The problem asks for the dimensional formula of the constant 'b' in the equation relating distance '$s$' and time '$t$':

$ s = at + bt^2 $

Dimensional Homogeneity Principle

According to the principle of dimensional homogeneity, every term in a physically consistent equation must have the same dimensions. We know the dimensions of distance '$s$' and time '$t$':

• Dimension of distance ($s$): $[s] = [\text{L}^1]$
• Dimension of time ($t$): $[t] = [\text{T}^1]$

Dimensional Formula of 'a'

Equating the dimensions of '$s$' and '$at$':

$ [s] = [a][t] $

$ [\text{L}^1] = [a][\text{T}^1] $

Solving for the dimensions of 'a':

$ [a] = \frac{[\text{L}^1]}{[\text{T}^1]} = [\text{L}^1\text{T}^{-1}] $

Dimensional Formula of 'b'

Equating the dimensions of '$s$' and '$bt^2$':

$ [s] = [b][t^2] $

$ [\text{L}^1] = [b][\text{T}^1]^2 $

$ [\text{L}^1] = [b][\text{T}^2] $

Solving for the dimensions of 'b':

$ [b] = \frac{[\text{L}^1]}{[\text{T}^2]} = [\text{L}^1\text{T}^{-2}] $

In terms of mass [M], length [L], and time [T], the dimensional formula is $[\text{M}^0\text{L}^1\text{T}^{-2}]$.