With usual notation, the following equation, said to give the distance covered in the `n"th"` second. `i.e.,
`S_(n)=u+`(a(2n-1)/2)` `is

Options

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

Answer

To determine whether the equation \( S_n = u + \frac{a(2n-1)}{2} \) is numerically and dimensionally correct, we will analyze the components of the equation step by step.

### Step 1: Understanding the Variables
- \( S_n \): Distance covered in the \( n \)-th second.
- \( u \): Initial velocity (in meters per second, m/s).
- \( a \): Acceleration (in meters per second squared, m/s²).
- \( n \): The \( n \)-th second (a dimensionless quantity).

### Step 2: Analyzing the Right Side of the Equation
The equation can be broken down into two parts:
1. The term \( u \) represents the distance covered due to initial velocity in the \( n \)-th second.
2. The term \( \frac{a(2n-1)}{2} \) represents the additional distance covered due to acceleration during that second.

### Step 3: Analyzing Dimensions
- The dimension of \( u \) is [L][T]⁻¹ (length/time).
- The dimension of \( a \) is [L][T]⁻² (length/time²).
- The term \( (2n-1) \) is dimensionless since \( n \) is a count of seconds.

Now, let's analyze the term \( \frac{a(2n-1)}{2} \):
- The dimension of \( a \) is [L][T]⁻².
- The factor \( (2n-1) \) does not affect the dimensions, so the dimension of \( \frac{a(2n-1)}{2} \) remains [L][T]⁻².

### Step 4: Combining the Terms
Now we combine the two parts:
- The dimension of \( u \) is [L][T]⁻¹.
- The dimension of \( \frac{a(2n-1)}{2} \) is [L][T]⁻¹ (because it is multiplied by time, which is [T]).

Thus, both terms on the right side of the equation have the same dimension of [L], which means they can be added together.

### Step 5: Conclusion
Since both terms on the right side have the same dimension and represent distances, we conclude that the equation \( S_n = u + \frac{a(2n-1)}{2} \) is both numerically and dimensionally correct.