The velocity (V) of a particle (in cm/s) is given in terms of time (t) in sec by the equation `V=at+(b)/(c+t)`. The dimensions of a, b and c are

Options

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

Answer

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

### Step-by-Step Solution:

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

2. **Analyze the term \( at \)**:
- The term \( at \) must also have the same dimensions as velocity.
- The dimension of time \( t \) is \( [t] = T \).
- Therefore, the dimension of \( a \) can be expressed as:
\[ [a][t] = [V] \implies [a] \cdot T = L T^{-1} \]
- Rearranging gives:
\[ [a] = \frac{L T^{-1}}{T} = L T^{-2} \]

3. **Analyze the term \( \frac{b}{c+t} \)**:
- The term \( c+t \) must also have the dimension of time since \( t \) has the dimension \( T \).
- Thus, the dimension of \( c \) must be the same as that of \( t \):
\[ [c] = T \]

4. **Determine the dimensions of \( b \)**:
- The term \( \frac{b}{c+t} \) must also have the same dimensions as velocity, which is \( L T^{-1} \).
- Since \( c+t \) has the dimension \( T \), we can write:
\[ \frac{[b]}{[c+t]} = [V] \implies \frac{[b]}{T} = L T^{-1} \]
- Rearranging gives:
\[ [b] = L T^{-1} \cdot T = L \]

5. **Summarize the dimensions**:
- We have found the following dimensions:
- \( [a] = L T^{-2} \)
- \( [b] = L \)
- \( [c] = T \)

### Final Answer:
- The dimensions are:
- \( a: L T^{-2} \)
- \( b: L \)
- \( c: T \)