The position of the particle moving along `Y`-axis is given as `y=At^(2)-Bt^(3)`, where `y` is measured in metre and `t` in second. Then, the dimensions of `B` are

Options

A. `["LT"^(-2)]`
B. `["LT"^(-1)]`
C. `["LT"^(-3)]`
D. `["MLT"^(-2)]`

Explanation

To find the dimensions of \( B \) in the equation \( y = At^2 - Bt^3 \), we will use the principle of homogeneity. This principle states that all terms in an equation must have the same dimensions.

### Step-by-Step Solution:

1. **Identify the dimensions of \( y \)**:
- The position \( y \) is measured in meters. Therefore, the dimension of \( y \) is:
\[ [y] = L \]
where \( L \) represents length.

2. **Identify the dimensions of \( t \)**:
- The time \( t \) is measured in seconds. Therefore, the dimension of \( t \) is:
\[ [t] = T \]
where \( T \) represents time.

3. **Analyze the term \( At^2 \)**:
- The term \( At^2 \) must also have the same dimensions as \( y \). Since \( t^2 \) has dimensions:
\[ [t^2] = T^2 \]
- Therefore, the dimensions of \( A \) can be expressed as:
\[ [A] \cdot [t^2] = L \implies [A] \cdot T^2 = L \implies [A] = \frac{L}{T^2} \]

4. **Analyze the term \( Bt^3 \)**:
- Similarly, for the term \( Bt^3 \), we have:
\[ [Bt^3] = L \]
- The dimensions of \( t^3 \) are:
\[ [t^3] = T^3 \]
- Therefore, the dimensions of \( B \) can be expressed as:
\[ [B] \cdot [t^3] = L \implies [B] \cdot T^3 = L \implies [B] = \frac{L}{T^3} \]

5. **Final result**:
- Thus, the dimensions of \( B \) are:
\[ [B] = LT^{-3} \]

### Conclusion:
The dimensions of \( B \) are \( LT^{-3} \).