Answer to Question #135298 in Mechanics | Relativity for Vanel

Question

The position of a particle moving under uniform acceleration is some function of time, t and the acceleration, a. Suppose we write this position as x = ka^m.(t^n) where k is a dimensionless constant.
Determine the values of m and n.
Can this analysis give the value of k? Explain.

Answer 1

Explanations & Calculations

• A proper equation is dimensionally correct & it's the same for this function.
• Consider the dimensions of all the quantities,

$\qquad\qquad \begin{aligned} \small [x] &= \small L \qquad \qquad [a] =LT^{-2}\qquad\qquad[t]=T\\ \end{aligned}$

• According to the requirement stated above,

$\qquad\qquad \begin{aligned} \small [x] &= \small [ka^mt^n] \\ &= \small [a^mt^n]\\ \small L&= \small (LT^{-2})^m \times(T)^n\\ \small L^1T^0&= \small L^{m}T^{(-2m+n)}\\ &.............................................\\ \small 1&=\small m\implies m=1\\ \small 0 &= \small -2m+n \implies n = 2\\ &.............................................\\ \therefore \bold{x} & =\small \bold{kat^{2}} \end{aligned}$

• Usually values of dimensionless constants cannot be found using this method. They are found experimentally.
• For this reason the validity of an equation cannot be fully assessed by this method.
• For an example $\small V = U +at$ and $\small V =\frac{1}{3}U+2at$ which are both dimensionally correct while only the first being the valid one.