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*

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What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Dimensional Analysis concept. You can view video lessons to learn Dimensional Analysis. Or if you need more Dimensional Analysis practice, you can also practice Dimensional Analysis practice problems.

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Based on our data, we think this problem is relevant for Professor Staff's class at Ryerson University.