With the usual notations the following equation `S_(t)=u+1/2a(2t-1)` is

Options

A. Only numerically correct
B. Only dimensionally correct
C. Both numerically and dimensionally correct
D. Neither numerically nor dimensionally correct

Answer

To analyze the equation \( S_t = u + \frac{1}{2} a (2t - 1) \) and check its dimensional consistency, we will follow these steps:

### Step 1: Identify the components of the equation
The equation represents the distance \( S_t \) covered by an object in the \( t \)-th second, where:
- \( S_t \) is the distance traveled in the \( t \)-th second.
- \( u \) is the initial velocity.
- \( a \) is the acceleration.
- \( t \) is the time.

### Step 2: Determine the dimensions of each term
1. **Distance \( S_t \)**: The dimension of distance is \( [L] \) (length).
2. **Initial velocity \( u \)**: The dimension of velocity is \( [L T^{-1}] \) (length per time).
3. **Acceleration \( a \)**: The dimension of acceleration is \( [L T^{-2}] \) (length per time squared).
4. **Time \( t \)**: The dimension of time is \( [T] \).

### Step 3: Analyze the term \( \frac{1}{2} a (2t - 1) \)
- The term \( (2t - 1) \) is dimensionless because it is a combination of time (which has dimension \( [T] \)) and a constant (1), which does not have dimensions.
- Therefore, the dimension of \( a(2t - 1) \) is the same as the dimension of \( a \), which is \( [L T^{-2}] \).

### Step 4: Combine the dimensions
Now, we can analyze the right-hand side of the equation:
- The term \( u \) has dimensions \( [L T^{-1}] \).
- The term \( \frac{1}{2} a (2t - 1) \) has dimensions \( [L T^{-2}] \).

### Step 5: Find a common dimension
To combine these terms, we need to express them in terms of the same dimension:
- The term \( u \) needs to be converted to a dimension of distance per time, so we can express the entire right-hand side in terms of distance:
- \( u + \frac{1}{2} a (2t - 1) \) should yield a dimension of \( [L] \).

### Step 6: Check dimensional consistency
1. The term \( u \) has dimensions \( [L T^{-1}] \).
2. The term \( \frac{1}{2} a (2t - 1) \) has dimensions \( [L T^{-2}] \cdot [T^2] = [L] \).

Thus, the right-hand side combines to give a dimension of \( [L] \), which is consistent with the left-hand side \( S_t \).

### Conclusion
Since both sides of the equation have the same dimension, the equation is dimensionally consistent.