With the usual notations, the following equation S = uₜ + a/2(2t-1) is

The correct option is "Both numerically and dimensionally correct".

Let us analyze this step-by-step to understand why this equation is both numerically and dimensionally correct.

1. Understanding the Equation
The given equation is:
$$S=u+\frac{a}{2}(2t-1)$$
Note: In the question, it is written as $u_t$ or $u$ representing the initial velocity, and the symbol $S$ represents the distance traveled in the $t^{\mathrm{th}}$ second (often written as $S_t$ or $S_n$).

2. Numerical Correctness
The distance covered in $t$ seconds is given by the standard equation of motion:
$$S_t=ut+\frac{1}{2}a{t}^2$$
Similarly, the distance covered in $(t-1)$ seconds is:
$$S_{t-1}=u(t-1)+\frac{1}{2}a{(t-1)}^2$$
The distance traveled specifically in the $t^{\mathrm{th}}$ second is the difference between these two distances:
$$S=S_t-S_{t-1}$$
Substituting the expressions:
$$S=[ut+\frac{1}{2}a{t}^2]-\left[u\right(t-1)+\frac{1}{2}a{(t-1)}^2]$$
Expanding and simplifying:
$$S=ut+\frac{1}{2}a{t}^2-ut+u-\frac{1}{2}a(t^2-2t+1)$$
$$S=u+\frac{1}{2}a{t}^2-\frac{1}{2}a{t}^2+at-\frac{a}{2}$$
$$S=u+at-\frac{a}{2}$$
Taking $\frac{a}{2}$ common:
$$S=u+\frac{a}{2}(2t-1)$$
Thus, the equation is numerically correct.

3. Dimensional Correctness
At first glance, it might appear that the dimensions do not match because the left-hand side is distance ($\left[L\right]$) and the right-hand side has velocity ($\left[L{T}^{-1}\right]$). However, we must note that $S$ is the distance traveled per unit time (in a specific 1-second interval).
Therefore, the actual physical quantity on the left-hand side is:
$$\left[\text{LHS}\right]=\frac{\text{Distance}}{\text{Time (1 second)}}=\left[L{T}^{-1}\right]$$
Now let us check the dimensions of each term on the right-hand side:
• Dimension of initial velocity $u$:
$$\left[u\right]=\left[L{T}^{-1}\right]$$
• Dimension of $\frac{a}{2}(2t-1)$:
Here, the number 1 inside the parenthesis represents a unit time interval of $1\text{second}$, and $t$ represents time. So the term $(2t-1)$ has the dimension of time $\left[T\right]$.
Since acceleration $a$ has dimensions $\left[L{T}^{-2}\right]$:
$$\left[\frac{a}{2}\right(2t-1\left)\right]=\left[L{T}^{-2}\right]\times \left[T\right]=\left[L{T}^{-1}\right]$$
Since all terms in the equation have the exact same dimensions ($\left[L{T}^{-1}\right]$), the principle of dimensional homogeneity is fully satisfied. Hence, the equation is also dimensionally correct.