Answer to Question #82923 in Mechanics | Relativity for Opone Nicholas

Question

In the equation V^n=K.a^j.x ,what number must n & j be to make the equation dimensionally correct? K= a dimensionless number, x = L , v= velocity and a= acceleration.

Answer 1

The given equation is:

$v^n = ka^jx.$

Let's start by writing the known dimensions (units) of the values:

$[k] = 1;~[x] = m;~[v] = \frac{m}{s};~[a] = \frac{m}{s^2}.$

Now let's rewrite the equation using only the units (e.g. assuming each value is equal to unit value) and rework the powers:

$(\frac{m}{s})^n = (\frac{m}{s^2})^jm; \\ m^ns^{-n} = m^{j+1}s^{-2j}.$

Now, for the equation to be correct, the powers of each unit (i.e. meters and seconds) have to match correspondingly:

$\begin{cases} n = j + 1; \\ -n = -2j. \end{cases}$

Now it's easy to solve these equations together for n and j:

$2j = j + 1; \implies j = 1; \\ n = 1 + 1 = 2.$

Therefore the correct form of the initial equation, with the found n and j is:

$v^2 = kax.$