# Problem: The position of a particle moving under uniform acceleration is some function of time and the acceleration. Suppose we write this position as x = kamtn where k is a dimensionless constant. (a) Show by dimensional analysis that this expression is satisfied if m = 1 and n = 2. (b) Can this analysis give the value of k?

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

###### FREE Expert Solution

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 (b) whether dimensional analysis can give us the value of k.

(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:

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

###### Problem Details

The position of a particle moving under uniform acceleration is some function of time and the acceleration. Suppose we write this position as x = kamtn where k is a dimensionless constant. (a) Show by dimensional analysis that this expression is satisfied if m = 1 and n = 2. (b) Can this analysis give the value of k?

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