The dimensionally correct formula is

Options

A. `S = u + 1/2 at^2`
B. `S = ut + 1/2 at^3`
C. `S = ut - at^2`
D. None of the above

Answer

To determine which of the given formulas is dimensionally correct, we need to analyze each option based on the dimensions of the quantities involved.

### Step-by-Step Solution:

1. **Identify the quantities and their dimensions**:
- Displacement \( s \): Dimensions = \( [L] \) (length)
- Initial velocity \( u \): Dimensions = \( [L][T]^{-1} \) (length per time)
- Acceleration \( a \): Dimensions = \( [L][T]^{-2} \) (length per time squared)
- Time \( t \): Dimensions = \( [T] \) (time)

2. **Analyze Option 1: \( s = u + \frac{1}{2} a t^2 \)**:
- Left-hand side (LHS): Dimensions of \( s = [L] \)
- Right-hand side (RHS):
- Dimensions of \( u = [L][T]^{-1} \)
- Dimensions of \( \frac{1}{2} a t^2 = \frac{1}{2} [L][T]^{-2} [T]^2 = [L] \)
- Therefore, RHS = \( [L] + [L] = [L] \)
- Since LHS = RHS, this equation is dimensionally correct.

3. **Analyze Option 2: \( s = ut + \frac{1}{2} a t^3 \)**:
- LHS: Dimensions of \( s = [L] \)
- RHS:
- Dimensions of \( ut = [L][T]^{-1} [T] = [L] \)
- Dimensions of \( \frac{1}{2} a t^3 = \frac{1}{2} [L][T]^{-2} [T]^3 = [L][T]^{1} \)
- Therefore, RHS = \( [L] + [L][T]^{1} \)
- Since LHS ≠ RHS, this equation is not dimensionally correct.

4. **Analyze Option 3: \( s = ut - \frac{1}{2} a t^2 \)**:
- LHS: Dimensions of \( s = [L] \)
- RHS:
- Dimensions of \( ut = [L][T]^{-1} [T] = [L] \)
- Dimensions of \( -\frac{1}{2} a t^2 = -\frac{1}{2} [L][T]^{-2} [T]^2 = -[L] \)
- Therefore, RHS = \( [L] - [L] = [L] \)
- Since LHS = RHS, this equation is dimensionally correct.

5. **Conclusion**:
- Options 1 and 3 are dimensionally correct.
- Therefore, the correct answer is that there are dimensionally correct formulas among the options provided.

### Final Answer:
The dimensionally correct formulas are:
- Option 1: \( s = u + \frac{1}{2} a t^2 \)
- Option 3: \( s = ut - \frac{1}{2} a t^2 \)