The dimension of a/b where x is the distance and t is the time is T
Given the expression
x = at + bt²
where
x is the distance
t is the time
Based on the homogeneity principle, the expression on the left-hand side must be equal to that on the right. Hence;
x = at
[tex]a = \frac{x}{t}[/tex]
Since x is the distance and distance is measured in metres, the dimension equivalent will be the length 'L'
Since t is the time and time is measured in seconds, the dimension equivalent will be the seconds 'T'
[tex]a=\frac{L}{T}[/tex]
Similarly;
x = bt²
[tex]b=\frac{x}{t^2}\\b=\frac{L}{T^2}[/tex]
Next is to get a/b;
[tex]\frac{a}{b} = \frac{L}{T} \div \frac{L}{T^2}\\\frac{a}{b} = \frac{L}{T}*\frac{T^2}{L} \\\frac{a}{b} =\frac{T^2}{T}\\\frac{a}{b} =T[/tex]
Hence the dimension of a/b is T