Since \( t \) represents time, the dimension of \( t \) is \([T]\).
- Therefore, the dimension of \( c \) is:
\[ [c] = [T] \]

4. **Determine the dimensions of \( b \):**
- The term \(\frac{b}{t + c}\) must have the same dimensions as velocity \( v \).
- Given \( v = \frac{b}{t + c} \), we have:
\[ [v] = \left[\frac{b}{t + c}\right] \]
\[ [L T^{-1}] = \left[\frac{b}{T}\right] \]
- Solving for the dimension of \( b \):
\[ [b] = [L T^{-1}][T] = [L] \]

### Final Dimensions:
- The dimension of \( a \) is \([L T^{-2}]\).
- The dimension of \( b \) is \([L]\).
- The dimension of \( c \) is \([T]\).

### Answer:
The dimensions of \( a \), \( b \), and \( c \) are respectively:
\[ [a] = [L T^{-2}], \quad [b] = [L], \quad [c] = [T] \]