In this problem, we're asked to look at dimensions of variables in an equation. The three main dimensions we work with in physics are mass [M], length [L], and time [T].

Remember that for an equation to work, both sides have to have the same dimensions. We can't equate a mass to a time, for example, because they're different physical quantities. Any *dimensionless factors* in an equation ** can't be derived or checked using dimensional analysis,** because they don't affect the dimensions on either side.

For this problem, we're asked to first (part **(a)**) __show that the dimensions work out if m = 1 and n = 2__, then

(a) Let's start by substituting each of the variables in the equation with their dimensions. The variable *x* usually represents length, written as [L] in dimensional analysis. The dimensions of acceleration, *a*, are [L T ^{-2}], and t represents time, written [T]. Substituting all those, we get:

$\begin{array}{lcl}\left[L\right]& =& k{\left[\frac{L}{{T}^{2}}\right]}^{m}{\left[T\right]}^{n}\end{array}$

The position of a particle moving under uniform acceleration is some function of time and the acceleration. Suppose we write this position as *x* = *ka ^{m}t*